Question

If the area of the triangle formed by the vertices A(-1,2) , B(k ,-2) and C(7, 4) (taken in order) is 22 sq. units, find the value of k

Answer 1

Solution :-

Given :-

The area of the triangle formed by the vertices A ( -1 , 2 ), B ( k , -2 ) and C ( 7 , 4 ) is 22 square units.

We know :-

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

Here :-

$\bullet \sf \ x_1=-1 \ , \ y_1= 2 \\\\ \bullet \ x_2= k \ , \ y_2= -2 \\\\\bullet \ x_3 = 7 \ , \ y_3 = 4$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -1(-2-4)+k(4-2)+7(2-(-2)) \ ] = 22 \\\\\longrightarrow \dfrac{1}{2} \times [ \ (-1\times - 6 )+2k+( 7 \times 4 ) \ ] = 22 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 6+2k+28 \ ] = 22 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 34+2k \ ] = 22 \\\\\longrightarrow 34+2k = 44 \\\\\longrightarrow 2k = 10 \\\\\longrightarrow\bf k = 5$

Value of k = 5