Question

Find the area of the triangle formed by the lines $Y-X=0,X+Y=0$ and $X-K=0$

Answer 1

Step 1: Solve for vertices of triangle

The given equation of the lines are:

$y-x=0$ …$\left[1\right]$

$x+y=0$ …$\left[2\right]$

$x-k=0$ …$\left[3\right]$

Solving equations $\left[1\right]$ and$\left[2\right]$, we get $x=0$ and $y=0$

Solving equations $\left[3\right]$ and $\left[1\right]$, we get $x=k$ and $y=k$

Solving equations $\left[2\right]$ and $\left[3\right]$, we get $x=k$ and $y=-k$

$\therefore$Vertices of triangle are $\left(0,0\right),\left(k,k\right)$ and $\left(k,-k\right)$

Step 2: Solve for area of triangle

Let the vertices of triangle be,

$$\begin{gathered} A\left(x_1,y_1\right)=\left(0,0\right) \\ B\left(x_2,y_2\right)=\left(k,k\right) \\ C\left(x_3,y_3\right)=\left(k,-k\right) \end{gathered}$$

Area of triangle$\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$\therefore$Area of $△ABC=\frac{1}{2}\left|0\left(k+k\right)+k\left(-k-0\right)+k\left(0-k\right)\right|$

$=\frac{1}{2}\left|0-k^2-k^2\right|=k^{2}sq.units$

Hence, area of the given triangle is $k^{2}sq.units$