Question

find k if the area of triangle whose vertices are A (4,k) , B (-5,-7) , C (-4,1) os 38 sq.units​

Answer 1

Answer:

$k=141$

Step-by-step explanation:

Given:-

Vertices of the triangle are A (4, $k$), B (-5, -7), C (-4, 1)

Area of the triangle ABC = 38 sq. units

To find:-

The value of $k$

Solution:-

$x_{1}=4, x_{2}=-5, x_{3}=-4, y_{1}=k,y_{2}=-7, y_{3}=1$

Area of the triangle ABC $=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$38=\frac{1}{2}[4(-7-1)+(-5(1-k))+(-4(k-(-7)))]$

$38=\frac{1}{2}[4(-8)-5+5k+(-4(k+7))]$

$38=\frac{1}{2}[-32-5+5k+(-4k-28)]$

$38=\frac{1}{2}[-37+5k-4k-28]$

$76=-65+k$

∴ $k=141$

∴ The value of $k$ is 141.