Mahesh

Solving these TS 10th Class Maths Bits with Answers Chapter 3 Polynomials Bits for 10th Class will help students to build their problem-solving skills.

Polynomials Bits for 10th Class

Question 1.
The coefficient of ‘z’ in the polynomial
z⁵ – 3z² + 4 is
A) 1
B) -3
C) 4
D) 0
Answer:
D) 0

Question 2.
If p(x) = 3x² – x – 4, then p(-1) =
A) 2
B) 0
C) -2
D) 1
Answer:
B) 0

Question 3.
If p(x) = 4x² – 4x + 1; then p(0) =
A) 1
B) -1
C) 4
D) -4
Answer:
A) 1

Question 4.
The remainder when 2x² + 3x + 1 is divided by x + 2 is
A) 15
B) -15
C) 3
D) -3
Answer:
C) 3

Question 5.
The number of zeroes of the polynomial 2x + 1 is
A) 0
B) \(\frac{-1}{2}\)
C) 3
D) 2
Answer:
B) \(\frac{-1}{2}\)

Question 6.
The zeroes of the polynomial p(x) = x² – 3 are
A) 1, 3
B) 1, -3
C) \(\sqrt{3}\), 1
D) \(\sqrt{3}\), –\(\sqrt{3}\)
Answer:
D) \(\sqrt{3}\), –\(\sqrt{3}\)

Question 7.
If α, β are the zeroes of x² + 7x + 10, then αβ =
A) -7
B) 7
C) 10
D) -10
Answer:
C) 10

Question 8.
If α, β are the zeroes of the polynomial f(x) = x² + x + 1, then \(\frac{1}{a}\) + \(\frac{1}{b}\) =
A) -1
B) 1
C) 0
D) none of these
Answer:
A) -1

Question 9.
If α, β are the zeroes of the polynomial f(x) = x² + x + 1, then (α + 1) (β + 1) = ……
A) αβ + α + β
B) α + β + 1
C) 3
D) -1
Answer:
C) 3

Question 10.
If α, β are the zeroes of the quadratic poly-nomial p(t) = t² – 4t + 3, then the value of
A) 4
B) 3
C) 4/3
D) -4/3
Answer:
B) 3

Question 11.
If α, β, γ are the zeroes of 3x³ – 5x² – 11x – 3, then α + β + γ =
A) -5/3
B) 5/3
C) -11/3
D) 1
Answer:
B) 5/3

Question 12.
The degree of the polynomial
\(\sqrt{2}\)x² – 3x + 1 is
A) 2
B) 1
C) 72
D) 3
Answer:
A) 2

Question 13.
The zero of the linear polynomial px + q is
A) \(\frac{-\mathrm{q}}{\mathrm{p}}\)
B) \(\frac{\mathrm{p}}{\mathrm{q}}\)
C) \(\frac{-\mathrm{p}}{\mathrm{q}}\)
D) q
Answer:
A) \(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 14.
4x – 3 is a
A) linear polynomial
B) cubic polynomial
C) biquadratic polynomial
D) quadratic polynomial
Answer:
A) linear polynomial

Question 15.
The degree of a cubic polynomial is
A) 1
B) 2
C) 3
D) 0
Answer:
C) 3

Question 16.
The zero of p(x) = ax + b is (
A) -a/b
B) a/b
C) -b/a
D) b/a
Answer:
C) -b/a

Question 17.
The degree of the polynomial (x + 1) (x + 2) (x + 3) is
A) 3
B) 4
C) 1
D) 2
Answer:
A) 3

Question 18.
The product of the zeroes of 3x³ – 5x² – 11x – 3 is
A) 3
B) 1
C) -1
D) -3
Answer:
B) 1

Question 19.
The sum of the zeroes of 2x³ – 5x² – 14x + 8 is
A) 5/2
B) -7
C) 4
D) -14
Answer:
A) 5/2

Question 20.
If α, β are the zeroes of the quadratic poly-nomial 4x² – 1, find the value of α² + β²
A) 1/4
B) -1/4
C) 3/4
D) 1/2
Answer:
D) 1/2

Question 21.
If α, β, γ are the zeroes of the polynomial
p(x) = ax³ + bx² + cx + d then \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) = r
A) -c/d
B) -b/d
C) c/a
D) -b/a
Answer:
A) -c/d

Question 22.
If α, β are the zeroes of the polynomial x² – 16, find αβ (α + β)
A) -16
B) -8
C) 0
D) -1
Answer:
C) 0

Question 23.
In the following expressions which one is a polynomial ?
A) \(\frac{1}{x+1}\)
B) 2x² – \(\frac{3}{x}\) + 5
C) x² + xy + y²
D) x^1/2 + x + 3
Answer:
B) 2x² – \(\frac{3}{x}\) + 5

Question 24.
The degree of the polynomial
9xy³ + y⁴ + \(\frac{3}{4}\) x⁴ + 7x³y² is
A) 3
B) 4
C) 2
D) 5
Answer:
D) 5

Question 25.
x³ – 3x² + 4x – 5 is divided by (x + 1) then the remainder is
A) 0
B) 13
C) -13
D) none of these
Answer:
C) -13

Question 26.
A polynomial of degree 2 is called ____ poly-nomial.
A) linear
B) quadratic
C) cubic
D) biquadratic
Answer:
B) quadratic

Question 27.
p(x) = 2x + 1 then p(-1/2) =
A) 2
B) -2
C) 0
D) 1
Answer:
C) 0

Question 28.
p(x) = x² – 3x + 2 then p(0) =
A) 0
B) 2
C) -3
D) 1
Answer:
B) 2

Question 29.
The graph of y = ax+b is a straight line which intersects x-axis at
A) (0, -b/a)
B) (0, b)
C) (b, 0)
D) (-b/a, 0)
Answer:
D) (-b/a, 0)

Question 30.
Number of zeroes for any cubic polynomial are ( )
A) 1
B) 3
C) 2
D) 4
Answer:
B) 3

Question 31.
If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d = 0 then αβγ
A) -b/a
B) c/a
C) -d/a
D) d/a
Answer:
C) -d/a

Question 32.
Order of the polynomial 5x⁷ – 6x⁵ + 7x – 6 is …….. (T.S.Mar.15)
A) 4
B) 5
C) 6
D) 7
Answer:
D) 7

Question 33.
If the order of ax⁵ + 3x⁴ + 4x³ + 3x² + 2x + 1 is 4 then a = …………
A) 5
B) 4
C) 0
D) not possible
Answer:
C) 0

Question 34.
Sum of zeroes of the polynomial 2x² – 8x + 6 is …… (T.S. Mar.15)
A) 4
B) -4
C) 3
D) -3
Answer:
A) 4

Question 35.
Product fo zeroes of the cubic polynomial 3x³ – 5x² – 11x – 3 is ……….. (T.S. Mar.15)
A) 1
B) -1
C) \(\frac{5}{3}\)
D) \(\frac{-5}{3}\)
Answer:
A) 1

Question 36.
The value of p(x) = 4x² + 3x + 1 at x = -1 is (T.S. Mar.’15)
A) 4
B) 3
C) 2
D) 1
Answer:
C) 2

Question 37.
The zero value of polynomial px + q is ….. (June’15)
A) \(\frac{-\mathrm{q}}{\mathrm{p}}\)
B) \(\frac{\mathrm{p}}{\mathrm{q}}\)
C) \(\frac{-\mathrm{p}}{\mathrm{q}}\)
D) q
Answer:
A) \(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 38.
4y² – 5y + 1 is a ………. (June’15)
A) linear polynomial
B) cubic polynomial
C) constant polynomial
D) quadratic polynomial
Answer:
D) quadratic polynomial

Question 39.
4x + 6y = 18 doesn’t pass through origin. It indicates a ………. (June’15)
A) curved line
B) straight line
C) parabola
D) None
Answer:
B) straight line

Question 40.
If α, β are the zeroes of the polynomial x² – x – 6 then α²β² = ……… (June ’14)
A) 36
B) 6
C) -6
D) -36
Answer:
A) 36

Question 41.
When p(x) = x² – 8x + k leaves a remainder when it is divided by (x – 1) then k = ………. (A.P. Mar.15)
A) 13
B) 8
C) -5
D) 5
Answer:
A) 13

Question 42.
The zero value of linear polynomial ax – b = ………… (T.S.Mar.16)
A) \(\frac{b}{a}\)
B) \(\frac{a}{b}\)
C) –\(\frac{b}{a}\)
D) –\(\frac{a}{b}\)
Answer:
A) \(\frac{b}{a}\)

Question 43.
The product of zeroes of 2x² – 3x + 6 = …….. (A.P. Mar.’15)
A) 3
B) -3
C) 2
D) -2
Answer:
A) 3

Question 44.
Sum of zeroes of bx² + ax + c = ……… (A.P.Mar. 16)
A) \(\frac{-a}{a}\)
B) \(\frac{a}{b}\)
C) –\(\frac{b}{a}\)
D) \(\frac{b}{a}\)
Answer:
A) \(\frac{-a}{a}\)

Question 45.
If α, β, γ are zeroes of x³ + 3x² – x + 2 then αβγ = ……. (A.P.Mar.16)
A) 2
B) 3
C) 5
D) -2
Answer:
D) -2

Question 46.
The quadratic polynomial having 2, 3 as zeroes is ……. (T.S. Mar.15)
A) x² – 5x – 6
B) x² + 5x + 6
C) x² – 5x + 6
D) x² + 5x – 6
Answer:
C) x² – 5x + 6

Question 47.
Which of the following has only one zero. (T.S.Mar.15)
A) p(x) = 2x² – 3x + 4
B) p(x) = x² – 2x + 1
C) p(x) = 2x + 3
D) p(x) = 5
Answer:
C) p(x) = 2x + 3

Question 48.
Observe the given rectangular figure, then its area in polynomial function is ………. (T.S. Mar.15)

A) A(x) = x² + 7x + 30
B) A(x) = -x² + 7x + 30
C) A(x) = x² – 7x + 30
D) A(x) = -x² – 7x + 30
Answer:
B) A(x) = -x² + 7x + 30

Question 49.
The coefficient of x⁷ in the polynomial 7x⁷ – 17x¹¹ + 27x⁵ – 7 is ……….. (T.S.Mar.16)
A) -1
B) 0
C) 7
D) 17
Answer:
B) 0

Question 50.
A quadratic polynomial whose zeroes are 5 and -2 is
A) x² + 5x – 2
B) x² + 3x – 10
C) x² – 3x – 10
D) x² – 2x + 5
Answer:
C) x² – 3x – 10

Question 51.
If \(\sqrt{3}\) and –\(\sqrt{3}\) are fhe zeroes of a polynomial p(x), then p(x) is
A) x² – 9
B) 3x² – 1
C) x² + 3
D) x² – 3
Answer:
D) x² – 3

Question 52.
The quadratic polynomial whose zeroes are \(\sqrt{15}\) and –\(\sqrt{15}\) is
A) x² – 15
B) x² – 225
C) 15x² – 1
D) x² – \(\sqrt{15}\)
Answer:
A) x² – 15

Question 53.
If one zero of the quadratic polynomial 2x² + kx – 15 is 3, then the other zero is
A) \(\frac{-15}{2}\)
B) k
C) \(\frac{-5}{2}\)
D) -15
Answer:
C) \(\frac{-5}{2}\)

Question 54.
The maximum number of zeroes that a poly-nomial of degree 3 can have is
A) three
B) one
C) two
D) none
Answer:
A) three

Question 55.
The product and sum of the zeroes of the quadratic polynomial ax² + bx + c respectively are
A) \(\frac{c}{b}\), 1
B) \(\frac{-b}{a}\), \(\frac{c}{a}\)
C) \(\frac{c}{a}\), \(\frac{b}{a}\)
D) \(\frac{c}{a}\), \(\frac{-b}{a}\)
Answer:
D) \(\frac{c}{a}\), \(\frac{-b}{a}\)

Question 56.
The number of zeroes of the polynomial function p(x) whose graph is given below is ( )

A) 2
B) 2
C) 0
D) 3
Answer:
D) 3

Question 57.
The zeroes of the polynomial p(x) = 4x² – 12x + 9 are
A) \(\frac{-3}{2}\), \(\frac{-3}{2}\)
B) -3, -4
C) \(\frac{3}{2}\), \(\frac{3}{2}\)
D) 3, 4
Answer:
C) \(\frac{3}{2}\), \(\frac{3}{2}\)

Question 58.
If α and β are zeroes of the polynomial p(x) = x² – 5x + 6, then the value of α + β – 3αβ is
A) – 13
B) 6
C) 13
D) -5
Answer:
A) – 13

Question 59.
If 1 is the zero of the quadratic polynomial x² + kx – 5, then the value of k is
A) 0
B) 5
C) -4
D) 4
Answer:
D) 4

Question 60.
The number of zeroes lying between -2 and 2 of the polynomial f(x) whose graph is given below is

A) 3
B) 4
C) 2
D) 1
Answer:
C) 2

Question 61.
If both the zeroes of a quadratic polynomial ax² + bx + c are equal and opposite in sign, then b is
A) -1
B) 5
C) 1
D) 0
Answer:
D) 0

Question 62.
Sum and product of the zeroes of polynomial x² – 3 are respectively
A) 0, 3
B) 0, -3
C) -3, 0
D) 3, 0
Answer:
B) 0, -3

Question 63.
If the zeroes of a quadratic polynomial are equal in magnitude but opposite in sign, then
A) product of its zeroes is 0
B) sum of its zeroes is 0
C) there are no zeores of the polynomial
D) one of the zero is 0
Answer:
B) sum of its zeroes is 0

Question 64.
If one of the zeroes of the quadratic polynomial ax² + bx + c is 0, thenn the other zero
A) \(\frac{b}{a}\)
B) \(\frac{-c}{a}\)
C) \(\frac{-b}{a}\)
D) 0
Answer:
C) \(\frac{-b}{a}\)

Question 65.
The polynomial whose zeroes are -5 and 4 is
A) x² + x – 20
B) x² + 5x – 4
C) x² – 9x – 20
D) x² – 5x + 4
Answer:
A) x² + x – 20

Question 66.
If – 1 is a zero of the polynomial f(x) = x² – 7x – 8, then the other zero is
A) 8
B) -8
C) 1
D) 6
Answer:
A) 8

Question 67.
If one zero of the quadratic polynomial x² – 5x – 6 is 6, then the other zero is ( )
A) 1
B) – 5
C) -1
D) – 6
Answer:
C) -1

Question 68.
If a and p are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) as its zeros is ……….
A) qx² + px + 1
B) qx² + px + 10
C) px² + qx + 1
D) none
Answer:
A) qx² + px + 1

Question 69.
If f(x) = ax² + bx + c has no real zeros and a + b + c < 0 then ……….
A) c = 0
B) c < 9
C) c < 0 D) c > 0
Answer:
C) c < 0

Question 70.
If α, β, γ are the zeroes of the polynomial f(x) = ax³ + bx² + cx + d then \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) + \(\frac{1}{\gamma}\) = ………..
A) \(\frac{1}{d}\)
B) \(\frac{1}{c}\)
C) \(\frac{c}{d}\)
D) –\(\frac{c}{d}\)
Answer:
D) –\(\frac{c}{d}\)

Question 71.
If the product of zeros of the polynomial f(x) = ax³ – 6x² + 11x – 6 is 4 then a = ……..
A) \(\frac{2}{3}\)
B) \(\frac{3}{2}\)
C) -1
D) 9
Answer:
B) \(\frac{3}{2}\)

Question 72.
If α, β are the zeros of the polynomial f(x) = x² + x + k is reciprocal of the other then k = ……..
A) 5
B) -5
C) 1
D) none
Answer:
D) none

Question 73.
If one root of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other then k = ………..
A) 5
B) – 5
C) 1
D) none
Answer:
A) 5

Question 74.
If α, β are the zeros of polynomial f(x) = x² – p (x + 1) – C then (α + 1) (β + 1) = ……….
A) a
B) 1 + c
C) 1 – c
D) c
Answer:
C) 1 – c

Question 75.
If the sum of the zeros of the polynomial f(x) = 2x³ – 3kx² + 4x – 5 is 6 then k =
A) – 1
B) 9
C) 0
D) 4
Answer:
D) 4

Question 76.
If α, β, γ are the zeros of the polynomial f(x) = ax³ + bx² + cx + d then α² + β² + γ² = ………
A) \(\frac{\mathrm{b}^2+4 \mathrm{ac}}{2}\)
B) \(\frac{b^2-2 a c}{a^2}\)
C) \(\frac{b+2 a c}{a^2}\)
D) none
Answer:
B) \(\frac{b^2-2 a c}{a^2}\)

Question 77.
If the polynomial f(x) = ax³ – bx – a is divisible by the polynomial g(x) = x² + bx + c then ab = ………
A) 1
B) 7
C) -1
D) 0
Answer:
A) 1

Question 78.
The product of the zeros of x³ + 4x² + x – 6 is ……….
A) 8
B) 7
C) 6
D) – 6
Answer:
C) 6

Question 79.
A quadratic polynomial, the sum of whose zeros is 0 and one zero is 3 is
A) x² + 3
B) x – 3
C) x² – 9
D) x² – 3
Answer:
C) x² – 9

Question 80.
If \(\sqrt{5}\) and –\(\sqrt{5}\) are two zeros of the polynomial x³ + 3x² – 5x – 15 then its third zero is ………
A) 7
B) 3
C) -3
D) none
Answer:
C) -3

Question 81.
If α, β, γ are the zeros of the polynomial f(x) = x³ – px² + qx – r then \(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}\) = …………
A) \(\frac{r}{p}\)
B) \(\frac{p}{r}\)
C) -r
D) none
Answer:
B) \(\frac{p}{r}\)

Question 82.
If x + 2 is a factor of x² + ax + 2b and a + b = 4 then a = ………
A) 3
B) 2
C) -1
D) 4
Answer:
A) 3

Question 83.
In the above problem b = ………
A) 9
B) 0
C) -1
D) 1
Answer:
D) 1

Question 84.
If one zero of the polynomial f(x) = (k² + 4)x² + 13x + 4k is reciprocal of the other then k =
A) -2
B) 2
C) 9
D) 1
Answer:
B) 2

Question 85.
If two zeros of x³ + x² – 5x – 5 are \(\sqrt{5}\) and –\(\sqrt{5}\) then its third zero is …………
A) -3
B) 2
C) -1
D) none
Answer:
C) -1

Question 86.
If zeros of the polynomial f(x) = x³ – 3px² + qx – r are in AP then 2p³ = …………..
A) pq – r
B) p – r
C) pq + 1
D) p + q – r
Answer:
A) pq – r

Question 87.
If α, β are the zeros of the polynomial f(x) = ax² + bx + c then \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) = ……….
A) b² – 2ac
B) \(\frac{b+4 a c}{c}\)
C) \(\frac{b+4 a c}{c^2}\)
D) \(\frac{b^2-2 a c}{c^2}\)
Answer:
D) \(\frac{b^2-2 a c}{c^2}\)

Question 88.
What should be subtracted from the polynomial x² – 16x + 30 so that 15 is the zero of the resulting polynomial ?
A) 15
B) -1
C) -15
D) none
Answer:
A) 15

Question 89.
The number of zeros of the polynomial in the graph is …………

A) 2
B) -1
C) 4
D) 1
Answer:
D) 1

Question 90.
The number of zeros of the polynomial in the graph is …….

A) 1
B) -2
C) 0
D) 4
Answer:
C) 0

Question 91.
The below graph represents

A) polynomial
B) not a polynomial
C) two zeros
D) none
Answer:
B) not a polynomial

Question 92.
ax + b represents ………. polynomial.
A) quadratic
B) cubic
C) linear
D) none
Answer:
C) linear

Question 93.
ax² + bx + c is a …… polynomial.
A) quadratic
B) linear
C) cubic
D) fourth
Answer:
A) quadratic

Question 94.
Which of the following is a cubic polyno-mial in general form ?
A) bx² + c³x + x + 1
B) ax³ + bx² + cx + d
C) x² + 2³
D) none
Answer:
B) ax³ + bx² + cx + d

Question 95.
The degree of the polynomial ax⁴ + bx³ + cx² + dx + e is ……..
A) 5
B) 4
C) 6
D) 0
Answer:
B) 4

Question 96.
a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ………. a_nxⁿ is polyno-mial of degree ……….
A) 1
B) n – 2
C) n
D) n²
Answer:
C) n

Question 97.
If a < 0 then the shape of ax² + bx + c = 0 is ……..

Answer:
D) None

Question 98.
If α, β, γ are roots of a cubic polynomial α + β + γ = ………
A) \(\frac{c}{a}\)
B) \(\frac{b}{a}\)
C) \(\frac{-b}{a}\)
D) none
Answer:
C) \(\frac{-b}{a}\)

Question 99.
If α, β, γ are roots of a cubic polynomial then αβ + βγ + γα = ……
A) \(\frac{-c}{a}\)
B) \(\frac{c}{a}\)
C) \(\frac{-d}{a}\)
D) none
Answer:
B) \(\frac{c}{a}\)

Question 100.
If α, β, γ are roots of a cubic polynomial then αβγ = ……
A) \(\frac{c}{a}\)
B) \(\frac{d}{a}\)
C) \(\frac{-d}{a}\)
D) none
Answer:
C) \(\frac{-d}{a}\)

Question 101.
f(x) = 3x – 2 then zero of f(x) is ………
Answer:
A) \(\frac{1}{2}\)
B) \(\frac{1}{3}\)
C) \(\frac{2}{-3}\)
D) \(\frac{2}{3}\)
Answer:
D) \(\frac{2}{3}\)

Question 102.
p(t) = t³ – 1, p(-2) = ………..
A) -9
B) -4
C) 1
D) 0
Answer:
A) -9

Question 103.
p(x) = x² + 5x + 6 then zeros of p(x) = ……….
A) -2, -3
B) 3, -2
C) 4, 1
D) 1, 8
Answer:
A) -2, -3

Question 104.
p(x) = 4x² + 3x – 1 then p(\(\frac{1}{4}\)) = ……..
A) 1
B) 0
C) -1
D) 12
Answer:
B) 0

Question 105.
x² + 7x + 10 = ……..
A) (x + 3)²
B) (x + 2)²
C) (x – 2) (x – 3)
D) (x + 2) (x + 5)
Answer:
D) (x + 2) (x + 5)

Question 106.
p(x) = x³ + 4x² + 5x – 2 then p(1) = ………
A) 8
B) 7
C) 3
D) none
Answer:
A) 8

Question 107.
p(x) = 3x³ – 2x² + 6x – 5 then p(2) = ……..
A) 19
B) 10
C) 12
D) 23
Answer:
D) 23

Question 108.
If m and n are zeros of the polynomial 3x² + 11x – 4 then the value of \(\frac{m}{n}\) + \(\frac{n}{m}\) = ………..
A) \(\frac{4}{11}\)
B) \(\frac{4}{7}\)
C) \(\frac{11}{4}\)
D) none
Answer:
C) \(\frac{11}{4}\)

Question 109.
Sum of the zeros of x² + 7x + 10 is ……..
A) 7
B) -3
C) 4
D) none
Answer:
D) none

Question 110.
If p and q are the zeros of the polynomial t² – 4t + 3 then \(\frac{1}{p}\) + \(\frac{1}{q}\) – 2pq + \(\frac{14}{3}\) = ……..
A) 0
B) -1
C) 2
D) 3
Answer:
A) 0

Question 111.
A polynomial of degree 3 is called …….. polynomial.
A) zero
B) order
C) qudratic
D) cubic
Answer:
D) cubic

Question 112.
The quotient when x⁴ + x³ + x² – 2x – 3 is divided by x² – 2 is ……..
A) x² + x + 3
B) x – 2
C) x² + 3x + 1
D) none
Answer:
A) x² + x + 3

Question 113.
In the above problem remainder is ……..
A) 1
B) 4
C) 3
D) -3
Answer:
C) 3

Question 114.
One zero of the polynomial 2x² + 3x + k is \(\frac{1}{2}\) then k = ………
A) 4
B) 1
C) 2
D) – 2
Answer:
D) – 2

Question 115.
In the above problem other zero is …….
A) -2
B) 2
C) 3
D) 4
Answer:
A) -2

Question 116.
Sum of the zeros of 6x² = 1 is ……
A) 3
B) 2
C) 0
D) -1
Answer:
C) 0

Question 117.
What must be subtracted or added to p(x) = 8x⁴ + 14x³ – 2x² + 8x – 12 so that 4x² + 3x – 2 is a factor of p(x) ?
A) 5x – 3
B) 15x – 1
C) 5x – 2
D) 15x – 14
Answer:
D) 15x – 14

Question 118.
The quadratic polynomial whose zeros are 4 + \(\sqrt{5}\) and 4 – \(\sqrt{5}\) is ……..
A) x² – 8x + 11
B) x² – 11x + 1
C) x² + 8x + 3
D) none
Answer:
A) x² – 8x + 11

Question 119.
If one of the zeros of the quadratic polyno-mial f(x) = 14x² – 42k²x – 9 is negative of the other then k = ……….
A) 3
B) -1
C) 0
D) none
Answer:
C) 0

Question 120.
If one zero of a polynomial 3x² – 8x + 2k + 1 = 0 is seven times the other then k = ……..
A) \(\frac{2}{3}\)
B) \(\frac{1}{3}\)
C) 1
D) none
Answer:
A) \(\frac{2}{3}\)

Question 121.
If 2x + 3 is a factor of 2x³ – x – b + 9x² then the value of b is
A) 3
B) 7
C) 10
D) 15
Answer:
D) 15

Question 122.
Divide (x³ – 6x² + 11x – 12) by (x² – x + 2) then quotient is …………
A) x + 5
B) x – 5
C) x + 1
D) none
Answer:
A) x + 5

Question 123.
In the above problem remainder is ……….
A) 2x – 1
B) x + 1
C) 4x – 2
D) none
Answer:
C) 4x – 2

Question 124.
Product of zeros of 3x² = 1 is ……….
A) -1
B) -2
C) 3
D) \(-\frac{1}{3}\)
Answer:
D) \(-\frac{1}{3}\)

Question 125.
Degree of 5x⁷ – 6x⁵ + 7x + 1 is …….
A) 4
B) 1
C) 7
D) 3
Answer:
C) 7

Question 126.
(3x – 4) (x + 1) = …….
A) x² – 3x + 1
B) x² – x + 4
C) x² – 3x + 7
D) 3x² – x – 4
Answer:
D) 3x² – x – 4

Question 127.
(x² – 3x – 28) + (x + 4) = ……….
A) x – 7
B) x + 7
C) x + 3
D) x – 1
Answer:
A) x – 7

Question 128.
In the above problem f(\(\frac{1}{2}\)) = ……..
A) 1
B) 7
C) x + 3
D) 0
Answer:
D) 0

Question 129.
(x + \(\sqrt{5}\)) (x – 3\(\sqrt{5}\)) = ………
A) x² – 2\(\sqrt{5}\) x + 15
B) x² – 2\(\sqrt{5}\)x – 15
C) x² – \(\sqrt{5}\)x + 15
D) none
Answer:
B) x² – 2\(\sqrt{5}\)x – 15

Question 130.
The remainder when 3x³ + x² + 2x + 5 is divided by x² + 2x + 1 is ……….
A) 9x + 10
B) x + 10
C) x – 1
D) 9x – 1
Answer:
A) 9x + 10

Question 131.
α = a – b, β = a + b then the quadratic polynomial is ……….
A) x³ – a²x + b²
B) x² – a³x + a²
C) x² – ax + a² + b²
D) x² – 2ax + a² – b²
Answer:
D) x² – 2ax + a² – b²

Question 132.
If the product of zeros of 9x² + 3x + p is 7 then p = ………
A) 14
B) -63
C) 63
D) 70
Answer:
C) 63

Question 133.
Degree of (x – 1) (x – 3) is ……..
A) 3
B) 2
C) 1
D) 7
Answer:
B) 2

Question 134.
The value of x¹⁵ – 1 at x = 0 is ……..
A) 3
B) 9
C) 7
D) -1
Answer:
D) -1

Question 135.
Zeros of the polynomial x² – 4x + 3 are 1 and p then p = ……..
A) 7
B) 3
C) 1
D) none
Answer:
B) 3

Question 136.
Degree of a linear polynomial is ……..
A) 2
B) 3
C) 7
D) 1
Answer:
D) 1

Question 137.
(a + 1)² = ………..
A) a² + 1
B) a² + 2 + a
C) a² + 2a + 1
D) none
Answer:
C) a² + 2a + 1

Question 138.
(x³ – 8) + x⁴ + 2x³ – 8x – 16
A) x
B) x – 1
C) x + 2
D) none
Answer:
C) x + 2

Question 139.
p(x) = \(\frac{x+1}{1-x}\) then P(0) = ………
A) 1
B) -1
C) 2
D) 3
Answer:
A) 1

Question 140.
(x – \(\sqrt{3}\)) (x + \(\sqrt{3}\)) = ……
A) x + 3
B) x² – 3
C) x + 7
D) none
Answer:
B) x² – 3

Question 141.
If one zero of the polynomial is \(\sqrt{2}\) – 1 then other zero may be ……
A) 1 + \(\sqrt{3}\)
B) \(\sqrt{2}\) + 1
C) \(\sqrt{2}\) – 2
D) all
Answer:
B) \(\sqrt{2}\) + 1

Question 142.
Binomial contains almost ….. terms.
A) 5
B) 4
C) 1
D) 2
Answer:
D) 2

Question 143.
p(x) = \(\frac{12}{x-3}\), p(3) = ……..
A) 1
B) 12
C) 0
D) not defined
Answer:
D) not defined

Question 144.
p(-3) = 0 then p(x) = ……
A) x – 7
B) x + 1
C) x – 4
D) x + 3
Answer:
D) x + 3

Question 145.
Number of constant terms in the polynomial x² + 7x – 7 is ……
A) 2
B) 1
C) 3
D) 2
Answer:
B) 1

Question 146.
In the product (x + 4) (x + 2) the constant term is ……..
A) 7
B) -3
C) 6
D) 8
Answer:
D) 8

Question 147.
p(\(\frac{\mathbf{a}}{\mathbf{b}}\)) = 0 then p(x) = ……….
A) ax – b
B) bx – a
C) ax
D) bx
Answer:
B) bx – a

Question 148.
a(a + 1) (a + 2) (a + 3) + a(a + 3) =
A) (a + 1) (a + 2)
B) (a + 1)²
C) (a + 2)²
D) none
Answer:
A) (a + 1) (a + 2)

Question 149.
(x² – 8x + 12) + (x – 6) = ………
A) x + 3
B) x + 1
C) x + 2
D) x – 2
Answer:
D) x – 2

Question 150.
x(3x² – 108) + 3x(x – 6) = …….
A) x + 3
B) x – 6
C) x + 6
D) x – 7
Answer:
C) x + 6

Question 151.
(p + 4) (p – 4) (p² + 16) = ……..
A) p⁴ – 16
B) p⁴ + 256
C) p³ – 100
D) p⁴ – 256
Answer:
D) p⁴ – 256

Question 152.
36(x + 4) (x² + 7x + 10) + 9 (x + 4) = …….
A) 4(x + 5) (x + 2)
B) (x + 5) (x – 7)
C) (x + 5) (3x – 1)
D) none
Answer:
A) 4(x + 5) (x + 2)

Question 153.
The degree of the polynomial 9xy³ + 10y⁴ + \(\frac{5}{4}\) x⁴ + \(\frac{7}{3}\) x³y² is (T.S.Mar.15)
A) 3
B) 2
C) 4
D) 5
Answer:
D) 5

Question 154.
The zero of the linear polynomial 2x + 3 is (T.S.Mar.15)
A) 0
B) \(\frac{-3}{2}\)
C) -1 \(\frac{1}{2}\)
D) B and C
Answer:
B) \(\frac{-3}{2}\)

Question 155.
The product of the zeroes of the polynomial 3x³ – 5x² – 10x + 15 is ……. (T.S. Mar.’15)
A) – 5
B) 5
C) \(\frac{5}{3}\)
D) \(\frac{-10}{3}\)
Answer:
D) \(\frac{-10}{3}\)

Question 156.
The quadratic polynomial with zeros 2 and 3 is (T.S. Mar.’15)
A) x² – 5x + 6
B) (x – 2) (x – 3)
C) 2x² – 10x + 12
D) All the above
Answer:
B) (x – 2) (x – 3)

Question 157.
If a fraction becomes 2 when 9 is added to its numerator and 1 when 2 is substracts from its denominator then the fraction is ………… (AP-SA-I-2016 )
A) 5/8
B) 8/5
C) 5/7
D) 7/9
Answer:
C) 5/7

Question 158.
The graph of y = ax + b is a straight line which intersects X-axis at exactly one point namely, ……
A) (0, \(\frac{b}{a}\))
B) (\(\frac{b}{a}\), 0)
C) (0, \(\frac{-b}{a}\))
D) (\(\frac{-b}{a}\), 0)
Answer:
D) (\(\frac{-b}{a}\), 0)

Question 159.
Coefficient of x in a polynomial ax² + bx + c is ‘0’. Then its zeroes are ………
A) equal
B) additive inverses to one another
C) multiplicative inverses to one another
D) none
Answer:
B) additive inverses to one another

Solving these TS 10th Class Maths Bits with Answers Chapter 2 Sets Bits for 10th Class will help students to build their problem-solving skills.

Sets Bits for 10th Class

Question 1.
The symbol for a null set is
A) ∈
B) ∪
C) ϕ
D) \(\not \subset\)
Answer:
C) ϕ

Question 2.
C = {x : x is a circle in a given plane} is
A) finite set
B) infinite set
C) universal set
D) void set
Answer:
B) infinite set

Question 3.
Which of the following set is infinite?
A) The set of natural numbers ≤ 100
B) The set of even natural numbers between 50 and 100.
C) The set of points on a circle.
D) The set of prime numbers between 10 and 50.
Answer:
C) The set of points on a circle.

Question 4.
The number of elements in the set D = {x : x is a day of the week} is
A) 6
B) 4
C) 5
D) 7
Answer:
D) 7

Question 5.
The symbol for “implies” is
A) ⇒
B) ⇔
C) ⊂
D) \(\supset\)
Answer:
A) ⇒

Question 6.
If A = {1, 2, 3} the number of subsets of A is
A) 3
B) 8
C) 9
D) 6
Answer:
B) 8

Question 7.
Two sets A and B are said to be disjoint if
A) A∪B = ϕ
B) A – B = ϕ
C) B – A = ϕ
D) A ∩ B = ϕ
Answer:
D) A ∩ B = ϕ

Question 8.
For every set A, A ∩ ϕ =
A) ϕ
B) A
C) 0
D) 1
Answer:
A) ϕ

Question 9.
For every set A, A ∩ A =
A) 1
B) 0
C) A
D) ϕ
Answer:
C) A

Question 10.
If A and B are two sets, then x ∈ A ∩ B ⇒
A) x ∈ A or x ∈ B
B) x ∈ A and x ∈ B
C) x ∈ A and x \(\notin\) B
D) x\(\notin\)A and x ∈ B
Answer:
B) x ∈ A and x ∈ B

Question 11.
If A and B are two sets, then x ∈ A∪B ⇒
A) x ∈A and x\(\notin\)B
B) x\(\notin\)A and x∈B
C) x ∈ A or x ∈ B
D) x\(\notin\)A and x\(\notin\)B
Answer:
C) x ∈ A or x ∈ B

Question 12.
If A and B are two sets, then x ∈ A – B ⇒
A) x ∈ A or x ∈ B
B) x ∈ A and x ∈ B
C) x\(\notin\)A and x∈B
D) x\(\notin\)A and x\(\notin\)B
Answer:
D) x\(\notin\)A and x\(\notin\)B

Question 13.
If V = {a, e, i, o, u} and B = {a, i, k, u} then V – B =
A) {a, e, i, o, u}
B) {a, e, o, u}
C) {e, o}
D) {a, e, o}
Answer:
C) {e, o}

Question 14.
The shaded portion in the adjacent figure represents

A) A ∩ B
B) A ∪ B
C) A – B
D) A ⊂ B
Answer:
B) A ∪ B

Question 15.
The shaded portion in the adjacent figure represents

A) A ∩ B
B) A ∪ B
C) A ⊂ B
D) B ⊂ A
Answer:
A) A ∩ B

Question 16.
The shaded portion in the adjacent figure represents
A) A ⊂ B
(B) B ⊂ A
(C) B – A
(D) A – B
Answer:
(C) B – A

Question 17.
Which of the following sets are equal ?
A) E = {1, 0}, H = {a, b)
B) A = {0, a}, C = {b, 0}
C) D = {4, 8, 12}, F = {8, 12, 4}
D) G = {1, 5, 7, 11}, I = {1, 2, 3, 4}
Answer:
C) D = {4, 8, 12}, F = {8, 12, 4}

Question 18.
If A = {1, 3} and B = {1, 5, 9} then A – B =
A) {3}
B) {1}
C) {5, 9}
D) {1, 3, 5, 9}
Answer:
A) {3}

Question 19.
Which of the following sets in infinite ?
A) {x : x ∈ N and (x – 1 )(x – 2) = 0}
B) {x : x ∈ N and x² = 9}
C) {x : x ∈ N and 2x – 1 = 0}
D) {x : x ∈ N and x is odd}
Answer:
D) {x : x ∈ N and x is odd}

Question 20.
Which of the following sets finite ?
A) {x : x ∈ N and x is prime}
B) {x : x ∈ N and x is an even number less than 20}
C) {x : x² = 4 and x is an odd integer}
D) {x : x² is an even prime greater than 2}
Answer:
B) {x : x ∈ N and x is an even number less than 20}

Question 21.
Set of human being that reside on moon is ………
A) finite set
B) null set
C) infinite set
D) universal set
Answer:
B) null set

Question 22.
Which of the following is true in the following venn diagram ……..

A) A ∪B = ϕ
B) A ∪ B = µ
C) A ∩ B = µ
D) A ∩ B = ϕ
Answer:
D) A ∩ B = ϕ

Question 23.
Which of the following is an example for finite set
A) {x / x ∈ N and x² = 9}
B) Set of rational numbers in between 2 and 3
C) Multiples of even primes
D) Set of all primes
Answer:
A) {x / x ∈ N and x² = 9}

Question 24.
The number of subsets of the null set ϕ is ……….
A) 0
B) 1
C) 3
D) 4
Answer:
B) 1

Question 25.
If A = (a, b, c, d), how many subsets does the set ‘A’ have
A) 5
B) 6
C) 16
D) 65
Answer:
C) 16

Question 26.
A = {1, 2, 3, 7, 8}; B = {4, 5, 6, 7} find A ∩ B
A) {1, 2, 3, 4, 5, 6, 7, 8}
B) 7
C) ϕ
D) {7, 8}
Answer:
B) 7

Question 27.
n(A) = 14; n (B) = 11; n(A ∩ B) = 19 then n(A ∪B) = ……
A) 6
B) 16
C) 22
D) 25
Answer:
A) 6

Question 28.
The shaded area in the figure shows.

A) A – B
B) B – A
C) A ∆ B
D) (A ∪ B) (A ∩ B)
Answer:
C) A ∆ B

Question 29.
If ‘A’ and ‘B’ are two sets such that A ⊂ B then A ∪ B =
A) A
B) B
C) A ∩ B
D) None
Answer:
B) B

Question 30.
If ‘A’ and ‘B’ are disjoint sets then n(A ∩ B) = …….
A) 1
B) ϕ
C) 0
D) { }
Answer:
D) { }

Question 31.
Match the following :
Group -1 Group – II

Answer:
A) L → (iii), M → (i), N → (ii), O → (iv)
B) L → (i), M → (ii), N → (iii), O → (iv)
C) L → (iii), M → (i), N → (iv), O → (ii)
D) L → (iii), M → (ii), N → (ii), O → (iv)
Answer:
A) L → (iii), M → (i), N → (ii), O → (iv)

Question 32.
A ∩ ϕ = ………
A) A
B) ϕ
C) ϕ – A
D) { }
Answer:
A) A

Question 33.
Let A = {1, 2, {1}, {1, 2}, 3, 4}, then which of the following is true ?
A) {3} ∈ A
B) {1, 3} ∈ A
C) {1, 2} ∈ A
D) None
Answer:
C) {1, 2} ∈ A

Question 34.
Which of the following is false ?
A) {1} ∈ A
B) {1, 2} ⊆ A
C) {1, 2} ∈ A
D) None
Answer:
D) None

Question 35.
If A = {1, 2, 3, 4}, B = {2, 4,6, 8} then A – B =
A) {6, 8}
B) {1, 2}
C) {1, 3}
D) None
Answer:
C) {1, 3}

Question 36.
If n(A ∪ B) = 8, n (A) = 6, n(B) = 4 then n (A ∩ B) = …….
A) 2
B) 4
C) 6
D) 8
Answer:
A) 2

Question 37.
Let A, B are two sets such that n(A) = 5, n(B) = 7 then the maximum number of elements in A ∪ B is
A) 7
B) 9
C) 12
D) None
Answer:
C) 12

Question 38.
If A = {1, 2, 3, 4 }, then the cardinality of set A is
A) 3
B) 4
C) 5
D) 6
Answer:
B) 4

Question 39.
If A, B are disjoint sets such that n(A) = 4 and n(A∪B) = 7, then n(B) = ……..
A) 4
B) 11
C) 3
D) 20
Answer:
C) 3

Question 40.
An object of a set is called
A) Subject
B) Number
C) Alphabet
D) Element
Answer:
D) Element

Question 41.
The symbol used for ‘belongs to’ is
A) ⊂
B) ⊆
C) ∈
D) \(\notin\)
Answer:
C) ∈

Question 42.
The set of all real numbers is
A) ϕ
B) Finite set
C) Infinite set
D) None
Answer:
C) Infinite set

Question 43.
The number of elements in the empty set is
A) 0
B) ϕ
C) 1
D) ∞
Answer:
A) 0

Question 44.
If A = {1, 2, 2, 1, 3, 4, 3, 4}, then n(A) =
A) 0
B) 4
C) 8
D) 20
Answer:
B) 4

Question 45.
If A ⊂ B, then A ∪ B = ………..
A) ϕ
B) μ
C) A
D) B
Answer:
D) B

Question 46.
A ∩ ϕ = ………
A) ϕ
B) µ
C) A
D) A^c
Answer:
A) ϕ

Question 47.
The German mathematician who developed the theory of sets ………
A) Bhaskara
B) Cayley
C) George Cantor
D) None
Answer:
C) George Cantor

Question 48.
A set is a …….. of objects.
A) well defined collection
B) collection
C) elements
D) none
Answer:
A) well defined collection

Question 49.
The objects in the set are called ……….. of the set.
A) elements
B) members
C) both A & B
D) false
Answer:
C) both A & B

Question 50.
Roster form of the set of natural numbers less then 6 is ……….
A) {4, 5, 6}
B) {1, 2, 3}
C) {2, 3, 4}
D) {1,2, 3, 4, 5}
Answer:
D) {1,2, 3, 4, 5}

Question 51.
The set formed from the letters of the word “SCHOOL” is ……
A) {S, O, H}
B) {H, O, L}
C) {S, C, H}
D) {S, C, H, O, L}
Answer:
D) {S, C, H, O, L}

Question 52.
Roster form is also called ……. form.
A) list
B) set
C) number
D) none
Answer:
A) list

Question 53.
Describing a set by same property common to all its elements is called …….. or ………
A) set builder form
B) rule form
C) both A & B
D) none
Answer:
C) both A & B

Question 54.
A = {2, 4, 6, 8, 10} then its rule form is ……..
A) A = {x² / x ∈ N}
B) A = {2x / x is odd, x ≤ 20}
C) A = {x³ / x ∈ N}
D) A = {x / x is an even number, x ≤ 10}
Answer:
D) A = {x / x is an even number, x ≤ 10}

Question 55.
If B = {1, 7, 2, 0, 6} then n(B) = …….
A) 5
B) 6
C) 7
D) 9
Answer:
A) 5

Question 56.
n (ϕ) = ……….
A) n
B) ϕ
C) 0
D) 9
Answer:
C) 0

Question 57.
Every set is ………. of itself.
A) subset
B) proper set
C) power set
D) none
Answer:
A) subset

Question 58.
If A ⊂ B and A ≠ B then ‘A’ is called the ……… of B.
A) subset
B) proper subset
C) power set
D) none
Answer:
B) proper subset

Question 59.
In set Builder form, the letter x denotes any …….. that belongs to the set.
A) constant
B) element
C) arbitrary element
D) none
Answer:
C) arbitrary element

Question 60.
In the rule form, the sland bar stands for ……….
A) subset
B) such that
C) belongs
D) all
Answer:
B) such that

Question 61.
2 is ……….. of set of natural numbers.
A) power
B) proper
C) subset
D) an element
Answer:
D) an element

Question 62.
-3 is …….. of the set of whole numbers.
A) proper
B) power
C) element
D) not an element
Answer:
D) not an element

Question 63.
0 ……….. to set of whole numbers.
A) does not belong
B) belong
C) subset
D) power set
Answer:
B) belong

Question 64.
A = {1, 2, 7, 10} then 7 …….. A.
A) ⊂
B) ∈
C) \(\notin\)
D) none
Answer:
B) ∈

Question 65.
A = {1, 2, 7,10} then 4 …… A.
A) \(\supset\)
B) ∈
c) ⊂
D) \(\notin\)
Answer:
D) \(\notin\)

Question 66.
“0 does not belong to the set of natural numbers” we write the statement symbolically as ………..
A) 0 \(\notin\) N
B) 0 ∈ N
C) 0 ⊂ N
D) none
Answer:
A) 0 \(\notin\) N

Question 67.
Set builder form of D = {1, \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\), \(\frac{1}{5}\), \(\frac{1}{6}\)} is ………..

Answer:
B) k = {\(\frac{x}{x}\) = \(\frac{1}{n^3}\), n ∈ N}

Question 68.
B = {\(\frac{x}{x}\) + 3 = 6}, B = …………
A) {0, 1, 3}
B) {7, 0}
C) {0, 3}
D) {3}
Answer:
D) {3}

Question 69.
A ……….. is a set with no elements in it.
A) inifinite set
B) finite set
C) null set
D) none
Answer:
C) null set

Question 70.
The null set is sometimes denoted as …………
A) empty set
B) void set
C) both A & B
D) none
Answer:
C) both A & B

Question 71.
Empty set is denoted by ………..
A) {ϕ}
B) {0}
C) N
D) ϕ
Answer:
D) ϕ

Question 72.
{0} is a set contains the element
A) 0
B) ϕ
C) {ϕ}
D) none
Answer:
A) 0

Question 73.
A set with only are element is known as ……. set.
A) Double
B) Singleton
C) Tri
D) None
Answer:
B) Singleton

Question 74.
Number of elements in a singleton set is …….
A) 0
B) 2
C) 7
D) 1
Answer:
D) 1

Question 75.
A = {x/x + 4 = 4} then A = ……
A) {4}
B) {0}
C) {0, 4}
D) {0, 7}
Answer:
B) {0}

Question 76.
B = {x / x ∈ N and x < 1000} is a …….. set.
A) finite
B) infinite
C) singleton
D) empty
Answer:
A) finite

Question 77.
If in two sets A and B, every element of A is in B and every element of B is in A. Then we write it as ………..
A) A ≠ B
B) A < B C) A > B
D) A = B
Answer:
D) A = B

Question 78.
A ≠ B means, set A and B do not contains same elements. This is
A) false
B) true
C) can’t be determined
D) none
Answer:
B) true

Question 79.
The number of elements in a set is called the ……. of the set.
A) cardinal
B) ordinal
C) true
D) all the above
Answer:
A) cardinal

Question 80.
If B = {1, 7, 2, 0, 6} then n(B) = ……..
A) 7
B) 0
C) 6
D) 5
Answer:
D) 5

Question 81.
If every element of A is also an element of B then we write this as ……
A) A < B
B) B < A
C) A ⊂ B
D) B ⊂ A
Answer:
C) A ⊂ B

Question 82.
If A = {1, 2, 3} and B = {1, 2, 3, 4} then we say A is a …… of B.
A) subset
B) superset
C) equal
D) none
Answer:
A) subset

Question 83.
In the above question B is ……… of A.
A) subset
B) superset
C) superset
D) all the above
Answer:
C) superset

Question 84.
A is not a subset of B if A contains ……… which is not in B.
A) equal
B) at least one element
C) 2
D) none
Answer:
B) at least one element

Question 85.
{x/x is a student of your school) is in ………. form.
A) Roster
B) Singleton
C) Set Builder
D) None
Answer:
C) Set Builder

Question 86.
{2, 4, 6, 8, 10} is an example of …….. set.
A) finite
B) infinite
C) singleton
D) two
Answer:
A) finite

Question 87.
{x/x is a natural number} is a ……….. set.
A) finite
B) infinite
C) singleton
D) none
Answer:
B) infinite

Question 88.
{x/x≠x} is a ……. set.
A) empty
B) infinite
C) singleton
D) none
Answer:
A) empty

Question 89.
A = {1, 2, 3}, B = {3, 4, 5} then A ∩ B = ……….
A) 3
B) {1, 2}
C) {4, 5}
D) {3}
Answer:
D) {3}

Question 90.
A = {a, b, c}, B = {c, a, b} then …………
A) A≠B
B) A = B
C) A ⊂ B
D) none
Answer:
B) A = B

Question 91.
A = {1, 2, 7}, B = {2, 1} then ………
A) A ⊂ B
B) B ⊂ A
C) A = B
D) none
Answer:
B) B ⊂ A

Question 92.
A ⊂ B then A – B = ……..
A) ⊂
B) B
C) A
D) ϕ
Answer:
D) ϕ

Question 93.
A – ϕ = ……….
A) A
B) ϕ
C) μ
D) none
Answer:
A) A

Question 94.
A ∪ A’ = ………..
A) A
B) ϕ
C) A
D) A’
Answer:
B) ϕ

Question 95.
μ’ = …………
A) A
B) μ
C) ϕ
D) none
Answer:
C) ϕ

Question 96.
A = {1, 2, 3}, B = {12, 0, 5} then A – B = ………
A) B
B) A
C) {5}
D) none
Answer:
B) A

Question 97.
A ∪ ϕ = ………….
A) A
B) B
C) ϕ
D) μ
Answer:
A) A

Question 98.
ϕ’ = ………..
A) B
B) A
C) μ
D) 0
Answer:
B) A

Question 99.
{2, 6, 10} ∩ {8, 9, 11, 12, 13} = ………..
A) {2}
B) {1, 2}
C) {13, 1}
D) ϕ
Answer:
D) ϕ

Question 100.
n(A) = 4 then n(p(A)) = …….
A) 12
B) 13
C) 15
D) 16
Answer:
D) 16

Question 101.
A – (A – B) = ……….
A) A ∩ B
B) ϕ
C) A ∪ B
D) B
Answer:
A) A ∩ B

Question 102.
(A’)’ = …………
A) A’
B) A
C) ϕ
D) none
Answer:
B) A

Question 103.
If A ⊂ B, then A – B = ………..
A) μ
B) B
C) A
D) ϕ
Answer:
D) ϕ

Question 104.
If A ⊂ B then A ∪ (B – A) = ………
A) B
B) A
C) ϕ
D) none
Answer:
A) B

Question 105.
W – {0} = ……..
A) R
B) A
C) Z
D) Q
Answer:
B) A

Question 106.
If A ⊂ B, B ⊂ C then ………
A) B = C
B) A = B
C) C ⊂ A
D) A ⊂ C
Answer:
D) A ⊂ C

Question 107.
Cardinal number of null set is ……
A) 4
B) ϕ
C) 0
D) none
Answer:
C) 0

Question 108.
A’ – B’ = ……..
A) A – B’
B) A’ – B
C) B – A
D) A – B
Answer:
C) B – A

Question 109.
If A = {1, 2, 3}, B = {3, 4, 5} then A ∆ B = ……….
A) {0}
B) {1, 2}
C) {7}
D) none
Answer:
D) none

Question 110.
A = ϕ, B = ϕ then A ∪ B = …………..
A) μ
B) ϕ
C) cant be determined
D) None
Answer:
B) ϕ

Question 111.
A ∩ B = ϕ then n (A ∩ B) = ………..
A) 7
B) 9
C) 3
D) none
Answer:
D) none

Question 112.
A ∪ B = A ∩ B then ……………….
A) A = B
B) A ≠ B
C) A ⊂ B
D) B ⊂ C
Answer:
A) A = B

Question 113.
μ’ = ϕ is called ……….. law.
A) Identity
B) Associative
C) Inverse
D) Complementary
Answer:
D) Complementary

Question 114.
A’ = B then A ∪ B = …….
A) A
B) μ
C) ϕ
D) none
Answer:
D) none

Question 115.
ϕ ∆ ϕ = ………..
A) μ
B) ϕ
C) {0}
D) none
Answer:
B) ϕ

Question 116.
A ∪ A = A is called ……… law.
A) idempotent
B) inverse
C) complete
D) identity
Answer:
A) idempotent

Question 117.
A ∪ B = B if …………
A) A \(\supset\)
B) A ⊂ B
C) A = B
D) none
Answer:
B) A ⊂ B

Question 118.
A = ϕ, B = ϕ then A ∩ B = ………
A) {6, 1}
B) {0}
C) μ
D) ϕ
Answer:
D) ϕ

Question 119.
n(A) = 10, n(B) = 4, n(A∩B) = 2 then n(A∪B) = ……….
A) 11
B) 16
C) 10
D) 12
Answer:
D) 12

Question 120.
(A ∪ B)’ = ………..
A) A’ ∩ B’
B) A’∪B
C) A’ ∩B
D) A ∩ B
Answer:
A) A’ ∩ B’

Question 121.
(A – B) ∪ {A – C) = ……..
A) (A – B) ∪ C
B) (A – B) ∩ C
C) (A – B) – C
D) none
Answer:
D) none

Question 122.
n(A) = 3 then number of proper subsets of A is ………
A) 10
B) 9
C) 7
D) 8
Answer:
C) 7

Question 123.
A ∩ B = ϕ then B ∩ A’ = ……….
A) μ
B) A
C) ϕ
D) B
Answer:
D) B

Question 124.
A ∪ (B ∩ C) = ……..
A) (A ∪ B) ∩ (A ∪ C)
B) (A∩B)∪(A∩C)
C) (A∪B)∩C
D) none
Answer:
A) (A ∪ B) ∩ (A ∪ C)

Question 125.
A = {all primes less than 20}
B = {all whole numbers less than 10} then
A∩B = ………….
A) {2, 3, 5, 7, 10}
B) {2, 8, 9}
C) {2, 3, 5, 7}
D) {2, 4, 6}
Answer:
C) {2, 3, 5, 7}

Question 126.
In the above problem A – B = ……….
A) {11, 13, 17, 19}
B) {1, 3, 7, 19}
C) {13, 17, 10}
D) none
Answer:
A) {11, 13, 17, 19}

Question 127.
N(A∪B) = 51, n(A) = 20, n(A∩B) = 13, then n(B) =
A) 80
B) 44
C) 40
D) 39
Answer:
B) 44

Question 128.
The identity element under union of sets of ………..
A) μ
B) {0}
C) {ϕ}
D) none
Answer:
C) {ϕ}

Question 129.
μ ∪ ϕ = ………..
A) ϕ
B) {0}
c) {ϕ}
D) μ
Answer:
D) μ

Question 130.

This venn – diagram represents ……..
A) A ∩ B
B) A – B
C) A ∪ B
D) A Δ B
Answer:
A) A ∩ B

Question 131.
N ∩ W = ……
A) Q
B) W
C) N
D) {0}
Answer:
C) N

Question 132.
If A and B are disjoint sets then n (A ∪ B) = ………
A) n(A) – n(B)
B) n(A) + n(B)
C) \(\frac{n(A)}{n(B)}\)
D) none
Answer:
B) n(A) + n(B)

Question 133.
From the venn diagram A ∪ B = ………..

A) {1, 2, 3}
B) {1, 2, 4, 5}
C) {6, 7, 10}
D) none
Answer:
D) none

Question 134.
Identity element under intersection of sets is ………
A) {0}
B) μ
C) ϕ
D) none
Answer:
B) μ

Question 135.
Which of the following is true ?
A) A – B ≠ B – A
B) A ∪ ϕ = A
C) μ’ = μ
D) all
Answer:
A) A – B ≠ B – A

Question 136.
A is the set of factors of 12. Which are of the following is not a member of A ?
A) 9
B) 10
C) 12
D) 5
Answer:
D) 5

Question 137.
If the number of proper subsets of a given set is 31 then the set contains elements.
A) 7
B) 6
C) 5
D) 10
Answer:
C) 5

Solving these TS 10th Class Maths Bits with Answers Chapter 4 Pair of Linear Equations in Two Variables Bits for 10th Class will help students to build their problem-solving skills.

Pair of Linear Equations in Two Variables Bits for 10th Class

Question 1.
Which of the following pairs of equations represent inconsistent system ?
A) 2x + 3y = 8
5x – 4y = 3
B) 6x + 3y = 9
x – 8y = 0
C) 2x + 5y = 11
4x + 10 y = 21
D) 3x – 4y = 6
6x – 8y = 12
Answer:
C) 2x + 5y = 11
4x + 10 y = 21

Question 2.
The pair of linear equations 3x + 4y + 5 = 0 and 12x + 16 y + 15 = 0 have ……… solution (s).
A) Unique
B) Many
C) Two
D) No
Answer:
D) No

Question 3.
The pair of linear equations px + 2y = 5 and 3x + y = 1 has unique solutions if
A) p ≠ 6
B) p = 6
C) p = 5
D) p ≠ 5
Answer:
A) p ≠ 6

Question 4.
The lines represented by 8x + 2pγ = 2 and 2x + 5y + 1 = 0 are parallel if p =
A) \(\frac{-5}{4}\)
B) \(\frac{2}{7}\)
C) 10
D) \(\frac{3}{8}\)
Answer:
C) 10

Question 5.
Solution of the equations \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 and \(\sqrt{3}\)x – \(\sqrt{8}\)y = 0
A) x = 1
y = 0
B) x = 0
y = 1
C) x = 1
y = 1
D) x = 0
y = 0
Answer:
D) x = 0
y = 0

Question 6.
The pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has a unique solution, then
A) \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) ≠ \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\)
B) \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
C) \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
D) \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
Answer:
A) \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) ≠ \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\)

Question 7.
If x = 1, then the value of y satisfying the equation \(\frac{5}{x}\) + \(\frac{3}{y}\) = 6
A) 3
B) +\(\frac{1}{3}\)
C) –\(\frac{1}{3}\)
D) 1
Answer:
A) 3

Question 8.
The value of y when \(\frac{x+y}{x y}\) = 2 and \(\frac{x-y}{x y}\) = 6 is
A) \(\frac{1}{4}\)
B) –\(\frac{1}{2}\)
C) –\(\frac{7}{4}\)
D) \(\frac{5}{4}\)
Answer:
A) \(\frac{1}{4}\)

Question 9.
If ax + by = c and px + qy = r has unique solution, then
A) \(\frac{a}{b}\) = \(\frac{p}{q}\)
B) ab = pq
C) \(\frac{a}{q}\) = \(\frac{b}{p}\)
D) aq ≠ bp
Answer:
A) \(\frac{a}{b}\) = \(\frac{p}{q}\)

Question 10.
If 5x + py + 8 =0 and 10x + 15y + 12 = 0 has no solution, then p =
A) 7\(\frac{1}{2}\)
B) 6\(\frac{1}{2}\)
C) 7
D) 4
Answer:
A) 7\(\frac{1}{2}\)

Question 11.
y = 5x is a line
A) parallel to x – axis
B) parallel to y – axis
C) parallel to x = 5y
D) passes through the origin
Answer:
D) passes through the origin

Question 12.
The point (-3,-8) is in the ____ quadrant.
A) Q₁
B) Q₂
C) Q₃
D) Q₄
Answer:
C) Q₃

Question 13.
The point (7, -5) is in the _____ quadrant.
A) I
B) II
C) III
D) IV
Answer:
D) IV

Question 14.
Find the value of x if y = \(\frac{3}{4}\) x and 5x + 8y = 33
A) 2
B) 3
C) 4
D) -3
Answer:
B) 3

Question 15.
\(\frac{120}{x}\) + \(\frac{12}{x}\) = 11, then x =
A) 132
B) 11
C) 12
D) 13
Answer:
C) 12

Question 16.
Which of the following equations is not a linear equation ?
A) x – 3y = 2x + y + 8
B) 4x² – 6y = y + 6
C) 8x – 3 = 5y + 7
D) 3x = 4y
Answer:
B) 4x² – 6y = y + 6

Question 17.
Which of the following number is a solution for the equation 3(7 – 3y) + 4y = 16 ?
A) 0
B) -1
C) 1
D) 2
Answer:
C) 1

Question 18.
When two lines in the same plane may in-tersect ?
A) At two points
B) At many points
C) At only one point
D) None
Answer:
C) At only one point

Question 19.
In the following values of x and y satisfies the equation 4x + 3y = 110
A) (10, 20)
B) (15, 30)
C) (10, 30)
D) (20, 10)
Answer:
D) (20, 10)

Question 20.
3x + 2y – 80 = 0; 4x + 3y – 110 = 0 solution for this linear equations is
A) Unique
B) Infinite
C) Two
D) No
Answer:
A) Unique

Question 21.
x + 2y – 30 = 0; 2x + 4y – 66 = 0 these lines represent
A) Intersecting lines
B) Parallel lines
C) Coincident lines
D) None of these
Answer:
B) Parallel lines

Question 22.
The point of intersection of x + y = 6 and x – y = 4 is ………. (A.P. Mar. 16)
A) (5, 1)
B) (1, 5)
C) (2, 4)
D) (4, 6)
Answer:
A) (5, 1)

Question 23.
A pair of linear equations in two variables are 2x – y = 4 and 4x – 2y = 6. The pair of equations are …….. (T.S. Mar’15)
A) Consistent
B) Dependent
C) Inconsistent
D) Cannot say
Answer:
C) Inconsistent

Question 24.
The value of ‘x’ in the equation 3x – (x – 4) = 3x + 1 is …….. (T.S. Mar’16)
A) -3
B) 0
C) 3
D) 10
Answer:
C) 3

Question 25.
The sum of squares of two consecutive positive even numbers is 340, then the numbers are …… (A.P. Mar. 15)
A) 12, 14
B) 10, 12
C) 14, 16
D) 16, 18
Answer:
A) 12, 14

Question 26.
Length of the dark line given in the; diagram (A.P. Mar. 16)

Answer:
C) \(\sqrt{l^2+\mathrm{b}^2+\mathrm{h}^2}\)

Question 27.
500x + 240y = 8, 130x + 240y = \(\frac{43}{10}\) then x = ………..
A) \(\frac{9}{200}\)
B) \(\frac{7}{20}\)
C) \(\frac{1}{100}\)
D) \(\frac{1}{10}\)
Answer:
C) \(\frac{1}{100}\)

Question 28.
In the above problem y = ……….
A) 1
B) 0
C) 10
D) none
Answer:
D) none

Question 29.
If the equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are consistent then ……….
A) \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)
B) \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\)
C) \(\frac{a_1}{a_2}\) = 1
D) none
Answer:
A) \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)

Question 30.
Sita has pencils and pens which are together 40 in number. If she has 5 less pencils and 5 more pens the number of pens become four times the number of pencils. Represent this situation in a linear equation form.  A) x – y = 40
B) x + y = 40
C) x – y = 7
D) all
Answer:
B) x + y = 40

Question 31.
The lines represented by 5x + 7y – 14 = 0 and 10x + 3y – 8 = 0 are ………. lines.
A) coincident
B) vertical
C) parallel
D) consistent
Answer:
D) consistent

Question 32.
The standard form of a linear equation is ……………
A) xa + y = 0
B) ax + by
C) ax + b = 0
D) ax + by + c = 0
Answer:
D) ax + by + c = 0

Question 33.
The lines 3x + 8y – 13 = 0 and -6x – 16y + 23 = 0 are ………. lines.
A) coincident
B) parallel
C) circular
D) none
Answer:
B) parallel

Question 34.
The lines represented by 5x + 3y – 7 = 0 and 6y + 10x – 14 = 0 are lines.
A) coincident
B) parallel
C) intersecting
D) none
Answer:
A) coincident

Question 35.
The pairs of equations 4x – 2y + 6 = 0 and 2x – y + 8 = 0 has …… solutions(s).
A) 1
B) 12
C) No solution
D) 10
Answer:
C) No solution

Question 36.
The number of solutions to the pair of equations 6x – 7y + 8 = 0 and 12x – 14y + 10 = 0 is
A) 1
B) 20
C) 3
D) infinite
Answer:
D) infinite

Question 37.
The number of solutions to the pair of equations 11x – 7y = 6 and 4x + 9y = 8 is
A) 4
B) 3
C) 7
D) 1
Answer:
D) 1

Question 38.
If the pair of equations kx + 14y + 8 = 0 and 3x + 7y + 6 = 0 has a unique solution then
A) k ≠ 6
B) k = 0
C) k = 7
D) none
Answer:
A) k ≠ 6

Question 39.
a₁x + b₁y₁ + c₁ = 0 and a₂x + b₂y + c₂ = 0 are ……… equations.
A) parallel
B) pair of linear
C) consistent
D) none
Answer:
B) pair of linear

Question 40.
If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) then the lines will have …….. solutions.
A) infinite
B) 2
C) 3
D) 7
Answer:
A) infinite

Question 41.
3x – 2y + 6 = 0, 6x – 4y + 8 = 0 represents ……. lines
A) consistent
B) inconsistent
C) circle
D) parallel
Answer:
D) parallel

Question 42.
5x – 2y – 10 = 0,10x – 4y – 20 = 0 these are ……… lines.
A) consistent
B) parallel
C) intersecting
D) none
Answer:
A) consistent

Question 43.
The number of solutions to 4x + 6y – 7 = 0 and 8x + 5y – 8 = 0 is ………..
A) 14
B) 3
C) 4
D) 1
Answer:
D) 1

Question 44.
\(\frac{2}{x}\) + \(\frac{3}{y}\) = 2, \(\frac{12}{x}\) – \(\frac{9}{y}\) = 3 then x = ……….
A) 1
B) 4
C) 2
D) none
Answer:
C) 2

Question 45.
In the above problem y = ………
A) 2
B) -1
C) 7
D) 3
Answer:
D) 3

Question 46.
If 2^x + 3^y = 17 and 2^x+2 – 3^y+1 = 5, then y = ……….
A) 2
B) 3
C) 1
D) 7
Answer:
A) 2

Question 47.
In the above problem x = ………
A) 2
B) 4
C) 3
D) none
Answer:
C) 3

Question 48.
\(\frac{\mathbf{a x}}{\mathbf{b}}\) – \(\frac{\mathbf{b y}}{\mathbf{a}}\) = a + b, ax – by = 2ab then x = ……….
A) 3b
B) \(\frac{-3}{b}\)
C) 1
D) -2a
Answer:
A) 3b

Question 49.
In the above problem y = ………
A) – a
B) 2a
C) -a²
D) 3b – a
Answer:
A) – a

Question 50.
Slope of the line ax + by + c = 0 is ………
A) \(\frac{b}{a}\)
B) \(\frac{1}{a}\)
C) \(\frac{a}{b}\)
D) \(\frac{-a}{b}\)
Answer:
D) \(\frac{-a}{b}\)

Question 51.
The line ax + by + c = 0 does not passes through ………..
A) (0, 0)
B) (a, 0)
C) both A & B
D) none
Answer:
C) both A & B

Question 52.
If x + y = 7, x – y = 1 then 2x = ……….
A) 3
B) 4
C) 7
D) 8
Answer:
D) 8

Question 53.
In the above problem y = ………..
A) 3
B) 7
C) 1
D) 4
Answer:
A) 3

Question 54.
Slope of the line y = x is ………
A) 2
B) – 1
C) 1
D) none
Answer:
C) 1

Question 55.
The line x = 2015 is ……….
A) slope not defined
B) parallel to Y – axis
C) both A & B
D) none
Answer:
C) both A & B

Question 56.
x + \(\frac{6}{y}\) = 6, 3x – \(\frac{8}{y}\) then y = ………..
A) – 2
B) 4
C) 1
D) 2
Answer:
D) 2

Question 57.
In the above problem x = ……….
A) 3
B) 2
C) -1
D) 9
Answer:
A) 3

Question 58.
3x – 5y = -1, – y + x = -1 then (x, y) = ……..
A) (-2, -1)
B) (2, -1)
C) curve
D) none
Answer:
A) (-2, -1)

Question 59.
The value of y in -5x + 10y = 100 at x = 0 is ……….
A) 12
B) 9
C) -10
D) 10
Answer:
D) 10

Question 60.
If x + y = 36, then at y = -1, x = ………..
A) 38
B) 37
C) 80
D) 12
Answer:
B) 37

Question 61.
x + y = 10, x – y = – 4 then x = …………
A) 4
B) 3
C) 5
D) none
Answer:
B) 3

Question 62.
In the above problem y = ……….
A) 9
B) 4
C) 7
D) 13
Answer:
C) 7

Question 63.
Solution to 2x – 2y – 2 = 0,4x – 4y – 5 = 0 is …….
A) (1, 4)
B) (2, -1)
C) (8, –\(\frac{1}{4}\))
D) no solution
Answer:
D) no solution

Question 64.
The two lines 2x + y – 6 = 0 and 4x – 2y – 4 = 0 intersect at ……….
A) (2, 2)
B) (3, 2)
C) (1, -4)
D) (1, 4)
Answer:
A) (2, 2)

Question 65.
Solution to x – y = 1, 2x – 2y = 7 is ……..
A) (1, 1)
B) (1, 9)
C) (8, 4)
D) no solution
Answer:
D) no solution

Question 66.
Perimeter of rectangle = ……..
A) l + b
B) l – b
c) 2(l + b)
D) \(\frac{l+\mathrm{b}}{2}\)
Answer:
c) 2(l + b)

Question 67.
x + y = 2015 has …… number of solutions.
A) 10
B) 2014
C) 20
D) infinite
Answer:
D) infinite

Question 68.
px + 3y – (p – 3) = 0, 12x + py – p = 0 has infinitely many solutions then p = ………..
A) 7
B) 9
C) ± 71
D) ± 6
Answer:
D) ± 6

Question 69.
If 3x + 4y + 2 = 0, 9x + 12y + k = 0 represent coincident lines, then k = ………….
A) 12
B) 9
C) 6
D) 7
Answer:
C) 6

Question 70.
2x + 3y = 1, 3x – y = 7 then (x, y) = ……..
A) (2, -1)
(B) (-2, 1)
(C) (8, \(\frac{1}{4}\))
(D) (0, 3)
Answer:
A) (2, -1)

Question 71.
Slope of the line x = 2y is ………
A) 2
B) -2
C) 1
(D) \(\frac{1}{2}\)
Answer:
B) -2

Question 72.
If 7x – 8y = 9, then y = …………
A) 9 + 7x
B) \(\frac{9-7 x}{8}\)
C) \(\frac{9-7 x}{6}\)
D) none
Answer:
D) none

Question 73.
Slope of X – axis is ……….
A) 0
B) 1
C) -1
D) 2
Answer:
A) 0

Question 74.
Angle between any two parallel lines is …………
A) 70°
B) 0°
C) 100°
D) 180°
Answer:
B) 0°

Question 75.
4m – 2n = 2, 6m – 5n = 9 then n = ………..
A) 5
B) 4
C) 1
D) – 3
Answer:
D) – 3

Question 76.
In the above problem m = …………
A) – 1
B) 4
C) -31
D) 7
Answer:
A) – 1

Question 77.
2u + 3y = 2, 4u – 6y = 0 then y = ……….
A) \(\frac{1}{2}\)
B) 1
C) \(\frac{1}{31}\)
D) \(\frac{1}{3}\)
Answer:
D) \(\frac{1}{3}\)

Question 78.
In Q. No. 77 the value of u = …………
A) \(\frac{1}{21}\)
B) \(\frac{1}{2}\)
C) 2
D) 4
Answer:
B) \(\frac{1}{2}\)

Question 79.
Area of rectangle = ………….
A) l²b
B) \(\frac{l}{b}\)
C) lb
D) none
Answer:
C) lb

Question 80.
\(\frac{x+3}{2}\) – y = 2, \(\frac{x-3}{2}\) + 2y = 4 \(\frac{1}{2}\) then x = ………..
A) 1
B) 4
C) 51
D) none
Answer:
D) none

Question 81.
In the above problem y = ……….
A) 2
B) – 2
C) 7
D) 12
Answer:
A) 2

Question 82.
If ax + b = 0, then x = ………..
A) -b
B) –\(\frac{\mathrm{b}}{\mathrm{a}}\)
C) \(\frac{\mathrm{b}}{\mathrm{a}}\)
D) none
Answer:
B) –\(\frac{\mathrm{b}}{\mathrm{a}}\)

Question 83.
2x – 3y = -12 then at x = 0, y = ……….
A) 4
B) 6
C) 8
D) 12
Answer:
A) 4

Question 84.
Two parallel lines differ by ……….
A) circle
B) triangle
C) constant
D) none
Answer:
C) constant

Question 85.
x = 1 and y = –\(\frac{1}{2}\) then x – y = ………..
A) – 1
B) 1
C) –\(\frac{1}{2}\)
D) 7
Answer:
D) 7

Question 86.
If 99x + 101y = 499, 101x + 99y = 510 then x = ………
A) -1
B) 3
C) 4
D) 2
Answer:
B) 3

Question 87.
In the above problem y = ………
A) 3
B) 4
C) 2
D) 8
Answer:
C) 2

Question 88.
141x + 93y = 189, 93x + 141y = 45 then y = …….
A) – 1
B) 4
C) -2
D) 2
Answer:
A) – 1

Question 89.
In the above problem x = …….
A) 3
B) 4
C) 1
D) 2
Answer:
D) 2

Question 90.
If -x + y = – 10, then x = …………….
A) y – 3
B) y² – 1
C) y – 10
D) y + 10
Answer:
D) y + 10

Question 91.
(-a, -b) ∈ ………..
A) Q₂
B) Q₃
C) Q₁
D) Q₄
Answer:
B) Q₃

Question 92.
(2, 0) lies on ……….
A) Q₁
B) Q₂
C) x axis
D) y axis
Answer:
C) x axis

Question 93.
The line x – y = 8 intersects X – axis at ……….
A) (2, 3)
B) (1, 1)
C) (0, 8)
D) (8, 0)
Answer:
D) (8, 0)

Question 94.
The values of k for which the pair of linear equations 3x – 2y = 7 and 6x + ky + 11 = 0 has a unique solution is
A) all numbers except 4
B) all numbers expect – 4
C) 4
D) – 4
Answer:
B) all numbers expect – 4

Question 95.
The pair of linear equations -3x + 4y = 7 and \(\frac{9}{2}\) x – 6y + \(\frac{21}{2}\) = 0 has ……………….
A) infinite number of solutions
B) no solution
C) two solutions
D) unique solution
Answer:
A) infinite number of solutions

Question 96.
If ad ≠ bc then the pair of linear equations ax + by = p and cx + dy = q has ……….. solutions.
A) no
B) unique
C) 2
D) none
Answer:
C) 2

Question 97.
A line parallel to the line x + 2y + 1 = 0 is ………
A) x + y + 3 = 0
B) 2x + 4y + 1 = 0
C) x – y + 1 = 0
D) all
Answer:
B) 2x + 4y + 1 = 0

Question 98.
The pair of equations x = 3, y = 2 graphi-cally represent lines which ………
A) intersect at (3, 4)
B) intersect at (4, 3)
C) parallel
D) coincident
Answer:
A) intersect at (3, 4)

Question 99.
If the pair of equations 2x + y = 7 and 6x – py – 21 =0 has infinite number of solutions then p = ………..
A) 3
B) 4
C) 5
D) none
Answer:
D) none

Question 100.
The value of k for which the system of equations kx + 3y = 1,12x + ky = 2 has no solution is ………..
A) k = -1
B) k = 3
C) k = 2
D) k = -6
Answer:
D) k = -6

Question 101.
The line 3x + y = 7 intersects X- axis at ………..
A) (\(\frac{7}{2}\), 0)
B) (0, \(\frac{7}{2}\))
C (0, 1)
D) (0, 3)
Answer:
A) (\(\frac{7}{2}\), 0)

Question 102.
For what value of k, 2x + 3y = 4 and (k + 2)x + 6y = 3k + 2 will have infinitely many solutions ?
A) k = -1
B) k = 2
C) k = 7
D) none
Answer:
B) k = 2

Question 103.
\(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 9, \(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 8 then x =
A) 13
B) 17
C) -3
D) 10
Answer:
A) 13

Question 104.
In the above problem y = ……….
A) 4
B) – 3
C) 9
D) 7
Answer:
D) 7

Question 105.
If the equations (2m – 1)x + 3y – 5 = 0, 3x + (n – 1)y – 2 = 0 has infinite number of solutions then n = …………
A) 1
B) \(\frac{5}{11}\)
C) \(\frac{1}{5}\)
D) \(\frac{11}{5}\)
Answer:
D) \(\frac{11}{5}\)

Question 106.
In the above problem m = ………..
A) \(\frac{17}{4}\)
B) \(\frac{7}{4}\)
C) \(\frac{1}{2}\)
D) \(\frac{8}{3}\)
Answer:
A) \(\frac{17}{4}\)

Question 107.
A fraction becomes \(\frac{9}{11}\) if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator it be-comes is …………
A) \(\frac{17}{4}\)
B) \(\frac{7}{4}\)
C) \(\frac{1}{2}\)
D) \(\frac{8}{3}\)
Answer:
A) \(\frac{17}{4}\)

Question 108.
The ratio of incomes of two persons is 11:7 and the ratio of their expenditures is 9 : 5. If each of them manages to save ₹ 400 per month then the monthly income of first person is ………..
A) ₹ 2,200
B) ₹ 1,200
C) ₹ 800
D) ₹ 1010
Answer:
A) ₹ 2,200

Question 109.
In the above problem income of second person is ……….
A) ₹ 8,001
B) ₹ 1,100
C) ₹ 1,400
D) ₹ 4,100
Answer:
C) ₹ 1,400

Question 110.
The age of a father 8 years ago was 5 time that of his son 8 years. Hence, his age will be 8 years more than twice the age of his son. Then the present age of father is years. ( )
A) 80
B) 92
C) 24
D) 48
Answer:
D) 48

Question 111.
In the above problem age of son is ……… years.
A) 16
B) 96
C) 12
D) none
Answer:
A) 16

Question 112.
The two lines 2x – y = 1, x + 2y = 13 will intersect at …………
A) (5, 3)
B) (3, 5)
C) (1, 3)
D) (3, 9)
Answer:
B) (3, 5)

Question 113.
Identify parallel lines
A) 2x + 3y = 6, 8x + 12y = 9
B) x + y = 7, x – y = 1
C) 2x + y = 7, x – y = 1 :
D) 2x + y = 7, 3x – y = 7 .
Answer:
A) 2x + 3y = 6, 8x + 12y = 9

Question 114.
Solution to 2x + 3y = 12, 2y – 1 = x is ……….
A) (8, -1)
B) (3, 8)
C) (3, 2)
D) (1, -1)
Answer:
C) (3, 2)

Question 115.
The lines x – y = 1; 2x + y = 8 intersects at ……….
A) (1, 9)
B) (9, 3)
C) (3, 4)
D) none
Answer:
D) none

Question 116.
\(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2, \(\frac{6}{x-1}\) + \(\frac{-3}{y-2}\) = 1 If then x = ……….
A) 4
B) 7
C) -1
D) 3
Answer:
A) 4

Question 117.
In the above problem y = …………
A) 1
B) -1
C) 5
D) 9
Answer:
C) 5

Question 118.
If 2x + 3y = 10 and 2x – 3y = 6 then (x, y) = …………
A) (2, 5)
B) (\(\frac{2}{3}\), 4)
C) (0, 0)
D) (4, \(\frac{2}{3}\))
Answer:
D) (4, \(\frac{2}{3}\))

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry InText Questions

Do This

Question 1.
i) From the figure write co-ordinates of the points A, B, C, D, E, F, G, H. (Page No. 159)

Solution:
Given : knight is at the origin (0, 0)
i.e., 36 + 25 + y² – 10y = 16 + 9 + y² – 6y
4y = 36
y = 9
So, the required point is (0, 9).
let us check our solution:
AP = \(\sqrt{(6-0)^2+(5-9)^2}\)
BP = \(\sqrt{(-4-0)^2+(3-9)^2}\)
= \(\sqrt{36+16}\) = \(\sqrt{52}\)
= \(\sqrt{16+36}\) = \(\sqrt{52}\)
So, (0, 9) is equidistant from (6, 5) and (-4, 3)

Therefore A(- 1, 2), B(l, 2), C(2, 1), D(2, -1), E(1, -2), F(-1, -2), G(-2, -1), H(-2, 1).

ii) Find the distance covered by the knight in each of its 8 moves i.e., find the distance of A, B, C, D, E, F, G, H and from the origin. (Page No. 159)
Solution:
Origin (0, 0)
Points A, B, C, D, E, E G, H
Distance of any point P(x, y) from the origin is

iii) What is the distance between two points H and C ? And also find the distance between two points A and B. (Page No. 159)
Solution:
Given : H(-2, 1), C(2, 1), A(-1, 2), B(1, 2)
Distance between any two points
P(x₁, y₁) Q(x₂, y₂) is
\(\overline{\mathrm{PQ}}\) = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
∴ Distance between H and C is
HC = \(\sqrt{[2-(-2)]^2+[1-1]^2}\)
= \(\sqrt{4^2+0}\) = \(\sqrt{16}\) = 4 units.
Distance between A and B is
AB = \(\sqrt{[1-(-1)]^2+(2-2)^2}\)
= \(\sqrt{2^2+0^2}\)
= \(\sqrt{4+0}\) = 2 units.

Question 2.
Where do these following points lie (-4, 0) (2, 0) (6, 0) (-8, 0) on co-ordinate plane ? (Page No. 160)
Solution:
Given points are (-4, 0), (2, 0), (6, 0), (-8, 0) these all points have their y-co-ordinate = 0
∴ These points lie on X – axis.

Question 3.
What is the distance between points (-4, 0) and (6, 0) on co-ordinate plane ? (Page No. 160)
Solution:
Given points = (- 4, 0) (6, 0).
These two points lie on the X – axis.
∴ Distance between them = | x₂ – x₁ |
= | 16 – (-4)| = |6 + 4| = 10

Question 4.
Find the distance between the following points.

i) (3, 8) (6, 8). (Page No. 162)
Solution:
Given points = A(3, 8), B(6, 8)
These two points lie on X – axis Distance between A(3, 8) & B(6, 8)
= | x₂ – x₁|
= |6 – 3| = 3 units.

ii) (-4, – 3) (-8,-3)
Solution:
Given points A(- 4, -3) & B(-8, -3)
These two points lie on X – axis
∴ Distance between
A(-4, -3) and B(-8, -3) = |x₂ – x₁|
= |-8 – (-4)|
= |-8 + 4|
= 4 units.

iii) (3, 4) (3, 8)
Solution:
Given points A(3, 4) & B(3, 8)
These two points lie on Y-axis
∴ Distance between A(3, 4) & B(3, 8) = |(y₂ – y₁)|
= | 8 – 4|
= 4 units.

iv) (-5, -8) (-5, -12)
Solution:
Given points A(-5, -8), B(-5, -12)
These two points lie on Y – axis
Distance between
A(-5, -8) & B(-5, -12) = | y₂ – y₁|
= |-12 + 8|
= 4 units.

Question 5.
Find the distance between the following points. (Page No. 162)

i) A = (2, 0) and B (0, 4)
Solution:
Given points = A(2, 0), B(0, 4)

ii) P(0, 5) and Q(12, 0)
Solution:
Given points P(0, 5), Q(12, 0)
PQ = \(\sqrt{(12-0)^2+(0-5)^2}\)
= \(\sqrt{144+25}\)
= \(\sqrt{169}\) = 13 units.

Question 6.
Find the distance between the following pair of points. (Page No. 164)
i) (7, 8) and (-2, 3)
Solution:
Given points = (7, 8) & (-2, 3)
distance formula

ii) (-8, 6) and (2, 0)
Solution:
Given points = (-8, 6) and (2, 0)

Try This

Question 1.
Where do these following points lie (0, -3), (0, -8), (0, 6), (0, 4) ? (Page No. 161)
Solution:
As the X-co-ordinate of all these points is zero, all points lie on Y-axis.

Question 2.
What is the distance between (0, -3), (0, -8) and justify that the distance between two points on Y- axis is | y₂ – y₁ | on Co-ordinate plane ? (Page No. 161)
Solution:
As the given two points lie on Y-axis, distance between them is | y₂ – y₁ | = | -3 + 8| = 5 units.
Let (0, y₁) and (0, y₂) be any two points on Y-axis, then distance between them
= \(\sqrt{(0-0)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
= \(\sqrt{0+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
= \(\sqrt{\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
= |y₂ – y₁|
[∵ distance can’t be negative].

Question 3.
Find the distance between points “O (ori-gin)” and A(7, 4). (Page No. 162)
Solution:
Given : Origin and a point (7, 4)
Distance of a point (x, y) from the origin is = \(\sqrt{x^2+y^2}\)
= \(\sqrt{7^2+4^2}\) = \(\sqrt{49+16}\) = \(\sqrt{65}\) units.

Question 4.
Find the distance between A(1, -3), B(-4, 4) and rounded to two decimal. (Page No. 164)
Solution:
Given: A(1, -3), B(-4, 4)
AB = \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
= \(\sqrt{(-4-1)^2+(4+3)^2}\)
= \(\sqrt{25+49}\)
= \(\sqrt{74}\)
= 8.602 = 8.60

Think – Discuss

Question 1.
How will you find the distance between two points in which x or y co-ordinates are same but not zero ? (Page No. 161)
Solution:
Let the points be A(2, 3), B(2, 5)
Here the x-co-ordinates are same, then the dis-tance between the points, A and B is | y₂ – y₁ | = |5 – 3| = 2 units.
It the points P(4, 3), Q(-8, 3), Here the y-co- ordinates are same. In such a case, the distance is given by |x₂ – x₁| = |-8 – 4| = |-12| = 12 units.
i.e.,

Question 2.
Ramu says the distance of a point P(x₁, y₁) from the origin O(0, 0) is \(\sqrt{\mathbf{x}_1^2+\mathbf{y}_1^2}\). Do you agree with Ramu or not ? Why ? (Page No. 163)
Solution:
Yes, The distance between 0(0,0) and P(x₁, y₁) is \(\sqrt{\left(x_1-0\right)^2+\left(y_1-0\right)^2}\) ⇒ \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}\)

Question 3.
Ramu also writes the distance formula as AB = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) why ? (Page No. 163)
Solution:
(x₁ – x₂)² is same as (x₂ – x₁)² and (y₁ – y₂)² is same as (y₂ – y₁)² i.e.,
= \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
= \(\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}\)

Question 4.
Sridhar calculated the distance between T(5, 2) and R(-4, -1) to the nearest decimal is 9.5 units.
Now you find the distance between P(4, 1) and Q(-5, -2). Do you get the same answer that Sridhar got ? Why ? (Page No. 164)
Solution:
Distance between T(5, 2) and R(-4, -1) to the nearest decimal = 9.5 units.
(According to Sridhar’s calculation)
We find the distance between P(4, 1) and (-5, -2)
= \(\sqrt{(4+5)^2+(1+2)^2}\)
= \(\sqrt{(9)^2+(3)^2}\) = \(\sqrt{81+9}\)
= \(\sqrt{90}\) = 9.48 = 9.5
because the points P(4, 1), Q(-5, 2), R(-4, -1) and T(5, 2) are almost have same signs.

Do This

Question 1.
Find the point which divides the line seg¬ment joining the points (3, 5) and (8, 10) internally in the ratio 2 : 3. (Page No. 171)
Solution:
Let P(x, y) be the required point then

Question 2.
Find the midpoint of the line segment joining the points (2, 7) and (12, -7). (Page No. 171)
Solution:
Midpoint of the line joining the points (x₁, y₁), (x₂, y₂) is

Question 3.
Find the trisectional points of line joining (2, 6) and (-4, 8). (Page No. 172)
Solution:
A(2, 6), B(-4, 8) be the given points.
Let P Q divide the line joining of \(\overline{\mathrm{AB}}\) in the ratio 1: 2 & 2: 1
section formula (x, y)
= \(\left[\frac{\mathrm{m}_1 \mathrm{x}_2+\mathrm{m}_2 \mathrm{x}_1}{\mathrm{~m}_1+\mathrm{m}_2}, \frac{\mathrm{m}_1 \mathrm{y}_2+\mathrm{m}_2 \mathrm{y}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right]\)
P(x, y) in the ratio 1 : 2

for Q(x, y) in the ratio 2 : 1

Question 4.
Find the trisectional points of line joining (-3, -5) and (-6, -8). (Page No. 172)
Solution:
Given : A(-3, -5) B(-6, -8). Let P, Q be the points of trisection of \(\overline{\mathrm{AB}}\), then P divides \(\overline{\mathrm{AB}}\) in the ratio 1: 2

Q(x, y) divides AB in the ratio 2: 1
Q(x, y)

∴ The points of trisection are P(-4, -6), Q(-5, -7)

Question 5.
Find the centrold of the triangle whose vertices are (-4, 6), (2, -2) and (2, 5) respectively. (Page No. 174)
Solution:
Given points: (-4, 6), (2, -2), (2, 5)
The co-ordinates of the centroid

∴ The centroid is (0, 3)

Try This

Let A (4, 2) B (6, 5) C (1, 4) be the vertices of Δ ABC.

Question 1.
The median from A meets BC at D. Find the co-ordinates of the point D. (Page No. 173)
Solution:
D is the mid point of BC

Question 2.
Find the co-ordinates of the point P on AD such that AP : PD = 2: 1 (Page No. 173)
Solution:
P is a point on AD which divides AD in the ratio 2 : 1
∴ P(x, y) = \(\left[\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right]\)

Question 3.
Find the co-ordinates of points E and F on medians BE and CF. (Page No. 173)
Solution:
E is the mid point of \(\overline{\mathrm{AC}}\)

similarly, F is the mid point of \(\overline{\mathrm{AB}}\)

Question 4.
Find the points which divide the line segment BE in the ratio 2: 1 and also that divide the line segment CF in the ratio
2: 1. (Page No. 173)
Solution:
Given: B(6, 5), E[\(\frac{5}{2}\), 3], ratio 2: 1
Let it be P(x, y) =

similarly, let P divide C(1, 4) and [5, \(\frac{7}{2}\)] in the ratio 2: 1 then

Question 5.
What do you observe? Justify the point that divides each median in the ratio 2: 1 is the centroid of a triangle. (Page No. 173)
Solution:
From the above problems, we conclude that the point “P” divides each median in the ratio 2 : 1 i.e., the three medians are concurrent at P, which is called centroid. A centroid divides each median in the ratio 2: 1.

Think – Discuss

Question 1.
The line joining points A(6, 9) and B(-6, -9) are given. (Page No. 175)

a) In which ratio does origin divide \(\overline{\mathbf{A B}}\) ? ‘And what it Is called for \(\overline{\mathbf{A B}}\) ?
Solution:
Given : A(6, 9) B(-6, -9)
Let origin 0(0, 0) divides AB in the ratio k : 1 internelly

∴ Ratio is 1 : 1
Here the origin bisects \(\overline{\mathbf{A B}}\).
∴ Origin is called the mid point of \(\overline{\mathbf{A B}}\).

b) In which ratio does the point P(2, 3) divide \(\overline{\mathbf{A B}}\) ?
Solution:
Given : A(6, 9) B(-6, -9) and P(2, 3) divide
\(\overline{\mathbf{A B}}\) internally in the ratio say k : 1 then
P(2, 3) = \(\left[\frac{\mathrm{k}(-6)+6}{\mathrm{k}+1}, \frac{\mathrm{k}(-9)+9}{\mathrm{k}+1}\right]\)
⇒ 2 = \(\frac{-6 k+6}{k+1}\) and 3 = \(\frac{-9 k+9}{k+1}\)
⇒ 2k + 2 = -6k + 6 and 3k + 3
⇒ -9k + 9 = 8k ⇒ 6 – 2 and 3k + 9k
⇒ 9 – 3
⇒ 8k = 4 and 12k = 6;
k = \(\frac{4}{8}\) and k = \(\frac{6}{12}\)
k = \(\frac{1}{2}\) and k = \(\frac{1}{2}\)
∴ The ratio (k : 1) = [\(\frac{1}{2}\) : 1] = [1 : 2].

c) In which ratio does the point P(-2, -3) divide \(\overline{\mathbf{A B}}\) ?
Solution:
Let Q divide \(\overline{\mathbf{A B}}\) in the ratio say k : 1 internally, then,
P(-2, -3) = \(\left[\frac{\mathrm{kx}_2+\mathrm{x}_1}{\mathrm{k}+1}, \frac{\mathrm{ky}_2+\mathrm{y}_1}{\mathrm{k}+1}\right]\)
(-2, -3) = \(\left[\frac{\mathrm{k}(-6)+6}{\mathrm{k}+1}, \frac{\mathrm{k}(-9)+9}{\mathrm{k}+1}\right]\)
-2 = \(\frac{-6 \mathrm{k}+6}{\mathrm{k}+1}\) and \(\frac{-9 \mathrm{k}+9}{\mathrm{k}+1}\) = -3
-2 = \(\frac{-6 \mathrm{k}+6}{\mathrm{k}+1}\)
-6k + 6 = -2k – 2
-6k + 2k = -2 – 6
-4k = -8
k = \(\frac{-8}{-4}\) k = 2
and \(\frac{-9 \mathrm{k}+9}{\mathrm{k}+1}\) = -3
-9k + 9 = -3k – 3
-9k + 3k = -3 – 9
-6k = -12
k = \(\frac{-12}{-6}\) k = 2
∴ Ratio is k : 1 = 2 : 1

d. In how many equal parts is \(\overline{\mathbf{A B}}\) divided by P & Q ?
Solution:
Since P, Q divided \(\overline{\mathbf{A B}}\) in the ratio 1 : 2 & 2 : 1
\(\overline{\mathbf{A B}}\) is divided into 3 equal parts by P & Q.

e. What do we call P & Q for \(\overline{\mathbf{A B}}\) ?
Solution:
P & Q are the points of trisection of \(\overline{\mathbf{A B}}\).

Try This

Question 1.
Take a point A on X—axis and B on Y—axis and find area of the triangle AOB. Discuss with your friends what did they
do. (Page No. 178)

Solution:
[∵ Axes are ⊥^er to each other].
consider the points A(5, 0), B(0, 6)
ΔAOB = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × 6 × 5 = 15 sq. units.
Area of Δ^le A (x, 0), 0(0, 0), B(0, y) is \(\frac{1}{2}\)xy.

Question 2.
Find the area of the square formed by (0, -1) (2, 1) (0, 3) and (-2, 1) taken are as vertices. (Page No. 181)

Solution:
Let A(0, -1), B(2, 1), C(0, 3) and D(-2, 1) are the vertices of square.
Area of the square ABCD = side² = AB²
But AB = \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
= \(\sqrt{(2-0)^2+(1+1)^2}\)
= \(\sqrt{2^2+2^2}\)
= \(\sqrt{4+4}\) = \(\sqrt{8}\)
∴ Area of square = \(\sqrt{8}\) × \(\sqrt{8}\)
= 8 sq. units

Think – Discuss

Question 1.
Let A(x₁, y₁), B(x₂, y₂) C(x₃, y₃). Then find the area of the following triangles in a plane. And discuss with your friends in groups about the area of that triangle. (Page No. 178)
i)

Solution:
Given ΔAOB where A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) But we know that the origin O is (0, 0) which is given as C
Now \(\overline{\mathrm{CB}}\) = x₂ – x₃ = x₂ – 0 = x₂
\(\overline{\mathrm{AB}}\) = y₁ – y₂ = y₁ – 0 = y₁
∴ Area of ΔABC = \(\frac{1}{2}\)bh
Δ = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × BC × AB
= \(\frac{1}{2}\) |(x₁ – x₃) × (y₂ – y₁)| sq. units

Question 2.
Find the area of the triangle formed by the following points. (Page No. 181)
i) (2, 0) (1, 2) (1, 6)
ii) (3, 1) (5, 0) (1, 2)
iii) (-1. 5, 3) (6, 2) (-3, 4).
What do you observe?
Solution:
i) (2, 0) (1, 2) (1, 6)
Take the third point as (-1, 6)
∆ = \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
∆ = \(\frac{1}{2}\) |2(2 – 6) + 1(6 – 0) – 1(0 – 2)|
= \(\frac{1}{2}\) |-8 + 6 + 2| = 0

ii) (3, 1) (5, 0) (1, 2)
Solution:
∆ = \(\frac{1}{2}\) | 3(0 – 2) + 5(2 – 1) + 1(1 – 0)|
= \(\frac{1}{2}\) |-6 + 5 + 1|
=0

iii) (-1.5, 3) (6, 2) (-3, 4)
Solution:
∆ = \(\frac{1}{5}\) | -1.5(2 – 4) + 6(4 – 3) – 3(3 – 2)|
= \(\frac{1}{2}\) | 3 + 6 – 3| = \(\frac{6}{2}\)
= 3 sq.units.

What do you observe?
Solution:
We observe that the area formed by above (i) & (ii) triangles is zero.

Plot these points on three different graphs. What do you observe?
Solution:

Solution:
We observe that the points are collinear.

Can we have a triangle with ‘0’ square units. (Page No. 181)
Solution:
No.

What does it mean?
Solution:
If the area of the triangle formed by any three points is zero, it means the points are collinear.

Do This

Question 1.
Find the area of the triangle whose vertices are
i) (5, 2), (3, -5), (-5, -1)
ii) (6, -6), (3, -7) and (3, 3) (Page No. 180)
Solution:
Given : The vertices of the triangle are (5, 2) (3, -5) (-5, -1)
Area of the triangle,
∆ = \(\frac{1}{2}\) | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= \(\frac{1}{2}\) | 5(-5 – (-1)) + 3((-1) – 2) + (-5)(2 – (-5))|
= \(\frac{1}{2}\) |5 × (-4) + 3 × (-3) + (-5) × 7|
= \(\frac{1}{2}\) |-20 – 9 – 35|
= \(\frac{1}{2}\) |-64|
= \(\frac{64}{2}\)
= 32 sq. units.

ii) (6, -6), (3, -7) and (3, 3)
Solution:
Given : The vertices of a triangle are (6, -6), (3, -7), (3, 3)
Area of a triangle,
∆ = \(\frac{1}{2}\) | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= \(\frac{1}{2}\) |6(-7 – 3) + 3(3 + 6) + 3(-6 + 7)|
= \(\frac{1}{2}\) |-60 + 27 + 3|
= \(\frac{1}{2}\) |-30|
= \(\frac{30}{2}\) = 15 sq. units

Question 2.
Verify whether the following points are collinear or not. (Page No. 182)

i) (1, -1), (4, 1), (-2, -3)
Solution:
Given : Three points
(1, -1), (4, 1), (-2, -3)
Area of ∆ = \(\frac{1}{2}\) | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= \(\frac{1}{2}\) |1(1 + 3) + 4(-3 + 1) + (-2)(-1 – 1)|
= \(\frac{1}{2}\)|4 – 8 + 4| = 0
As the area of the triangle is “O”, the three points are collinear.

ii) (1, -1) (2, 3) (2, 0)
Solution:
Given points are (1, -1) (2, 3) (2, 0)
∆ = \(\frac{1}{2}\) | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= \(\frac{1}{2}\) |1(3 – 0) + 2(0 + 1) + 2(-1 – 3)|
= \(\frac{1}{2}\)|3 + 2 – 8|
= \(\frac{1}{2}\)|-3|
= \(\frac{3}{2}\) ≠ 0
∆ ≠ 0
Hence the points are not collinear.

iii) (1, -6) (3, -4) (4, -3)
The given points are (1, -6) (3, -4) (4, -3)
Area of a triangle
∆ = \(\frac{1}{2}\) | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= \(\frac{1}{2}\)|1(-4 + 3) + 3(-3 + 6) + 4(-6 + 4)|
= \(\frac{1}{2}\)|-1 + 9 – 8| = = \(\frac{1}{2}\)|0| = 0
∆ = 0, the points are collinear.

Question 3.
Find the area of the triangle whose lengths of sides are 15m, 17m, 21m (use Herons formula) (Page No. 183)
Solution:
Given : The sides of a triangle
a = 15m, b = 17m, c = 21m
S = \(\frac{a+b+c}{2}\)
= \(\frac{15+17+21}{2}\)
= \(\frac{53}{2}\)
Heron’s formula.
∆ = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Question 4.
Find the area of the triangle formed by the points (0, 0) (4, 0) (4, 3) by using Heron’s formula. (Page No. 183)
Solution:
The given points are 0(0, 0), A(4, 0), B(4, 3) then the sides

Let a = 4, b = 5, c = 3 then s = \(\frac{a+b+c}{2}\)
= \(\frac{4+5+3}{2}\) = \(\frac{12}{2}\) = 6
Heron’s formula
∆ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{6 \times(6-4)(6-5) \times(6-3)}\)
= \(\sqrt{6 \times 2 \times 1 \times 3}\)
= \(\sqrt{36}\)
= 6 sq. units.

Do This

Question 1.
Plot these points on the co-ordinate plane and join them
i) A(1, 2) B(-3, 4) C(-7, -1)
ii) P(3, -5), Q(5, -1), R(2, 1), S(1, 2)
Which gives a straight line?
Which does not? Why?
Solution:
i) A(1, 2), B(-3, 4), C(-7, -1) ABC gives the straight line because the points lie on same line.

ii) P(3, -5) Q(5, -1) R(2, 1) S(1, 2) PQRS doesn’t give straight line because the points are not lie on same line.

Question 2.
Find the slope of with the given end points.
i) A(4, -6), B(7, 2)
Solution:
Slope m = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{2+6}{7-4}\) = \(\frac{8}{3}\)

ii) A(8, -4), B(-4, 8)
Solution:
Slope m = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{8+4}{-4-8}\) = \(\frac{12}{-12}\) = -1

iii) A (-2, -5), B(1, -7)
Solution:
Slope m = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-7+5}{1+2}\) = \(\frac{-2}{3}\)

Think — Discuss

Question 1.
Is y = x + 7 represent a straight line? Draw the line on the co-ordinate plane. At which point does this line intersect Y—axis? How much angle does it make with X — axis? Discuss with your friends. (Page No. 185)
Solution:
Yes, y = x + 7 represents a straight line.


tan θ = 1 = tan 45°
∴ θ = 45°
Angle made by y = x + 7 with X — axis is 45°
[∵ (0, 7) & (-7, 0) are equidistant from the origin and hence the triangle formed is right isosceles triangle.

Question 2.
Find the slope AB with the points lying on A(3, 2) B(-8, 2). when the line \(\overline{\mathbf{A B}}\) parallel to X—axis ? Why ? Think and discuss with our friends in groups. (Page No. 188)
Solution:
Given : A(3, 2) B(-8, 2) then slope
m = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{2-2}{-8-3}\)
= \(\frac{0}{-11}\) = 0
Yes, the line is parallel to X – axis as the points are of the form (x₁, k) (x₂, k).

Try This

Find the slope of \(\overline{\mathrm{AB}}\) with the points lying on (Page No. 188)

Question 1.
A(2, 1), B(2, 6)
Solution:
m = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{6-1}{2-2}\) = \(\frac{5}{0}\) = ∞

Question 2.
A(-4, 2) B(-4, -2)
Solution:
m = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{-2-2}{-4+4}\) = \(\frac{-4}{0}\) = ∞

Question 3.
A(-2, 8) B(-2, -2)
Solution:
m = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{-2-8}{-2+2}\) = \(\frac{-10}{0}\) = ∞

Question 4.
Justify that the line \(\overline{\mathbf{A B}}\) line segment formed by given points in the above three
examples is parallel to Y—axis. What can you say about their slope ? Why?
Solution:
In the above problems, all points are of the form (k, y) where k is fixed number and y is variable.
All lines in the above problems are parallel to Y-axis.

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Optional Exercise

Question 1.
Centre of the circle Q is on the Y-axis. And the circle passes through the points (0, 7) and (0, -1). Circle intersects the positive X-axis at (P, 0). What is the value of P ?
Solution:
Let Q be the centre of the circle. It lies on Y- axis. So, let the co-ordinates of the Q be (0, y). R(0, 7) is a point on the circle.

∴ Radius of the circle QR = 7 – y —- (1)
S(0, – 1) is also a point on the circle.
∴ Radius of the circle QS = y + 1 —– (2)
From (1) & (2), we get
y + 1 = 7 – y
⇒ y + y = 7 – 1
⇒ 2y = 6
⇒ y = 6/2 = 3
Hence centre of the circle is (0, 3).
Radius of the circle = 7 – 3 = 4 units —- (3)
PQ denotes the radius of the circle
∴ Distance between (p, 0) and (0, 3)
= \(\sqrt{(0-p)^2+(3-0)^2}\) = \(\sqrt{\mathrm{p}^2+9}\)
From (3), \(\sqrt{\mathrm{p}^2+9}\) = 4
Squaring on both sides,
\(\left(\sqrt{\mathrm{p}^2+9}\right)^2\) = 4²
⇒ P² + 9 = 16
P² = 16 – 9 = 7
Hence p = \(\sqrt{7}\)

Question 2.
The triangle ΔABC is formed by the points A(2, 3), B(-2, -3), C(4, -3). What is the point of intersection of side BC and angular bisector of A ?
Solution:
In ΔABC, AD is the angular bisector of ∠A in-tersecting BC at D. The vertices of A, B and C are (2, 3), (-2, -3) and (4, -3) respectively.
We know that angular bisector of A divides BC in the ratio of the other two sides (i.e)
AB : AC
Length of AB

Let the co-ordinates of D be (x, y)

Question 3.
The side of BC of an equilateral triangle ΔABC is parallel to X-axis. Find the slopes of line along sides BC, CA and AB.
Solution:
ΔABC is an equilateral triangle. BC is parallel to X-axis.
The angle made by AB with X-axis is 60°.
∴ Slope of AB = tan θ = tan 60° = \(\sqrt{3}\)
BC is parallel to X-axis
∴ Slope of BC = 0
The angle made by AC with X-axis is 60°
∴Slope of AC = tan 60°= \(\sqrt{3}\)

Question 4.
A right triangle has sides ‘a’ and ‘b’ where a > b. If the right angle is bisected then find the distance between orthocentres of the smaller triangles using coordinate geometry.
Solution:

A triangle is formed with a cm hypotenuse (a > b)
∠B = 90°. A line is drawn from B to D the line divides the triangle into two smaller triangles. If we want to determine the distance between orthocentres, we have to find the distance between B and D.
∴ The distance between orthocentres = The distance between the points B and D.
\(\sqrt{\left(x_4-x_2\right)^2+\left(y_4-y_2\right)^2}\)

Question 5.
Find the centroid of the triangle formed by the line 2x + 3y – 6 = 0 with the co-ordinate axis.
Solution:
The given line is 2x + 3y – 6 = 0
If we want to find the X-intersection, we have to substitute (x, 0) in the given line.
2x + 3y – 6 = 0
2x + 3.0 – 6 = 0
2x = 6

∴ x = 3
∴ The intersection point of X-axis = (3, 0)
If we want to find the Y-intersection, we have to substitute (0, y) in the given line
2x + 3y – 6 = 0
2.0 + 3.y – 6 = 0
3y – 6 = 0; 3y = 6

∴ The intersection point of Y-axis = (0, 2)
The coordinates of origin = (0, 0)
∴ Sides of the triangle = (0, 0), (3, 0), (0, 2)
∴ Centroid = \(\left[\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3}, \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}\right]\)
= \(\left[\frac{0+3+0}{3}, \frac{0+0+2}{3}\right]\) = [1, \(\frac{2}{3}\)]

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2

Question 1.
Find the co-ordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 : 3.
(A.P. Mar. 16)
Solution:
Given points P(-1, 7) and Q (4, -3).
Let ‘R’ be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 2 : 3 then

Question 2.
Find the co-ordinates of the points of tri-section of the line segment joining (4, -1) and (-2, -3). (A.P.June ’15)
Solution:
Given points A(4, -1) and B(-2, -3)
Let P and Q be the points of trisection of \(\overline{\mathrm{AB}}\), then AP = PQ = QB

∴P divides AB internally in the ratio 1:2.

P(x, y) = (2, -5/3)
Also, Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.
Q(x, y) = \(\left[\frac{-4+4}{3}, \frac{-6-1}{3}\right]\) = (0, -7/3)
Q(x, y) = (0, -7/3)

Question 3.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given : A(-3, 10) and B(6, -8)
Let P(-1, 6) divides \(\overline{\mathrm{AB}}\) in the ratio k : 1 internally.
By section formula

6k – 3 = -k – 1 and -8k + 10 = 6k + 6
-7k = – 2 and 14k = 4 4 14
∴ The ratio is 2/7 : 1 or 2 : 7

Question 4.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:

Given : ABCD is a parallelogram where
A(1, 2), B(4, y), C(x, 6) and D(3, 5).
In a parallelogram, diagonals bisect each other, i.e., the mid points of the diagonals coincide with each other.
i.e., mid point of AC = mid point of BD

1 + x = 7 and ⇒ 8 = y + 5
x = 6 and y = 8 – 5 = 3
∴ x = 6 and y = 3.

Question 5.
Find the co-ordinates of a point A, where AB is the diameter of a circle. Whose centre is (2, -3) and B is (1, 4).
Solution:

Given : A circle with centre ‘C’ (2, -3)
\(\overline{\mathrm{AB}}\) is diameter where B = (1, 4); A = (x, y)
∴ C is the mid point of AB
[∵ centre of a circle is the midpoint of the diameter].

Question 6.
If A and B are (-2, -2) and (2, -4) respectively. Find the co-ordinates of such that AP = \(\frac{3}{7}\) AB and P lies on the segment AB.
Solution:
Given : A(-2, -2) and B(2, -4)
P lies on AB such that AP = \(\frac{3}{7}\) AB

⇒ \(\frac{A P}{A B}\) ⇒ \(\frac{3}{7}\) = \(\frac{A P}{P B}\) = \(\frac{3}{7-3}\) = \(\frac{3}{4}\)
i.e., P divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 4 By section formula.

Question 7.
Find the co-ordinates of points which divide the line segment joining A(-4, 0) and B(0, 6) into four equal parts.
Solution:
Given, A(-4, 0) and B(0, 6)
Let R Q and R be the points which divide \(\overline{\mathrm{AB}}\) into four equal parts.

P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3
Q → 1 : 1 and R → 3 : 1 use section formula to find P, Q and R.
The Q is the mid point of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 8.
Find the co-ordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Given : A(-2, 2) and B(2, 8)
Let P, Q and R be the points which divide AB into four equal parts.

Then Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 9.
Find the co-ordinates of the point which divide the line segment joining the points (a + b, a – b) and (a – b, a + b) in the ratio 3 : 2 internally.
Solution:
Given : A(a + b, a – b) and B(a – b, a + b)
let P(x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 2 inter¬nally.
By using section formula,

Question 10.
Find the co-ordinates of centroid of the triangle with vertices following.

i) (-1, 3), (6, -3) and (-3, 6).
Solution:
Given : ΔABC in which A(-1, 3), B(6, -3) and C(-3, 6).
Centroid

ii) (6, 2), (0, 0) and (4, -7).
Solution:
Given : The three vertices of a triangle are A(6, 2), B(0, 0) and C(4, -7)
Centroid (x, y)

iii) (1, -1) (0, 6) and (-3, 0)
Solution:
Given: (1, -1), (0, 6) and (-3, 0) are the vertices of a triangle.
Centroid (x. y)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 7 Coordinate Geometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

Previous Exams Questions

Question 1.
Where do the points (0, -3) and (-8, 0) lie on co-ordinate axis ? (A.P. June ’15)
Solution:
The point (0, -3) lie on OY’
∴ Its X-co-ordinate is zero and Y-co-ordinate is negative and the point (-8, 0) lie on OX’
∵ Its Y-co-ordinate is zero and X-co-ordinate is negative.

Question 2.
Find the centroid triangle whose vertices are (3, 4) (-7, -2) and (10, -5). (T.S. Mar. 15)
Solution:
Centroid of the triangle whose vertices are (x₁, y₁) (x₂, y₂) (x₃, y₃)
= \(\left[\frac{X_1+X_2+X_3}{3}, \frac{Y_1+Y_2+Y_3}{3}\right]\)
Here x₁ = 3, y₁ = 4, x₂ = -7, y₂ = -2, and x₃ = 10, y₃ = -5
Now G = \(\left(\frac{3-7+10}{3}, \frac{4-2-5}{3}\right)\)
= \(\left(\frac{6}{3}, \frac{3}{3}\right)\) = (2, 1)

Question 3.
Show that the points A = (4, 2), B (7, 5) and C (9, 7) are collinear. (A.P. Mar. 15)
Solution:
To show that three points are collinear the area formed by the triangle is zero. Area of triangle
Δ = \(\frac{1}{2}\) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Here x₁ = 4, y₁ = 2, x₂ = 7, y₂ = 5,
x₃ = 9, y₃ = 7
Now Δ = \(\frac{1}{2}\) |4(5 – 7) + 7(7 – 2) + 9(2 – 5)|
= \(\frac{1}{2}\)|4(-2) + 7(5) + 9(-3)|
= \(\frac{1}{2}\)|-8 + 35 – 27|
= \(\frac{1}{2}\)|-35 + 35| = \(\frac{1}{2}\) |0| = 0
So the above three are collinear.

Question 4.
Name the type of the quadrilateral formed by joining the points A (-1, -2), B (1, 0) C (-1, 2) and D (-3, 0) are graph paper, justify your answer. (A.P. Mar. 15)
Solution:
Given points are A (-1, -2); B (1, 0); C (-1, 2) and D (-3, 0)

Distance between the two points

∴ ABCD is a square.
(OR)

From (1) and (2) ABCD is a square.
(OR)
From the graph
1) Diagonals are equal \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{BD}}\) = 4 units
2) Also diagonals \(\overline{\mathrm{AC}}\) and \(\overline{\mathrm{BD}}\) bisect each other. \(\overline{\mathrm{BD}}\) is a part of X-axis and \(\overline{\mathrm{AC}}\) is parallel to Y-axis.
∴ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\)
From (i), (ii) and (iii) ABCD is a square.

Question 5.
Find the mid point of the line segment formed by the points (-5, 5) and (5, -5) (A.P. Mar. 16)
Solution:
Formula for the mid points of line segment formed by (x₁, y₁) and (x₂, y₂) is
\(\left[\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right]\)
here x₁ = -5, y₁ = 5 and x₂ = 5, y₂ = -5
∴ mid point
= \(\left[\frac{-5+5}{2}, \frac{5-5}{2}\right]\) = \(\left[\frac{0}{2}, \frac{0}{2}\right]\) = [0, 0]

Question 6.
Show that the points A(-3, 3) B (0, 0) C(3, -3) are collinear. (T.S.Mar. 16)
Solution:
To show them as collinear the area formed by the triangle should be zero. Formula for area of triangle
∆ = \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁ + x₃(y₁ – y₂)|
Here x₁ = -3, y₁ = 3; x₂ = 0, y₂ = 0 and x₃ = 3, y₃ = -3
So ∆ = \(\frac{1}{2}\) | (-3) (3 – 0) + 0 (-3 – 3) + 3(3 – 0) |
= \(\frac{1}{2}\) |(-3)(3) + 0(-6) + 3(3) |
= \(\frac{1}{2}\) |-9 + 0 + 9|
= \(\frac{1}{2}\) |0| = 0 sq. units
Hence the above three points are collinear.

Question 7.
Find the trisection points of the line segment joined by the points (-3, 3) and (3, -3). (A.P. Mar. ’16)
Solution:
The points which divide the line segment by 1:2 and 2:1 ratio (internally) are called trisection points.
Formula for the points of trisection of the line segment joined by (x₁, y₂) and (x₂, y₂) are
= \(\left[\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}\right]\)
Where m = 2 and n = 1.
Here x₁ = -3, y₁ = 3 and x₂ = 3, y₂ = -3, then the point in the ratio 2 : 1 is

So (1, -1) is one trisection point. The point which is at 1 : 2 ratio is another trisection point.
So m = 1, n = 2.
Here x₁ = -3, = 3, x₂ = 3 and y₂ = -3
then \(\left[\frac{1(-3)+2(3)}{1+2}, \frac{1(-3)+2(3)}{1+2}\right]\)
= \(\left[\frac{-3+6}{3}, \frac{-3+6}{3}\right]\) = \(\left[\frac{3}{3}, \frac{3}{3}\right]\)
= (1, 1) is another trisection point.

P, Q are trisection points.

Additional Questions

Question 1.
Find the distance between the following pairs of points. (Each 1 Mark)

i) (3, 4) and (6, 2)
ii) (-4, 6) and (0, 2)
iii) (-a, b) and (a, -b)

i) (3, 4) and (6, 2)
Solution:
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(6-3)^2+(2-4)^2}\)
= \(\sqrt{9+4}\) = \(\sqrt{13}\) Units

ii) (-4, 6) and (0, 2)
Solution:
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(0+4)^2+(2-6)^2}\)
= \(\sqrt{16+(-4)^2}\)
= \(\sqrt{16+16}\) = \(\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= \(\sqrt{16}\) × \(\sqrt{2}\) = 4\(\sqrt{2}\) Units

iii) (-a, b) and (a, -b)
Solution:
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(a+a)^2+(-b-b)^2}\)
= \(\sqrt{(2 a)^2+(-2 b)^2}\)
= \(\sqrt{4 a^2+4 b^2}\) = \(\sqrt{4\left(a^2+b^2\right)}\)
= \(\sqrt{4} \sqrt{a^2+b^2}\) = \(2 \sqrt{a^2+b^2}\) Units

Question 2.
Verify Whether the points A(1, 3), B(3, 5) and C(-4, 2) are collinear or not.
Solution:
Given A = (1, 3), B = (3, 5) and C = (-4, 2)
Slope of AB = m₁ = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{5-3}{3-1}\) = \(\frac{2}{2}\) = 1
Slope of BC = m₂ = \(\frac{2-5}{-4-3}\) = \(\frac{-3}{-7}\) = \(\frac{3}{7}\)
Since m₁ ≠ m₂
∴ A, B, C are not collinear.

Question 3.
Show that (5, 3), (1, 2) and (-3, 1) are the vertices of an isosceles triangle.
Solution:
Let A = (5, 3), B = (1, 2) and C = (-3, 1)
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Since AB = BC
∴ ΔABC is an isosceles triangle.
Hence given points are the vertices of an isos- celes triangle.

Question 4.
Show that the points (0, -2), (3, 2), (0, 6) and (-3, 2) are the vertices of a square.
Solution:
Let A = (0, -2), B = (3,2), C = (0,6), D = (-3, 2)
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)


Sin AB = BC = CD = DA
∴ Given vertices form a square.

Question 5.
Show that the points (2, 4), (3, 7), (6, 8) and (5, 5) are the vertices of a Rhombus.
Solution:
Let A = (2, 4), B = (3, 7), C = (6, 8) and D = (5, 5)
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Since AB = BC = CD = DA = \(\sqrt{10}\) and AC ≠ BD
∴ Given points form a Rhombus.

Question 6.
Find the co-ordinates of the point which divides the line segment joining the points (-2, 6) and (3, -4) in the ratio 3 : 5.
Solution:
Given points P (-2, 6) and Q (3, -4)
Let R be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 3 : 5, then

Here P = (-2, 6) = (x₁, y₁),
Q = (3, -4) = (x₂, y₂)
and m₁ : m₂ = 3 : 5
Then R (x, y)

Question 7.
Find the ratio in which the X-axis divides the line segment joining the points (6, -2) and (4, 1) also find the point of intersection.
Solution:
Let the ratio be k : 1
given points are A = (6, -2) and B = (4, 1)
Then by section formula, the co-ordinates of the point which divide AB in the ratio k : 1 are
\(\left(\frac{\mathrm{k} \times 4+1 \times 6}{\mathrm{k}+1}, \frac{\mathrm{k} \times 1+1 \times-2}{\mathrm{k}+1}\right)\)
i.e., \(\left(\frac{4 k+6}{k+1}, \frac{k-2}{k+1}\right)\) ……. (1)
Since the point lies on the X-axis
∴ Its Y-co-ordinate is zero
i.e., \(\frac{k-2}{k+1}\) = 0 ⇒ k – 2 = 0 ⇒ k = 2
∴ The ratio is k : 1 = 2 : 1 ⇒ k = 2
Putting k = 2 in (1), we get, the point of intersection
= \(\left(\frac{4 \times 2+6}{2+1}, \frac{2-2}{2+1}\right)\)
= \(\left(\frac{8+6}{3}, \frac{0}{3}\right)\)
= \(\left(\frac{14}{3}, 0\right)\)

Question 8.
Find the ratio in which the Y-axis divides the line segment joining the points (9, -4) and (3, -2). Also find the point of intersection.
Solution:
Let the ratio be k : 1
Given points are A = (9, -4) and B = (3, -2)
Then by section formula, the co-ordinates of the point which divides AB in the ratio k : 1 are
\(\left(\frac{\mathrm{k} \times 3+1 \times 9}{\mathrm{k}+1}, \frac{\mathrm{k} \times(-2)+1 \times(-4)}{\mathrm{k}+1}\right)\)
i.e., \(\left(\frac{3 k+9}{k+9}, \frac{-2 k-4}{k+1}\right)\) —– (1)
Since the point lies on the Y-axis
∴ Its abscissa = 0
i.e., \(\frac{3 \mathrm{k}+9}{\mathrm{k}+9}\) = 0 ⇒ 3k + 9 = 0 ⇒ 3k = -9
⇒ k = \(\frac{-9}{3}\) = -3
∴ The ratio is k : 1 = -3 : 1 ⇒ k = -3
Putting k = -3 in (1), we get, the point of intersection
= \(\left(\frac{3 \times(-3)+9}{-3+9}, \frac{-2(-3)-4}{-3+1}\right)\)
= \(\left(\frac{-9+9}{6}, \frac{6-4}{-2}\right)\)
= \(\left(\frac{0}{6}, \frac{2}{-2}\right)\)
= (0, -1)

Question 9.
Find the centroid of the triangle whose vertices are given below.
i) (1, 6), (7, 4), (1, 5)
ii) (1, 0), (-3, -3), (-4, -3)

i) (1, 6), (7, 4), (1, 5)
Solution:
Given A = (1, 6) = (x₁, y₁)
B = (7, 4) = (x₂, y₂)
C = (1, 5) = (x₃, y₃)
Centroid of the triangle
= \(\left(\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3}, \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}\right)\)
∴ Centroid of the triangle
= \(\left(\frac{1+7+1}{3}, \frac{6+4+5}{3}\right)\)
= \(\left(\frac{9}{3}, \frac{15}{3}\right)\) = (3, 5)

ii) (1, 0), (-3, -3), (-4, -3)
Solution:
Given A = (1, 0), B = (-3, -3), C = (-4, -3)
Centroid of the triangle
= \(\left(\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3}, \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}\right)\)
= \(\left(\frac{1-3-4}{3}, \frac{0-3-3}{3}\right)\)
= \(\left(\frac{-6}{3}, \frac{-6}{3}\right)\)
= (-2, -2)

Question 10.
Find the area of the triangle whose vertices are
i) (3, 4), (0, 1), (3, -3)
ii) (2, 10), (4, 7), (0, -1)
iii) (3, 4), (0, 1), (3, -3)
Solution:
i)
Given A = (3, 4) = (x₁, y₁)
B = (0, 1) = (x₂, y₂)
C = (3, -3) = (x₃, y₃)
are the vertices of ∆ABC.
Area of ∆ABC
= \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\)|3(1 + 3) + 0(-3 – 4) + 3(4 – 1)|
= \(\frac{1}{2}\)|3(4) + 0(-7) + 3(3)|
= \(\frac{1}{2}\)|12 + 0 + 9|
= \(\frac{1}{2}\)|21|
= \(\frac{21}{2}\) sq. units

ii) (2, 10), (4, 7), (0, -1)
Solution:
Given A = (2, 10), B = (4, 7), C = (0, -1)
Area of ∆ ABC
= \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\)|(7 + 1) + 4(-1 – 10) + 0(10 – 7)|
= \(\frac{1}{2}\)|2(8) + 4(-11) + 0(3)|
= \(\frac{1}{2}\)|16 – 44 + 0|
= \(\frac{1}{2}\)|-28| = |-14|
= 14sq. units

Question 11.
Find the value of k’ for which the points are collinear.
i) (k, 9), (7, 7), (1, 5)
ii) (6, -1), (k, -6), (0, -7)
iii) (k, k), (1, 2), (3, -2)

i) (k, 9), (7, 7), (1, 5)
Solution:
Given A(k, 9) = (x₁, y₁)
B(7, 7) = (x₂, y₂)
C(1, 5) = (x₃, y₃) are collinear
∴ Area of Δ ABC = 0
⇒ \(\frac{1}{2}\) x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒| \(\frac{1}{2}\)|k(7 – 5) + 7(5 – 9) + 1(9 – 7)| = 0
⇒ |k(2) + 7(-4) + 1(2)| = 0
⇒ 2k – 28 + 2 = 0
⇒ 2k – 26 = 0
⇒ 2k = 26
k = \(\frac{26}{2}\) = 13
∴ k = 13

ii) (6, -1), (k, -6), (0, -7)
Solution:
Given A(6, -1), B(k, -6), C(0, -7) are collinear
∴ Area of Δ ABC = 0
⇒ \(\frac{1}{2}\)|6(-6 + 7) + k(-7 + 1) + 0(-1 + 6)| = 0
⇒ |6(1) + k(-6) + 0(5)| = 0
⇒ 6 – 6k + 0 = 0
⇒ -6k = -6
⇒ k = \(\frac{-6}{-6}\) = 1
∴ k = 1

iii) (k, k), (1, 2), (3, -2)
Solution:
Given A = (k, k), B(1, 2), C(3, -2) are collinear
∴ Area of Δ ABC = 0
⇒ \(\frac{1}{2}\)|k(2 + 2) + 1(-2 – k) + 3(k – 2)| = 0
⇒ \(\frac{1}{2}\)|k(4) + 1(-2 – k) + 3(k – 2)| = 0
⇒ 4k – 2 – k + 3k – 6 = 0
⇒ 6k – 8 = 0
⇒ 6k = 8
⇒ k = \(\frac{8}{6}\) = \(\frac{4}{3}\)
∴ k = \(\frac{4}{3}\)

Question 12.
Find the slope of the line joining the points.
i) (7, -5) and (8, 1)
ii) (4a, 6b) and (a, -b)
iii) (5, -1) and (-8, 0)

i) (7, -5) and (8, 1)
Solution:
Given A(7, -5) = (x₁, y₁)
B(8, 1) = (x₂, y₂)
Slope of line AB = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{1-(-5)}{8-7}\)
= \(\frac{1+5}{1}\) = \(\frac{6}{1}\) = 6

ii) (4a, 6b) and (a, -b)
Solution:
Given A(4a, 6b), B(a, -b)
Slope of line AB = \(\frac{-b-6 b}{a-4 a}\) = \(\frac{7 b}{3 a}\)

iii) (5, -1) and (-8, 0)
Solution:
Given A(5, -1) and B(-8, 0)
Slope of line AB = \(\frac{0-(-1)}{-8-5}\)
= \(\frac{0+1}{-13}\) = \(\frac{-1}{13}\)

Question 2.
Find the slope of a line whose inclination is

i) 60°
ii) 135°
iii) 120°
iv) 135°

i) 60°
Solution:
Given 60° = θ
Slope of the line = m
= tan θ = tan 60° = \(\sqrt{3}\)

ii) 135°
Solution:
Given 135° = θ
Slope of the line = m = tan θ = tan 135°
= tan (90° + 45°)
= -cot 45° = -1

iii) 120°
Solution:
Given 120° = θ
Slope of the line = m = tan θ = tan 120°
= tan (90° + 30°)
= -cot 30°
= –\(\sqrt{3}\)

iv) 135°
Slope of the line = m = tan θ = tan 135°
= tan (90° + 45°)
= -cot 45°
= -1

Question 3.
If (6, 7) is the mid point of the line segment joining A(7, 6) and B(5, y), find ‘y’.
Solution:
Given Mid point = (6, 7) = m and
A(7, 6) = (x₁, y₁)
B(5, y) = (x₂, y₂)
Mid point of A and B = (6, 7)
⇒ \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) = (6, 7)
⇒ \(\left(\frac{7+5}{2}, \frac{6+y}{2}\right)\) = (6, 7)
Equations Y-co-ordìnate on both sides
⇒ \(\frac{6+\mathrm{y}}{2}\) = 7 ⇒ 6 + y = 2 × 7
⇒ 6 + y = 2 × 7
⇒ 6 + y = 14
⇒ y = 14 – 6
y = 8
∴ y = 8

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.4

Question 1.
Find the slope of the line passing the two given points.

i) (4, -8) and (5, – 2)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) = \(\frac{-2+8}{5-4}\) = \(\frac{6}{1}\) = 6

ii) (0, 0) and (\(\sqrt{3}\), 3)
Solution:
Slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{3-0}{\sqrt{3}-0}\)

iii) (0, 0) and (\(\sqrt{3}\), 3)
Solution:
Slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-b-3 b}{a-2 a}\)

iv) (a, 0) and (0, b)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) = \(\frac{b-0}{0-a}\) = \(\frac{-b}{a}\)

v) A(-1.4, -3.7) and B(-2.4, 1.3)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) = \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\) = -5

vi) A(3,-2) and B(-6,-2)
Solution:
Slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-2+2}{-6-3}\) = 0

vii)

Solution:

viii) A(0, 4), B(4, 0)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{0-4}{4-0}\)

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.
Find the area of the triangle whose vertices are

i) (2, 3), (-1, 0), (2,-4)
Solution:
Given : A(2, 3), B(-1, 0), C(2, -4) are the vertices of a ΔABC
Area of the ΔABC
= \(\frac{1}{2}\)|(y₂ – y₃) + x₂ (y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\) |8 + 7 + 6|
= \(\frac{21}{2}\) = 10\(\frac{1}{2}\) sq. units.

ii) (- 5, – 1), (3, -5), (5, 2) (A.P. Mar. 15)
Solution:
Given A(-5, -1), B(3, -5), C(5, 2) are the ver-tices of a ΔABC
Area of the Δ ABC
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) | 135 + 9 + 20|
= \(\frac{64}{2}\) = 32 sq.units.

iii) (0, 0), (3, 0) and (0, 2)
Solution:
Given 0(0, 0), A(3, 0), B(0, 2) are the vertices of a ΔABC
Area of the ΔAOB
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) |0(0 – 2) + 3(2 – 0) + 0(0 – 0)|
= \(\frac{1}{2}\) |6| = 3 sq. units. (or)

ΔAOB = \(\frac{1}{2}\) × 3 × 2 = 3 sq. units.

Question 2.
Find the value of ‘K’ for which the points are collinear. (A.P. June ’15)

i) (7, -2), (5, 1), (3, K)
Solution:
Given : A(7, -2), B(5, 1), C(3, K) are collinear
∴ Area of ΔABC = 0
But area of triangle.
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
⇒ \(\frac{1}{2}\) | 7 – 7K + 5K + 10 – 9| = 0
⇒ |-2K + 8| = 0
⇒ – 2K = – 8
⇒ K = 4

ii) (8, 1), (K, -4), (2, -5)
Solution:
Given A(8, 1), B(k, -4), C(2, -5) are collinear
∴ Area of ΔABC = 0
⇒ \(\frac{1}{2}\) |8(- 4 + 5) + K(-5 – 1) + 2 (1 + 4) 1= 0
⇒ | 8 – 6K + 10| = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
∴ K = 3

iii) (K, K) (2, 3) and (4, -1)
Solution:
A(K, K), B(2, 3) & C(4, -1) are collinear
∴ Area of ∆ABC = O
⇒ \(\frac{1}{2}\) | K(3 + 1) + 2 (-1 – K) + 4(K – 3)| = 0
⇒ | 4K – 2 – 2k + 4K – 12| = 0
⇒ |6K – 14| = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\)
= \(\frac{7}{3}\)
⇒ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (0, -1), (2, 1)
and (0, 3). Find the ratio of this area to the area of the given triangle. (A.P. Mar.16)
Solution:

Given : A(0, -1), B(2, 1), C(0, 3) are the vertices of ∆ABC. Let D, E & F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) & \(\overline{\mathrm{AC}}\)
midpoint (x, y) = \(\left[\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right]\)
∴ D = \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) = (1, 0)
E = \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) = (1, 2)
F = \(\left(\frac{0+0}{2}, \frac{-1+3}{2}\right)\) = (0, 1)
Area of ∆DEF
= \(\frac{1}{2}\) |1(2 – 1) + 1(1 – 0) + 0(0 – 2)|
= \(\frac{1}{2}\) |1 + 1| = \(\frac{2}{2}\) = 1 sq. units.
Ratio of areas = ∆ABC : ∆DEF = 4 : 1
∆ADF \(\cong\) ∆BED \(\cong\) ∆DEF \(\cong\) ∆CEF
∴ ∆ABC : ∆DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)
Solution:

Given : A(-4, -2), B(-3, -5), C(3, -2) & D(2, 3) are the vertices of the quadrilateral ABCD.
Area of
ABCD = ∆ABC + ∆ACD
Area of triangle
= \(\frac{1}{2}\) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Area of ∆ABC
= \(\frac{1}{2}\) |(-4)(-5 + 2)(-3)(-2 + 2) + 3(-2 + 5) |
= \(\frac{1}{2}\)|12 – 0 + 9| = \(\frac{21}{2}\) sq.units.
Area of ∆ACD
= \(\frac{1}{2}\)|-4(-2 – 3) + 3(3 + 2) + 2(-2 + 2) |
= \(\frac{1}{2}\)|20 + 15 + 0 | = \(\frac{35}{2}\) sq. units.
∴ Area of ABCD = \(\frac{21}{2}\) + \(\frac{35}{2}\) = \(\frac{21+35}{2}\)
= \(\frac{56}{2}\) = 28 sq.units.

Question 5.
Find the area of the triangle formed by the points (8, -5), (-2, -7) & (5, 1) by using Herons formula.
Solution:
Given : A(8, -5), B(-2, -7) & C(5, 1) are the vertices of ∆ABC
Distance formula
= \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
AB = \(\sqrt{100+4}\) = \(\sqrt{104}\) = 10.19 = a
BC = \(\sqrt{49+64}\) = \(\sqrt{113}\) = 10.63 = b

These TS 10th Class Maths Chapter Wise Important Questions Chapter 3 Polynomials given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 3 Polynomials

Previous Exams Questions

Question 1.
Find p(3) if p(x) = x² – 5x + 6 is given.
Solution:
p(x) = x² – 5x + 6 (given)
then p(3) = 3² – 5(3) + 6
= 9 – 15 + 6 = 15 – 15 = 0
∴ p(3) = 0.

Question 2.
Find p(3) if p (x) = x² – 5x – 6 is given. (A.P.Mar.’16)
Solution:
p(x) = x² – 5x + 6 (given)
p (3) = 3² – 5(3) + 6
= 9 – 15 + 6 = 15 – 15 = 0
∴ p(3) = 0.

Question 3.
Find the quotient \(\frac{x^5+x^4+x^3+x^2}{x^3+x^2+x+1}\) when x ≠ 1.
Solution:
\(\frac{x^5+x^4+x^3+x^2}{x^3+x^2+x+1}\)
= \(\frac{x^2\left(x^3+x^2+x+1\right)}{\left(x^3+x^2+x+1\right)}\)
= x²
So the quotient is [x²]

Question 4.
We can write a trinomial having degree 7^th. Justify the given statement by giving one example. (T.S. Mar.15)
Solution:
Since the required is a trinomial, it should have 3 terms. As its degree is 7 the highest exponent (power) of the variable is 7. So the required will be
1) x⁷ – x⁶ + x⁵ (or) x⁷ – x + 10 (or) x⁷ – x⁵ – x (or) x⁷ + 5x⁶ + 6x⁵ or …

Question 5.
If we multiply or divide both sides of a linear equation by a non zero number, then the roots of linear equation will remain the same. Is it true ? If so justify with an example. (T.S. Mar. ’15)
Solution:
Let linear equation be 2x + 8 = 11
Its solution is 2x = 11 – 8 ⇒ 2x = 3
So x = \(\frac{3}{2}\)
Then 2x + 8 = 11 linear equation is divided by 10 on both sides.
\(\frac{2 x+8}{10}\) = \(\frac{11}{10}\) = 20x + 80 = 110
⇒ 20x = 110 – 80 = 30
then x = \(\frac{30}{2}\) = \(\frac{3}{2}\)
So its solution doesn’t change.
So the given statement is true.

Question 6.
Laxmi does not want to disclose the l, b, h of a cuboid of her project. She has constructed a polynomial x³ – 6x² + 11x – 6 by taking the values of l, b, h as its zeros. Can you open the secret ? (T.S. Mar. 15)
Solution:
P(x) = x³ – 6x² + 11x – 6
P(1) = (1)³ – 6(1)² + 11 (1) – 6
= 1 – 6 + 11 – 6
= 12 – 12 = 0
∴ for P(x), (x – 1) is a factor

x² – 5x + 6 = x² – 3x – 2x + 6
= x (x – 3) – 2(x – 3)
= (x – 3) (x – 2)
P(x) = (x – 1) (x – 2) (x – 3)
P(x) zeroes of the polynomial are 1, 2,3
∴ Measurements of the cuboid are 1,2 and 3 units.
Its solution doesn’t change
So it is true.

Question 7.
Draw a graph for the polynomial p(x) = x² + 3x – 4 and find its zeros from the graph.
Solution:
p(x) = x² + 3x – 4


∴ Zeroes of polynomial are -4 and 1

Question 8.
Given an example for a quadratic polynomial which has no zeroes.
Solution:
A quadratic polynomial is in the form of ax² + bx + c. As this has no zeroes. Its discriminant will not be a real number.
So b² – 4ac < 0
So we can choose certain a, b, c values where
b² – 4ac < 0
For examples a = 1, b = 4 and c = 9
So ax² + bx + c = 0
⇒ x² + 4x + 9 = 0 will not have zeroes.

Question 9.
The length of a rectangle is 5 more than its breadth. So express its perimeter in the form of polynomial.
Solution:
Let the breadth of rectangle = xm and its
length = x + 5m.
So the perimeter = 2(l + b)
= 2(x + 5 + x)
= 2(2x + 5)
= 4x + 10m

4x + 10 is the polynomial which represents the perimeter of above rectangle.

Question 10.
Draw the graph of polynomial
p(x) = x² – 3x + 2 and find its zeros.
Sol, let y = p(x) = x² – 3x + 2
If x = 0 then p(0) = 0 – 0 + 2
= 2 So (0, 2)
x = 1 then p(1) = 1² – 3(1) + 2
= 1 – 3 + 2
= 0 So (1, 0)
x = 2 then p(2) = 2² – 3(2) + 2
= 4 – 6 + 2
= 0 So (2, 0)
x = 3 then p(3) = 3² – 3(3) + 2
= 9 – 9 + 2
= 2 So (3, 2)
and if x = -1 then p(-1)
= (-1)² – 3(-1) + 2
= 1 + 3 + 2
= 6 So (-1, 6)
x = -2 then p(-2) = (-2)² – 3(2) + 2
= 4 + 6 + 2 = 12 So (-2, 12)
that means the graph of the polynomial
p(x) = x² – 3x + 2 passes through the points. (0, 2), (1, 0) (2, 0) (3, 2) (-1, 6) and (-2, 12)

So 1 and 2 are zeroes of the given polynomial.

Question 11.
Draw the graphs of the following two linear equations and find their solution.
Solution:
First we have to recognise the points through which the line 3x – 2y = 2 passes then after 2x + y = 6 pass. Let us find the points
3x – 2y = 2
So y = \(\frac{3 x-2}{2}\) —- (1)
Put x = 0 in above equation
∴ y = \(\frac{3(0)-2}{2}\) = \(\frac{0-2}{2}\) = 1
So (0, 1)
Now x = 1 then
y = \(\frac{3(1)-2}{2}\) = \(\frac{3-2}{2}\) = \(\frac{1}{2}\) So(1, \(\frac{1}{2}\))
Now x = 2 then
y = \(\frac{3(2)-2}{2}\) = \(\frac{6-2}{2}\) = \(\frac{4}{2}\) So(2, 2)
that means the line 3x – 2y = 2 passes through the points (0, -1), (1, \(\frac{1}{2}\)) and (2, 2)
Similarly
2x + y = 6
⇒ y = 6 – 2x —– (2)
Put x = 0 in the above equation (2) we get
y = 6 – 2(0) = 6 – 0 = 6 So(0, 6)
and x = 1 ⇒ y = 6 – 2(1) = 6 – 2 = 4
So(1, 4)
and x = 2 ⇒ y = 6 – 2(2) = 6 – 4 = 2
So (2, 2)
So the line 2x + y = 6 passes through the points (0, 6) (1, 4) and (2, 2)
Here we observe (2, 2) is the common point.
That means they intersert at (2, 2)
So x = 2 and y = 2 will be the solution of above the equations.

Additional Questions

Question 1.
If p(x) = 6x⁷ – 2x⁵ + 4x – 8, find
(i) coefficient of x⁵
(ii) degree of p(x)
(iii) constant term. (Each 1 Mark)
Solution:
If p(x) = 6x⁷ – 2x⁵ + 4x – 8,

i) coefficient of x⁵ = – 2
ii) degree of p(x) = highest degree of x – 7
iii) constant term = – 8

Question 2.
If p(t) = 3t³ + 4t – 5, find the values of p(1), p(-1), P(0), p(2), p(-2).
Solution:
given p(t) = 3t³ + 4t – 5
∴ p(1) = 3(1)³ + 4(1) – 5 = 3 + 4 – 5 = 2
p(-1) = 3(-1)³ + 4(-1) – 5 = -3 – 4 – 5 = -12
p(0) = 3(0)³ + 4(0) – 5 = 0 + 0 – 5 = -5
p(2) = 3(2)³ + 4(2) – 5 = 3(8) + 8 – 5
= 24 + 8 – 5 = 27
p(-2) = 3(-2)³ + 4 (-2) – 5
= – 24 – 8 – 5 = – 37

Question 3.
Check whether – 4 and 4 are the zeroes of the polynomial x⁴ – 256.
Solution:
given p(x) = x⁴ – 256
p(-4) = (-4)⁴ – 256 = 256 – 256 = 0
p(4) = (4)⁴ – 256 = 256 – 256 = 0
yes, -4 and 4 are zeroes of the given polynomial

Question 4.
Check whether 5 and -4 are the zeroes of the polynomial p(x) when p(x) = x² – x – 20
Solution:
given p(x) = x² – x – 20
p(5) = (5)² – 5 – 20 = 25 – 5 – 20 = 25 – 25 = 0
P(-4) = (-4)² – (-4) – 20
= 16 + 4 – 20 = 20 – 20 = 0
yes, 5 and – 4 are the zeroes of the polynormial
p(x) = x² – x – 20.

Question 5.
Check whether 3 and – 7 are the zeroes of the polynomial p(x) = x² + 4x – 21.
Solution:
given p(x) = x² + 4x – 21
p(3) = 3² + 4(3) – 21
= 9 + 12 – 21 = 21 – 21 = 0
p(-7) = (-7)² + 4(-7) – 21
= 49 – 28 – 21 = 49 – 49 = 0
yes, 3 and – 7 are the zeroes of the polynomial p(x) = x² + 4x – 21.

Question 6.
Find the zeroes of the given polynomials.
i) p(x) = 4x
ii) p(x) = x² – 9x + 20
iii) p(x) = (x + 5) (x + 6)
iv) p(x) = x⁴ – 81
Solution:
i) p(x) = 4x
p(0) = 4 × 0 = 0
∴ The zero of p(x) = 4x is 0.

ii) p(x) = x² – 9x + 20
= x² – 4x – 5x + 20
= x (x – 4) -5 (x – 4)
= (x – 4) (x – 5)
To find zeroes, let p(x) = 0
⇒ (x – 4) (x – 5) = 0
⇒ x – 4 = 0 or x – 5 = 0
⇒ x = 4 or x = 5
∴ The zeroes of x² – 9x + 20 are 4 nd 5.

iii) p(x) = (x + 5) (x + 6)
To find zeroes, let p(x) = 0
⇒ (x + 5) (x + 6) = 0
⇒ x + 5 = 0 or x + 6 = 0
⇒ x = -5 or x = -6
∴ The zeroes of (x + 5) (x + 6) are – 5 and – 6

iv) p(x) = x⁴ – 81.
To find zeroes, let p(x) = 0
⇒ x⁴ – 81 = 0
⇒ (x²)² – (9)² = 0
⇒ (x² + 9) (x² – 9) = 0
⇒ x² + 9 = 0 or x² – 9 = 0
x² = -9 or (x + 3) (x – 3) = 0
x = ±\(\sqrt{-9}\) = 9 or x + 3 = 0 or x – 3 = 0
x = -3 or x = 3
∴ The zeroes of the polynomial x⁴ – 81 are -3, 3 and ± \(\sqrt{-9}\).

Question 7.
Find the zeroes of the given polynomials.
i) p(x) = x² – x – 12
ii) p(x) = x² – 6x + 9
iii) x² p(x) = x² – 4x + 5
iv) p(x) = x² + 3x – 4
Solution:
i) p(x) = x² – x – 12
To find zeroes, let p(x) = 0
⇒ x² – x – 12 = 0
x² – 4x + 3x – 12 = 0
x(x – 4) + 3 (x – 4) = 0
(x – 4) (x + 3) = 0
⇒ x – 4 = 0 or x + 3 = 0
∴ The zeroes of x² – x – 12 are 4 and – 3

ii) p(x) = x² – 6x + 9
To find zeroes, let p(x) = 0
x² – 6x + 9 = 0
x² – 3x – 3x + 9 = 0
x(x – 3) – 3 (x – 3) = 0
⇒ (x – 3) (x – 3) = 0
⇒ x – 3 = 0 or x – 3 = 0
⇒ x = 3 or x = 3
∴ The zeroes of x² – 6x + 9 is 3.

iii) p(x) = x² – 4x – 5
To find zeroes, let p(x) = 0
x² – 4x – 5 = 0
x² – 5x + x – 5 = 0
⇒ x (x – 5) + 1(x – 5) = 0
⇒ (x – 5) (x + 1) = 0
⇒ x – 5 = 0 or x + 1 = 0
⇒ x = 5 or x = -1
∴ The zeroes of x² – 4x – 5 are 5 and – 1

iv) p(x) = x² + 3x – 4
To find zeroes, let p(x) = 0
x² + 3x – 4 = 0
⇒ x² + 4x – x – 4 = 0
⇒ x(x + 4) – 1(x + 4) = 0
⇒ (x + 4) (x – 1) = 0
⇒ x + 4 = 0 or x – 1 = 0
⇒ x = -4 or x = 1
∴ The zeroes of x² + 3x – 4 are -4 and 1.

Question 8.
Why are \(\frac{1}{3}\) and -2 zeroes of polynomial
p(x) = 3x² + 5x – 2 ?
Solution:
p(x) = 3x² + 5x – 2
Now p(\(\frac{1}{3}\)) = 3(\(\frac{1}{3}\))² + 5(\(\frac{1}{3}\)) – 2
= \(\frac{1}{3}\) + \(\frac{5}{3}\) – 2
= \(\frac{1+5-6}{3}\) = \(\frac{6-6}{3}\) = \(\frac{0}{3}\) = 0
p(-2) = 3 (-2)² + 5(-2) – 2
= 3(4) – 10 – 2
= 12 – 12 = 0
since p(\(\frac{1}{3}\)) and p(-2) are equal to zero
–\(\frac{1}{3}\) and – 2 are zeroes of the polynomial.

Question 9.
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients.
i) x² – 5x + 6
ii) 3x² – 7x + 4
iii) 5x² + 10x
iv) t² – 12
Solution:
i) Let p(x) = x² – 5x + 6
= x² – 3x – 2x + 6
= x(x – 3) -2 (x – 3)
= (x – 3) (x – 2)
To find zeroes, let p(x) = 0
Hence the zeroes of p(x) are 3 and 2
sum of the zeroes = 3 + 2 = 5 = –\(\frac{(-5)}{1}\)

Product of zeroes = 3 × 2 = 6 = \(\frac{6}{1}\) constant term

ii) 3x² – 7x + 4
Let p(x) = 3x² – 7x + 4
= 3x² – 3x – 4x + 4
= 3x (x – 1) -4 (x – 1)
= (x – 1) (3x – 4)
To find zeroes, let p(x) = 0
⇒ (x – 1) (3x – 4) = 0
x = 1 or x = 4/3
Hence, the zeroes of p(x) are 1 and \(\frac{4}{3}\).
sum of the zeroes = 1 + \(\frac{4}{3}\) = \(\frac{3+4}{3}\) = \(\frac{7}{3}\)

product of zeroes = 1 × \(\frac{4}{3}\) = \(\frac{4}{3}\)

iii) Let p(x) = 5x² + 10x
= 5x (x + 2)
To find zeroes, Let p(x) = 0
⇒ 5x (x + 2) = 0
⇒ x = 0 or x + 2 = 0
⇒ x = 0 or x = -2
Hence the zeroes of p(x) are 0 and -2.
sum of the zeroes = 0 + (-2) = -2

coefficient of x Product of zeroes = 0 x – 2 = 0

iv) t² – 12
let p(t) = t² – 12
to find zeroes, let p(t) = 0
⇒ t² – 12 = 0
⇒ t² = 12 ⇒ t = ±\(\sqrt{12}\) = \(\sqrt{12}\) and –\(\sqrt{12}\)
Hence the zeroes of p(t) are \(\sqrt{12}\) and –\(\sqrt{12}\)
sum of zeroes = \(\sqrt{12}\) + (-\(\sqrt{12}\))
= \(\sqrt{12}\) – \(\sqrt{12}\) = 0 = \(\frac{0}{1}\)

product of zeroes = \(\sqrt{12}\) × (-\(\sqrt{12}\))
= -12 = \(-\frac{12}{1}\)

Question 10.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes.
i) \(\frac{1}{3}\), -2
ii) 0, \(\sqrt{3}\)
iii) 3, 5
iv) –\(\frac{1}{2}\), \(\frac{1}{2}\)
Solution:
i) \(\frac{1}{3}\), -2
Let α, β be the zeroes of the quadratic polynomial sum of zeroes = α + β = \(\frac{1}{3}\)
product of zeroes = αβ = -2
The required quadratic polynomial will be
k [x² – x(α + β) + αβ] when k is a constant
⇒ k[x² – \(\frac{x}{3}\) – 2]
If k = 3, then the polynomial will be
3[x² – \(\frac{x}{3}\) – 2] = 3x² – x – 6

ii) 0, \(\sqrt{3}\)
Let α, β be the zeroes of quadratic polynomial.
sum of the zeroes = α + β = 0
product of the zeroes = αβ = \(\sqrt{3}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ] where k is a constant.
⇒ k [x² – x(0) + \(\sqrt{3}\))
⇒ k [x² – 0 + \(\sqrt{3}\)] ⇒ k[x² + \(\sqrt{3}\)
If k = 1, then the polynomial will be 1[x² + \(\sqrt{3}\)] = x² + \(\sqrt{3}\).

iii) 3, 5
Let α, β be the zeroes of quadratic polynomial.
sum of the zeroes = α + β = 3
product of the zeroes = αβ = 5
The required quadratic polynomial will be k[x² – x(α + β) + αβ] where k is a constant
⇒ k [x² – x(3) + 5]
⇒ k [x² – 3x + 5]
If k = 1, then the polynomial will be
1[x² – 3x + 5] = x² – 3x + 5

iv) –\(\frac{1}{2}\), \(\frac{1}{2}\)
Let α, β be the zeroes of quadratic polynomial sum of the zeroes = α + β
= –\(\frac{1}{2}\)
product of the zeroes αβ = \(\frac{1}{2}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ] when k is a constant
k[x² – x(-\(\frac{1}{2}\)) + \(\frac{1}{2}\)]
⇒ k[x² + \(\frac{x}{2}\) + \(\frac{1}{2}\)]
If k = 2, then the polynomial will be 2[x² + \(\frac{x}{2}\) + \(\frac{1}{2}\)] = 2x² + x + 1

Question 11.
Find the quadratic polynomial the zeroes α, β given in each case

i) 4, -3
ii) \(\sqrt{5}\) – \(\sqrt{5}\)
iii) \(\frac{1}{3}\), \(\frac{5}{3}\)
Solution:
i) 4, -3
Let the quadratic polynomial be ax² + bx + c, a ≠ 0,
and its zeroes be a and b.
then α = 4, β = – 3
sum of the zeroes = a + b = 4 – 3 = 1
product of the zeroes = a . b = 4 × -3
= -12
The requiral quadratic polynomial is k [x² – x (α + β) + αβ] where k is a constant
⇒ k[x² – x(1) + (-12)]
⇒ k[x² – x – 12]
When k = 1, the quadratic polynomial will be x² – x – 12.

ii) \(\sqrt{5}\) – \(\sqrt{5}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be a and b, Here α = \(\sqrt{5}\), β = –\(\sqrt{5}\)
sum of the zeroes = α + β = \(\sqrt{5}\) – \(\sqrt{5}\) = 0
product of the zeroes = α . β
= \(\sqrt{5}\) × –\(\sqrt{5}\) = -5
The required quadratic polynomial is
k² [x – x (α + β) + αβ] where k is a constant
⇒ k² [x² – x(0) + (-5)]
⇒ k[x² – 5]
wherek k = 1, the quadratic polynomial will be x² – 5

iii) \(\frac{1}{3}\), \(\frac{5}{3}\)
Let the quadrate polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α and β.
Then α = \(\frac{1}{3}\) and β = \(\frac{5}{3}\)
sum of the zeroes = α + β = \(\frac{1}{3}\) + \(\frac{5}{3}\)
= \(\frac{6}{3}\) = 2
Product of the zeroes = αβ = \(\frac{1}{3}\) . \(\frac{5}{3}\)
= \(\frac{5}{9}\)
The required quadratic polynomial is
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(2) + \(\frac{5}{9}\)]
⇒ k[x² – 2x + \(\frac{5}{9}\)]
Where k = 9, the quadratic polynomial will be a [x² – 2x + \(\frac{5}{9}\)] = 9x² – 18x + 5

Question 12.
Verify that 1, -2 and -3 are the zeroes of the cubic polynomial x³ + 4x² + x – 6 and check the relationship between zeroes and the coefficients.
Solution:
The given polynomial is x³ + 4x² + x – 6.
Comparing the given polynomial with
ax³ + bx² + cx + d, we get
a = 1, b = 4, c = 1, d = – 6
Let p(x) = x³ + 4x² + x – 6
p(1) = (1)³ + 4(1)² + 1 – 6
= 1 + 4 + 1 – 6 = 6 – 6 = 0
p(-2) = (-2)³ + 4(-2)² + (-2) – 6
= – 8 + 4(4) – 2 – 6
= – 8 + 16 – 2 – 6 = 16 – 16 = 0
p(-3) =(-3)³ + 4(-3)² + (-3) – 6
= – 27 + 4(9) – 3 – 6
= 36 – 36 = 0
∴ 1, -2 and – 3 are the zeroes of given polynomial. So, α = 1, β = -2, ∝= – 3

Question 13.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.

i) p(x) = x³ + 4x² + 5x + 6, g(x) = x² + 3
ii) p(x) = x⁴ + 3x² – 4x + 8
g(x) = x² + 2 – x
Solution:
i) p(x) = x³ + 4x² + 5x + 6,
g(x) = x² + 3

The degree of x² + 3 is 2. The degree of 2x – 6 is 1
Since degree of (2x – 6)² < degree of (x² + 3)
∴ we stop here.
So, the quotient is x + 4 and the remainder is 2x – 6.

ii) p(x) = x⁴ + 3x² – 4x + 8
g(x) = x² + 2 – x = x² – x + 2
(∴ writing it in standard form)

since degree of (-4x + 4) < degree of x² – x + 2
∴ we stop here
So, the quotient is x² + x + 2 and the remainder is (-4x + 4).

Question 14.
Check in each case the first polynomial is a factor of the second polynomial.
i) x² – 2, 2x⁴ – 3x³ – 3x² + 6x – 2
ii) x² – x + 1, x⁴ – 3x² + 4x – 3
Solution:
Hint : If the remainder is zero, then the first polynomial is a factor of the second one

i) x² – 2, 2x⁴ – 3x³ – 3x² + 6x – 2
The given polynomials are in standard form.

Since the remainder is zero, x² – 2 is the factor of 2x⁴ – 3x³ – 3x² + 6x – 2.

ii) x² – x + 1, x⁴ – 3x² + 4x – 3
The given polynomials is standard form.

Since the remainder is zero, x² – x + 2 is the factor of x⁴ – 3x² + 4x – 3.

Question 15.
Draw the graph of the polynomial x² + x – 6 and mark the zeroes by the …….
Solution:
given that y = x² + x – 6
= x² + x – 6


zeroes of the polynomial x² + x – 6 are the x – co-ordinates of the points on the – 3 and 2 are the zeroes of the given polynomial on graph.

Question 16.
Check whether \(\frac{1}{2}\) is the zero of the polynomial 2x² + x – 1 or not. (AP-SA-II : 2016)
Solution:
Let p(x) = 2x² + x – 1
we have p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))² + \(\frac{1}{2}\) – 1
= 2 × \(\frac{1}{4}\) + \(\frac{1}{2}\) – 1
= \(\frac{1}{2}\) + \(\frac{1}{2}\) – 1 = 1 – 1
= 0
∴\(\frac{1}{2}\) is the zero of the polynomial.

Question 17.
Verify the relationship between the zeroes and the coefficient of x² – 25 by finding its zeroes. (AP-SA-II : 2016)
Solution:
Given polynomial is x² – 25
we have x² – 25 = 0
⇒ (x + 5) (x – 5) = 0
⇒ x = -5 or x = 5
∴ The zeroes of x² – 25 are – 5 and 5
∴ The sum of zeroes = -5 + 5 = 0 coefficient of x

= –\(\frac{0}{1}\) = 0
And product of the zeroes = (-5) × (5) = – 25

= –\(\frac{25}{1}\) = -25
Hence verified.

Question 18.
Give two examples of the polynomials p(x) and g(x) satisfying division algori-tham. (AP-SA-II : 2016)
Solution:
Using division alogaritham,
We have p(x) = q(x) × g(x) + r(x)
Example (1) : on dividing 12x² + 8x + 24 by 3x² + 2x + 6, we get

Here p(x) = 12x² + 8x + 24
g(x) = 3x² + 2x + 6
g(x) = 4
r(x) = 0

Example – 2 : an dividing 3x³ + 6x² + 18x by x² + 2x + 6

Here p(x) = 3x² + 6x² + 18x
g(x) = x² + 2x + 6
g(x) = 3x
r(x) = 0

Question 19.
Verify that 4, -1, –\(\frac{1}{4}\) are the zeroes of the cubic polynomial 4x³ – 11x² – 19x – 4 and check. (AP-SA-II: 2016)
Solution:
Given polynomial is 4x³ – 11x² – 19x – 4
p(x) comparing the given polynomial with
ax³ + bx² + cx + d, we get
a = 4, b = – 11, c = -19, d = – 4
Further p(4) = 4(4)³ – 11(4)² – 19(4) – 4
= 4 × 64 – 11(16) – 19(4) – 4
= 256 – 176 – 76 – 4
= 256 – 256 = 0
p(-1) = 4(-1)³ – 11(-1)² – 19(-1) – 4
= 4(-1) – 11(1) + 19-4
= -4 – 11 + 19 – 4
= -19 + 19 = 0


Therefore 4, -1, –\(\frac{1}{4}\) are the zeroes of
4x³ – 11x² – 19x – 4.
We take α = 4, β = – 1 and γ = –\(\frac{1}{4}\)
now α + β + γ = 4 – 1 – \(\frac{1}{4}\) = 3 – \(\frac{1}{4}\) = \(\frac{11}{4}\)
= \(\frac{-(-11)}{4}\) = –\(\frac{b}{a}\)

Question 20.
Draw the graph of the polynomial x² + x – 6 and mark the zeroes of the polynomial
Solution:
Given that y = x² + x – 6


Scale : On X-axis :1 cm = 1 unit
On Y-axis :1 cm = 1 unit
Zeroes of the polynomial x² + x – 6 are the x – coordinates of the points on the x – axis.
-3 and 2 are the zeroes of the given polynomial on graph.

Question 21.
If α, β are zeroes of the polynomial 2x² + 7x + 5, find the value of α + β + αβ? (AP New SCERT Model Paper) Solution:
2x² + 7x + 5
α + β = \(\frac{-b}{a}\) = \(\frac{-7}{2}\)
α.β = \(\frac{\mathrm{c}}{\mathrm{a}}\) = \(\frac{5}{2}\)
∴ α + β + α.β = \(\frac{-7}{2}\) + \(\frac{5}{2}\) = \(\frac{-2}{2}\) = -1

Question 22.
If a, b and c are the zeroes of a polynomial of degree 3, then give the relations between the zeroes and the coefficients of the polynomial. (AP New SCERT Model Paper)
Solution:
α + β + γ = \(\frac{-b}{a}\)
αβ + βγ + γα = \(\frac{c}{a}\)
αβγ = \(\frac{-d}{a}\)

Question 23.
Solve the quadratic polynomial x² – 3x – 4 graphically. (AP New SCERT Model Paper)
Solution:
Let y = x² – 3x – 4

o.p. = (-2, 6), (-1, 0), (0, -4), (1, -6), (2, -6) (3, -4), (4, 0), (5, 6)
Scale : On X-axis 1 cm = 1 unit
On Y-axis 1 cm = 1 unit
The graph of y = x² – 3x – 4 intersects X-axis at (-1, 0) and (4, 0).
Hence the zeroes of p(x) are -1 and 4.