Question

The area of a triangle is $5sq.unit$. If two vertices of the triangle are $(2,1),(3,-2)$ and the third vertex is $(x,y)$ where $y=x+3$, then find the coordinates of the third vertex.

Answer 1

Given that,

Two vertices of two triangle are $Q(2,1)$ and $R(3,-2)$.

Let the third vertex be $P(x,y)$

Area of $△PQR=5sq.units$

Area of $△PQR=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$5=\frac{1}{2}\left[x\right(1-(-2))+2(-2-y)+3(y-1\left)\right]$

$10=3x-4-2y+3y-3$

$3x+y-7=10$

$3x+y-17=0.....\left(1\right)$

Given that the equation

$y=x+3.....\left(2\right)$

Put the value of $y$ in equation (1) and (2) we get,

$3x+y-17=0$

$3x+x+3-17=0$

$4x-14=0$

$4x=14$

$x=\frac{14}{4}$

$x=\frac{7}{2}$

Put the value of $x$ in equation (2)

$y=x+3$

$y=\frac{7}{2}+3$

$y=\frac{13}{2}$

Hence, point $P(x,y)=(\frac{7}{2},\frac{13}{2})$ the third vertex.