Question

Let $ABC$,be a triangle such that the coordinates of the vertex $A$ are $(-3,1)$.Equation of the median through $B$ is $2x+y-3=0$ and equation of the angular bisector of $C$ is $7x-4y-1=0$.Find the slope of line $BC$.

Answer 1

Consider the coordinates of C as $\left(a,b\right)$, and since the point C lies on the angular bisector of C, therefore the equation becomes,

$7a-4b=1$ (1)

The midpoint of AC will be $\left(\frac{a-3}{2},\frac{b+1}{2}\right)$ and as the median lies on the midpoint of B, therefore the equation becomes,

$2\left(\frac{a-3}{2}\right)+\left(\frac{b+1}{2}\right)=3$

$2a-6+b+1=6$

$2a+b=11$ (2)

From equation (1) and (2),

$a=3,b=5$

Slope of AC is,

$m_1=\frac{5-1}{3-\left(-3\right)}$

$=\frac{2}{3}$

The slope of bisector is,

$m_2=-\frac{Coeff.ofx}{Coeff.ofy}$

$=\frac{7}{4}$

The angle between AC and the bisector will be,

$\tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{\frac{2}{3}-\frac{7}{4}}{1+\left(\frac{2}{3}\right)\left(\frac{7}{4}\right)}\right|$

$=\frac{1}{2}$

Let $m$ be the slope of BC.

The angle between BC and bisector will be equal to the angle between AC and bisector.

$\left|\frac{m-m_2}{1+m{m}_2}\right|=\frac{1}{2}$

$\left|\frac{m-\frac{7}{4}}{1+m\left(\frac{7}{4}\right)}\right|=\frac{1}{2}$

$\frac{4m-7}{7m+4}=\pm \frac{1}{2}$

$m=\frac{2}{3};m=18$

When the slope of BC is $\frac{2}{3}$, then A,B,C will be collinear and it will form no triangle.

Therefore, the slope of BC is 18.