Chapter 4 Determinants Ex 4.3 Solutions
Question - 1 : - Find area of thetriangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4,3) (ii) (2, 7), (1, 1), (10, 8)
(iii) (−2, −3), (3, 2),(−1, −8)
Answer - 1 : -
(i) The area of thetriangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,
(ii) The area of thetriangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
(iii) The area of thetriangle with vertices (−2, −3), (3, 2), (−1, −8)
is given by therelation,
Hence, the area of thetriangle is
Question - 2 : - Show that points
are collinear
Answer - 2 : -
Area of ΔABC is given bythe relation,
Thus, the area of thetriangle formed by points A, B, and C is zero.
Hence, the points A, B,and C are collinear.
Question - 3 : - Find values of k ifarea of triangle is 4 square units and vertices are
(i) (k, 0), (4,0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)
Answer - 3 : -
We know that the area ofa triangle whose vertices are (x1, y1), (x2, y2), and
(x3, y3) is the absolutevalue of the determinant (Δ), where
It is given that thearea of triangle is 4 square units.
∴Δ = ± 4.
(i) The area of thetriangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
Δ =
∴−k + 4 = ± 4
When −k + 4= − 4, k = 8.
When −k + 4= 4, k = 0.
Hence, k =0, 8.
(ii) The area of thetriangle with vertices (−2, 0), (0, 4), (0, k) is given by therelation,
Δ =
∴k − 4 = ± 4
When k −4 = − 4, k = 0.
When k −4 = 4, k = 8.
Hence, k =0, 8.
Question - 4 : - (i) Find equation ofline joining (1, 2) and (3, 6) using determinants
(ii) Find equation ofline joining (3, 1) and (9, 3) using determinants
Answer - 4 : -
(i) Let P (x, y)be any point on the line joining points A (1, 2) and B (3, 6). Then, the pointsA, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
Hence, the equation ofthe line joining the given points is y = 2x.
(ii) Let P (x, y)be any point on the line joining points A (3, 1) and
B (9, 3). Then, thepoints A, B, and P are collinear. Therefore, the area of triangle ABP will bezero.
Hence, the equation ofthe line joining the given points is x − 3y = 0.
Question - 5 : - If area of triangle is35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
A. 12 B. −2 C. −12,−2 D. 12, −2
Answer - 5 : -
The area of the trianglewith vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
It is given that thearea of the triangle is ±35.
Therefore, we have:
When 5 − k =−7, k = 5 + 7 = 12.
When 5 − k =7, k = 5 − 7 = −2.
Hence, k =12, −2.
The correct answer is D.