Question

The area of triangle ABC whose vertices are (3, –5), (–2, k) and (1, 4) are 33/2 sq. unit find the value of k .

Answer 1

Answer:

k =16 is the answer for the given problem

Answer 2

Given :

• Coordinates of A = ( 3 , - 5 )

• Coordinates of B = ( - 2 , k )

• Coordinates of C = ( 1 , 4 )

• Area of triangle = 33/2 unit²

To Find :

• Vale of k

Solution :

$\large\sf area = \dfrac{1}{2} \bigg[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg]$

$\sf \implies \frac{33}{2} = \dfrac{1}{2} \bigg[3(k - 4) - 2(4 + 5) + 1( - 5 -k) \bigg] \\ \\\sf \implies \frac{33}{2} - \dfrac{1}{2} = 3(k - 4) - 2(9) + 1( - 5 -k) \\ \\\sf \implies \frac{33 - 1}{2} = 3k - 12 - 18 - 5 - k \\ \\\sf \implies 16 = 2k - 35 \\ \\\sf \implies 2k = 16 + 35 \\ \\\sf \implies 2k = 51\\ \\ \sf \implies k = \frac{51}{2}$