Question

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

Answer 1

Simplification of given data.

The equation of the given lines are
$y-x=0\dots \left(1\right)$
$x+y=0\dots \left(2\right)$
$x-k=0\dots \left(3\right)$

The point of intersection of lines $\left(1\right)$ and $\left(2\right)$:

From $\left(1\right)$, $y=x$

Putting it in $\left(2\right)$, $x+x=0\Rightarrow x=0$

Putting it in $\left(1\right)$, $y=0$

So, point of intersection is $(0,0)$

The point of intersection of lines $\left(2\right)$ and $\left(3\right)$:
From $\left(3\right)$, $x=k$
Putting it in $\left(2\right)$, $y+k=0\Rightarrow y=-k$
So, point of intersection is $(k,-k)$

The point of intersection of lines $\left(3\right)$ and $\left(1\right)$:
From $\left(3\right)$, $x=k$
Putting it in $\left(1\right)$, $y-k=0\Rightarrow y=k$
So, point of intersection is $(k,k)$

Thus, the vertices of the triangle formed by the three given lines are $(0,0),(k,-k),$ and $(k,k).$

Area of triangle
We know that area of triangle whose vertices are $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is

$=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Area of triangle $\Delta OAB$ whose vertices are $0(0,0),A(k,k)$ and $B(k,-k)$

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

$=\frac{1}{2}|0+k(-k)+k(-k)|$

$=\frac{1}{2}|-k^2-k^2|$

$=\frac{1}{2}|-2k^2|$

$=\frac{2k^2}{2}$

$=k^2\text{square units}$.

Final answer:
Therefore, required area of triangle is $k^2$ square units.