Question

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

Answer 1

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 (k, k).

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

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

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

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

$=k^2$square units