Question

Find the values of k so that the area of the triangle with vertices (1, -1), (-4,2k) and (-k, – 5) is 24 sq. units

Answer 1

Step-by-step explanation:

We need to find the value of k so that the triangle with vertices (1, -1), (-4,2k) and (-k, – 5) is 24 sq. units.

The area of triangle is given by :

$A=\dfrac{1}{2}\times [x_1(y_2-y_3)-x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here,

x₁ = 1, y₁ = -1, x₂ = -4, y₂ = 2k, x₃ = -k and y₃ = -5

So,

$24=\dfrac{1}{2}\times [1(2k-(-5))-(-4)((-5)-(-1))+(-k)((-1)-(2k))]$

We need to solve the above equation,

$48=[1(2k-(-5))-(-4)((-5)-(-1))+(-k)((-1)-(2k))]$

On solving the above equation we get the value of k as :

k = 4.7