Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Determinants Solutions Exercise 4b
I.
Question 1.
Find area of the triangle with vertices (1, 0), (6, 0), (4, 3)
Solution:
Area of the triangle with vertices A(x1, y1) = (1, 0), B(x2, y2) = (6,0), C(x3, y3) = (4, 3) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
1 & 0 & 1 \\
6 & 0 & 1 \\
4 & 3 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[1(0 – 3) – 0(6 – 4) + 1(18 – 0)]|
= \(\frac{1}{2}\)|[-3 + 18]|
= \(\frac{1}{2}\)[15] = \(\frac{15}{2}\) Sq.units
Question 2.
Find area of the triangle with vertices (2, 7), (1, 1), (10,8)
Solution:
Area of the triangle with vertices A(x1, y1) (2, 7), B(x2, y2) (1, 1), C(x3, y3)= (10, 8) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[2(1 – 8) – 7(1 – 10) + 1(8 – 10)]|
= \(\frac{1}{2}\) |[2(-7) – 7(-9) + 1(-2)]|
= \(\frac{1}{2}\)|-14 + 63 – 2| = \(\frac{1}{2}\)[47]
= \(\frac{47}{2}\) Sq.units
Question 3.
Find area of the triangle with vertices (-2, -3), (3, 2), (-1, -8)
Solution:
Area of the triangle with vertices A(x1, y1) (-2, -3), B(x2, y2) (3, 2), C(x3, y3)= (-1, -8) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)]|
= \(\frac{1}{2}\) |[-2(10) + 3(4) + 1(-22)]|
= \(\frac{1}{2}\)|[-20 + 12 – 22]| = \(\frac{1}{2}\)|-30|
= 15 Sq.units
Hence, area of the triangle is 15 Sq. units
Question 4.
Show that points A (a, h + c), B (b, c + a), C (c, a + b) are collinear.
Answer:
Area of the triangle with vertices A(x1, y1) = (a, b + c), B(x2, y2) (b,c + a), C (x3, y3) (c, a + b) is given by (We apply row operations to simplify easily)
∆ = \(=\frac{1}{2}\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
a & b+c & 1 \\
b-a & a-b & 0 \\
c-a & a-c & 0
\end{array}\right| \)
R2 → R2 – R1
R3 → R3 – R1
= \(\frac{1}{2}\)(a – b) (c – a) \(\left|\begin{array}{ccc}
a & b+c & 1 \\
-1 & 1 & 0 \\
1 & -1 & 0
\end{array}\right|\)
= \(\frac{1}{2}\)(a – b)(c – a)| (-1)(-1) – (1)(1)| = \(\frac{1}{2}\)(a – b)(c – a)(0) = 0
Thus, the area of the triangle formed by the given points is zero.
Hence, the given 3 points are collinear.
Question 5.
Find the values of k if area of triangle is 4 sq. units and vertices are (k, 0). (4, 0), (0, 2).
Solution:
Area of ∆ ABC with vertices A(x1, y1) (k, 0), B(x2, y2) = (4, 0), C(x3, y3) = (0, 2) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
\mathrm{x}_1 & \mathrm{y}_1 & 1 \\
\mathrm{x}_2 & \mathrm{y}_2 & 1 \\
\mathrm{x}_3 & \mathrm{y}_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
\mathrm{k} & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) |[k(0 – 2) – 0(4 – 0) + 1(8 – 0)]
= \(\frac{1}{2}\)|[-2k + 8]| = |-k + 4|
∴ |-k + 4| = 4 ⇒ -k + 4 = ± 4
-k + 4 = 4 ⇒ k = 4 + 4 = 8
when -k + 4 = 4 ⇒ k = 0
∴ k = 0, 8
Question 6.
Find values of k if area of triangle is 4 sq. units and vertices are (-2, 0), (0, 4), (0, k)
Solution:
Area of ∆ABC with vertices A(x1, y1) = (-2, 0), B(x2, y2) = (0, 4), C(x3, y3) = (0, k) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
\mathrm{x}_1 & \mathrm{y}_1 & 1 \\
\mathrm{x}_2 & \mathrm{y}_2 & 1 \\
\mathrm{x}_3 & \mathrm{y}_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
-2 & 0 & 1 \\
0 & 4 & 1 \\
0 & \mathrm{k} & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[-2(4 – k)]| = |k – 4|
∴ |k – 4| = 4 ⇒ k + 4 = ±4
When k – 4 = 4 ⇒ k = 4 + 4 = 8
When k – 4 = -4 ⇒ k = 0
∴ k = 0, 8
Question 7.
Kind equation of line joining (1, 2) and (3, 6) using determinants.
Solution:
Let P(x, y) be a point on the line joining points and A (x1, y1) = (1, 2) and B(x2, y2) = (3, 6).
Then, the points A,B and P are collinear.
Hence, the area of triangle ABP is zero.
∴ ∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 6 & 1 \\
x & y & 1
\end{array}\right|=\) = 0
⇒ \(\frac{1}{2}\)[1(6 – y) – 2(3 – x) + 1(3y – 6x)] = 0
⇒ 6 – y – 6 + 2x + 3y – 6x = 0
⇒ 2y – 4x = 0 ⇒ y = 2x
∴ The equation of the line joining the given points is y = 2x.
Question 8.
Kind equation of line joining (3, 1) and (9, 3) using determinants.
Solution:
Let P(x, y) be a point on the line joining points and A (x1, y1) = (3, 1) and B(x2, y2) =(9, 3).
Then, the points A,B and P are collinear.
Hence, the area of triangle ABP will be zero.
∴ ∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
3 & 1 & 1 \\
9 & 3 & 1 \\
x & y & 1
\end{array}\right|\) = 0
⇒ \(\frac{1}{2}\) |[3(3 – y) – 1(9 – x) + 1(9y – 3x)] = 0
⇒ 9 – 3y – 9 + x + 9y – 3x = 0
⇒ 6y – 2x = 0
⇒ x – 3y = 0
∴ The equation of the line joining the given points is x – 3y = 0