Question 10

We can find the coordinates of point $B$ by solving the given equations of lines.

Solving them we can find that $B (-3,-2)$ is the vertex of triangle.

Slope of line $AB = \dfrac{1}{5}$

Now we can find the angle between the line $AB$ and angle bisector of angle $B$ .

Slope of angle bisector = $\dfrac{3}{2}$ . $\tan\theta =\Bigg| \dfrac{\dfrac{1}{5} - \dfrac{3}{2}}{1+\dfrac{1.3}{2.5}}\Bigg|$ $\tan\theta = 1$ Hence angle $B$ is right angled and equation of $BC$ is line perpendicular to $AB$ and passing through $(-3,-2)$ .

Hence slope of $BC = -5$ .

Hence equation of $BC$ is :- $y+2 = -5(x+2)$ $5x+y+17=0$

Let a point $D(x,y)$ on $BC$ be the mirror-image of point $A(2,-1)$ across the angle-bisector of $\angle B$ given by the equaton : $3x-2y+5=0$ ,

Clearly $AD$ is bisected by the line $3x-2y+5=0$ . Let the Mid-point be at point $O$ .

Therefore, the co-ordinates of point $O$ are $O(\dfrac{x+1}{2},\dfrac{y-1}{2})$ . Also, $AD$ is perpendicular to line $3x-27y+5=0$ . From these two properties we can derive 2 equations:

$\text{Equation 1:}$ Since $O$ lies on $3x-2y+5=0$ , we have, $3(\dfrac{x+2}{2})-2(\dfrac{y-1}{2})+5=0$

$\text{Equation 2:}$ Clearly slope of line $OD$ is $m_{1}= \dfrac {y+1}{x-2}$ . Also slope of line $3x-2y+5=0$ is $m_{2}=\dfrac{3}{2}$

Therefore, $m_{1}\times m_{2}= \dfrac{3(y+1)}{2(x-2)}= (-1)$

Hence solving the 2 simultaneous linear equations , we get $x=-4, y=3\ or D(-4,3)$ Now , the equation of the altitude from $B$ is $7x-10y+1=0$ . So $BC$ has an equation of the form, $(3x-27+5 )+ \lambda(7x-10y+1)=0......(iii)$ for some constant $\lambda$ , or , $(3+7\lambda)x -(2+10\lambda)y+ (5+\lambda)=0$

Since $D$ lies on this line we can replace $x=-4, y=3$ , which gives $\lambda=\dfrac{-13}{57}$

Plugging the value of $\lambda$ in $Eq(iii)$ , we get the equation of $BC$ as : $5x+y+17=0 :)$

Use the fact image of any vertex of triangle with respect to angle bisectors always lies on opposite side. From this you will get a point lying on BC And other point is point of intersection of angle bisector and altitude. using two points we can find equation of line

The two lines pass through $B$ are:

$\begin{cases} 7x-10y+1 = 0 & \Rightarrow y = 0.7x+0.1 \\ 3x-2y+5 = 0 & \Rightarrow y = 1.5x + 2.5 \end{cases} \\ \Rightarrow 0.8x + 2.4 = 0 \quad \Rightarrow x = -3\quad \Rightarrow y = 0.7(-3)+0.1 = -2 \\ \Rightarrow B(-3,-2)$

Now let $\angle CBA = 2 \theta$ . This $\theta$ is also the angle between the angle-bisector $y = 1.5x+2.5$ and line $BA$ . Let the tangent of $y=1.5x+2.5$ and line $BA$ be $\tan{\alpha}$ and $\tan{\beta}$ respectively. Then, we have:

$\begin{cases} \tan{\alpha} = 2.5 \\ \tan{\beta} = \dfrac {y_B-y_A}{x_B-x_A} = \dfrac {-2+1}{-3-2} = 0.2\end{cases}$

Now, we have:

$\theta = \alpha-\beta \\ \Rightarrow \tan{\theta} = \tan{(\alpha-\beta)} = \dfrac {\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}} = \dfrac{1.5-0.2}{1+(1.5)(0.2)} = \dfrac {1.3}{1.3} = 1 \\ \Rightarrow \theta = 45^\circ \quad \Rightarrow \angle CBA = 90^\circ$

Therefore, the line $BC$ is perpendicular to line $BA$ and its gradient is $-\dfrac {1}{0.2} = -5$

The equation of $BC$ is given by:

$\dfrac {y-y_B}{x-x_B} = \dfrac {y+2}{x+3} = - 5\quad \Rightarrow y+2 = 5x + 15 \quad \Rightarrow \boxed {5x+y+17=0}$