Question

Find the area of the triangle formed by the lines $y-x=0,x+y=0$ and $x-k=0$

Answer 1

The equation of the given lines are
$y-x=0$....(1)
$x+y=$...(2)
$x-k=0$...(3)
The point of intersection of lines (1) and (2) is given by
$x=0$ and $y=0$
The point of intersection of lines (2) and (3) is given by
$x=k$ and $y=-k$
The point of intersection of lines(3) and (1) is given by
$x=k$ and $y=k$
Thus, the vertices of the triangle formed by the three given lines are $(0,0),(k,-k),$ and $(k,k)$

We know that the area of the triangle whose vertices are $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ is $\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Therefore, area of the triangle formed by the three given lines,

$=\frac{1}{2}\left|0\right(-k-k)+k(k-0)+k(0+k\left)\right|$square unts

$=\frac{1}{2}|k^2+k^2|$square units

$=\frac{1}{2}\left|2k^2\right|=k^2$ square units