Question

If the area of Δ ABC with vertices A(3, 1), B(−2, 1) and C(0, k) is 5 sq. units, then values of k are ______.

विकल्प

• 3, 1

• −1, 3

• −1, 2

• 0, 2

Answer 1

If the area of Δ ABC with vertices A(3, 1), B(−2, 1) and C(0, k) is 5 sq. units, then the values of k are −1 and 3.

Explanation:

Use the determinant formula for the area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):

Area = `1/2 |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|`

Vertices: A(3, 1), B(−2, 1) and C(0, k)

Area: 5 sq. units

5 = `1/2 |3(1 - k) + (-2)(k - 1) + 0(1 - 1)|`

5 = `1/2 |3 - 3k - 2k + 2 + 0|`

5 = `1/2 |5 - 5k|`

10 = |5 - 5k|   ...[Multiply both sides by 2]

The expression inside the absolute value can be either 10 or −10

Case 1:

5 − 5k = 10

−5k = 10 − 5

−5k = 5

k = `5/-5`

k = −1

Case 2:

5 − 5k = −10

−5k = −10 − 5

−5k = −15

k = `(−15)/(−5)`

k = 3

The values of k are −1 and 3.