 # Find the area of the triangle formed by the lines Question:

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

Solution:

The equations of the given lines are

y – x = 0 … (1)

x + y = 0 … (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 (kk).

We know that the area of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$ is $\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$.

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

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

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

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

$=k^{2}$ square units