Practice AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines MCQ to identify your strengths and weak areas.
AP Inter 1st Year Maths Straight Lines MCQ
Question 1.
The equation of the straight line making an intercept of 3 units on the Y -axis and inclined at 45° to the X -axis is
1) y = x – 1
2) y = x + 3
3) 45x + 3
4) y = x + 45
Answer:
2) y = x + 3
Explanation:
θ = 45° (inclined) m = tan 45° = 1;c = 3 ⇒ y = x + 3
Question 2.
The angle made by the line segment joining (5, 2), (6, -15) at (0, 0) is
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{4}\)
3) \(\frac{\pi}{2}\)
4) π
Answer:
3) \(\frac{\pi}{2}\)
Explanation:
Let O(0, 0), P(5, 2), Q(6, -15)
m1 = slope of OP = \(\frac{2}{5}\), m2 = slope of OQ = \(\frac{-15}{6}=\frac{-5}{2}\), m1m2 = -1 ⇒ θ = \(\frac{\pi}{2}\)
Question 3.
The lines 2x + 3y = 6, 2x + 3y = 8 cut the X – axis at A, B respectively. A line T drawn through the point (2, 2) meets the X- axis at ‘C’ in such a way that abscissae A,B,C are in arithmetic progression then the equation of the line T is
1) 2x + 3y = 10
2) 3x + 2y = 10
3) 2x – 3y = 10
4) 3x – 2y = 10
Answer:
1) 2x + 3y = 10
Explanation:
A = (3, 0), B = (4, 0) and let C = (a, 0)
⇒ \(\frac{a+3}{2}\) = 4
⇒ a = 5 (∵ A, B, C are in A.P)
∴ Equation of l is y – 2 = \(\frac{-2}{3}\) (x – 2)
Question 4.
The area (in sq.m) of the triangle formed by the lines x = 0; y = 0 and 3x + 4y = 12 is
1) 3
2) 4
3) 6
4) 12
Answer:
3) 6
Explanation:
\(\frac{C^2}{2|a b|}=\frac{144}{24}\) = 6
Question 5.
The area of the triangle formed by the axes and the line x cos a + y sec a = 2
1) 4
2) 3
3) 2
4) 1
Answer:
3) 2
Explanation:
\(\frac{\mathrm{C}^2}{2|\mathrm{ab}|}=\frac{4}{2|\cos \alpha \cdot \sec \alpha|}\) = 2
Question 6.
If the area of the triangle formed by the lines x =0, y = 0; 3z + 4y = a(a > O) is 1; then a =
1) √6
2) 2√6
3) 4√6
4) 6√2
Answer:
2) 2√6
Explanation:
\(\frac{1}{2}\) = 1
⇒ a = ± \(\sqrt{24}\)
⇒ a = 2√6
Question 7.
EquatIon of the straight line passing through the Image of (3. 6) in the line 2x – y = 5 and perpendlcular to the line 3z +4y = 15 is
1) 4x -3y = 10
2) 4x – 3y = 7
3) 4x – 3y = 5
4) 4x – 3y = 16
Answer:
4) 4x – 3y = 16
Explanation:
Image of (3, 6) is (7, 4)
⇒ Equation of straight line y – y1 = m(x – x1) ⇒ 4x – 3y – 16 = 0.
Question 8.
A straight line passing through the point A (3, 4) is such that its Intercept. between the axes bisected at A its equation is
1) 4x+ 3y = 24
2) 3x + 4y = 25
3) x + y = 7
4) 3x – 4y + 7 = 0
Answer:
1) 4x+ 3y = 24
Explanation:
\(\frac{x}{x_1}+\frac{y}{y_1}\) = 2
⇒ \(\frac{x}{3}+\frac{y}{4}\) = 2
⇒ 4x + 3y = 24
Question 9.
If the straight lines y = 4 – 3x; ay = x + 10 and 2y + bx +9 = 10 represet the three consective sides of a rectangle, then ab =
1) 18
2) -3
3) \(\frac{1}{2}\)
4) –\(\frac{1}{3}\)
Answer:
1) 18
Explanation:
m1 = -3, m2 = \(\frac{1}{a}\), m3 = –\(\frac{b}{2}\)
m1m2 = -1
⇒ \(\frac{-3}{a}\) = 1
⇒ a = 3;
⇒ m2 × m3 = -1
\(\left(\frac{1}{3}\right)\left(\frac{-\mathrm{b}}{2}\right)\) = -1
⇒ b = 6
∴ ab = (3) (6) = 18
Question 10.
The ends of a rod of length ‘1’ moves on the co-ordinate axes the locus of the point on the rod which divides It In the ratIo 1 : 2 is ( 3 )
1) 36x2 + 9y2 = 4l2
2) 36x2 + 9y2 = l2
3) 9x2 + 36y2 = 4l2
4) 9x2 + 36y2 = l2
Answer:
3) 9x2 + 36y2 = 4l2
Explanation:
Let A(a, 0), B(0, b) be the end points of the rod.
AB = 1 ⇒ a2 + b2 = 12 …………… (1)
Let (α, β) divides \(\overrightarrow{\mathrm{AB}}\) in the ratio 1: 2.
∴ (α, β) = \(\left(\frac{2 \mathrm{a}}{3}, \frac{\mathrm{~b}}{3}\right)\)
⇒ a = \(\frac{3 \alpha}{2}\), b = 3β
(1) ⇒ \(\left(\frac{3 \alpha}{2}\right)\)2 + (3β)2 = l2
∴ Locus of P(a, 13) is 9x2 + 36y2 = 4l2
Question 11.
Two equal sldes of an isoceles triangle are 7x – y + 3 = 0, x + y – 3 = 0 and its third side passes through the point (1, -10) the equation of the third side is
1) 3x + y + 7 = 0
2) x – 3y + 29 = 0
3) 3x + y + 3 = 0
4) 3x + y – 3 = 0
Answer:
1) 3x + y + 7 = 0
Explanation:
By verification (1, -10) satisfies 3x + y + 7 = 0
Question 12.
The perpendicular distance from (1, 2) to this straight line 12x + 5y = 7 is
1) \(\frac{15}{13}\)
2) \(\frac{12}{13}\)
3) \(\frac{5}{13}\)
4) \(\frac{7}{13}\)
Answer:
1) \(\frac{15}{13}\)
Explanation:
\(\frac{|12+10-7|}{13}=\frac{15}{13}\)
Question 13.
If p and q are the perpendicular distances from the origin to the straight lines x sec θ – y cosec θ = a and x cos θ + y sin θ = a cos 2θ then
1) 4p2 + q2 = a2
2) p2 + q2 = a2
3) p2 + 2q2 = a2
4) 4p2 + q2 = 2a2
Answer:
1) 4p2 + q2 = a2
Explanation:
Result only 4p2 + q2 = a2
Question 14.
The equation of the line parallel to 5x + 12y = 1 and at distance of 3 units from (1, 0) is
1) 5x + 12y + 34 = 0
2) 12x – 5y + 34 = 0
3) 12x – 5y – 44 = 0
4) 5x + 12y + 44 = 0
Answer:
1) 5x + 12y + 34 = 0
Explanation:
By verification (1) and (4) are parallel to the given line
(1) ⇒ \(\frac{|5+0+34|}{13}=\frac{39}{13}\) = 3
⇒ 5x + 12y + 34 = 0 is parallel line which is at a distance of 3 units.
Question 15.
The area (in sq. units) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is
1) 4π
2) 3π
3) 2π
4) π
Answer:
4) π
Explanation:
r = \(\frac{1}{2}\left(\frac{\left|\mathrm{c}_1-\mathrm{c}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right)=\frac{1}{2}\left(\frac{|-15+5|}{5}\right)=\frac{1}{2}\)(2) = 1
Area = π(1)2 = π
Question 16.
If A (2, -1) and B (6, 5) are two points, the ratio in which the foot of the perpendicular from (4, 1) to AB divides It in
1) 8 : 15
2) 5 : 8
3) -5 : 8
4) -8 : 5
Answer:
2) 5 : 8
Explanation:
Slope of AB = and Slope of the line ⊥r to \(\overline{\mathrm{AB}}=-\frac{2}{3}\)
Equation is y – 1 = (x – 4) ⇒ 2x + 3y – 11 = 0
Ratio = \(\frac{-(4-3-11)}{(12+15-11)}=\frac{10}{16}=\frac{5}{8}\) = 5 : 8
Question 17.
A straight line which makes equal intercepts on positive X and Y axes and which is at a distance 1 unit from the origin intersects the straight line
y = 2x + 3 + √2 at (x0, y0) then 2x0 + y0 =
1) 3 + √2
2) √2 – 1
3) 1
4) 0
Answer:
2) √2 – 1
Explanation:
Let 1 be the line y- a.
Perpendicular distance from origin is 1 ⇒ \(\frac{|\mathrm{a}|}{\sqrt{2}}\) = 1 a = √2
Now equation is x + y = √2 …………. (1) and Given line y = 2x + 3 + √2 …………. (2)
From (1) and (2) we get x = -1 and y = √2 + 1;
x0 = -1, y0 = √2 + 1 2x0 + y0 = -2 + √2 = √2 – 1
Question 18.
If a, b, c are in AP, the line ax + by + c = 0 passes through the fixed point
1) (1, 2)
2) (-1, 2)
3) (1, -2)
4) (-1, -2)
Answer:
3) (1, -2)
Explanation:
If a, b, c are in AP then \(\frac{\mathrm{a}+\mathrm{c}}{2}\) = b
a – 2b + c = 0 →, ax + by + c = 0; (x, y) = (1, -2)
Question 19.
If a, b, c are in AP then the line ax + by + c = 0
1) passes through a fixed point
2) form an equilateral triangle
3) form a rhombus
4) form a square
Answer:
1) passes through a fixed point
Explanation:
From (18) the line passing through fixed point.
Question 20.
The point on the line 2x – 3y = 5 which is equidistant from (1, 2) and (3, 4) is
1) (2, 3)
2) (4, 1)
3) (1, -1)
4) (4, 6)
Answer:
2) (4, 1)
Explanation:
By verification option (2) and (3) satisfies the 2x – 3y = 5 and distance from (4, 1) is
\(\sqrt{9+1}=\sqrt{10}\) and \(\sqrt{1+9}=\sqrt{10}\)
Option (2) satisfies.