Question

The area of the triangle formed by the lines $x+y=3$ and angle bisectors of the pair of straight lines $x^2-y^2+2y=1$ is?

Answer 1

Step 1: Finding the vertices of the triangle :

The given lines are

$x^2-y^2+2y=1$

$$\begin{gathered} \Rightarrow x^2=y^2-2y+1 \\ \Rightarrow x^2={\left(y-1\right)}^2 \\ \Rightarrow x^2-{\left(y-1\right)}^2=0 \end{gathered}$$

The given lines are $x–y+1=0$ and $x+y–1=0$whose angular bisectors are

$\left[\frac{x+y-1}{\sqrt{2}}\right]=\pm \left[\frac{x-y+1}{\sqrt{2}}\right]$

$y–1=0$ and $x=0$

Equation of angle bisectors of these lines $x=0$ and $y=1$

Thus, the vertices of the triangle are $(0,1),(0,3)$ and$(2,1)$.

Step 2: Finding the area :

We know that,

$Area=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

$\begin{array}{rcl}& =& \frac{1}{2}[0(3–1)+0(1–1)+2(1–3\left)\right]\\ & =& \frac{1}{2}\left(0+0-4\right)\\ & =& -2\end{array}$

Since the area is always positive. It can't be negative.

Thus, Area $=2squnits$

Hence, the area of the triangle is $2squnits$