Question

Two equal sides of an isoceles triangle are given by $7x-y+3=0$ and $x+y-3=0$ and the third side passes through the point $(1,10)$ then the slope $m$ of the third side is given by

• A. $3m^2-1=0$
• B. $m^2+1=0$
• C. $3m^2+8m-3=0$
• D. $m^2-3=0$

Answer 1

The correct option is C $3m^2+8m-3=0$

For equation $BC$.
$AB$ and $AC$ are isoceles, so $AB=AC$ and $\angle ABC=\angle ACB$ slope of $AB\left(m_1\right)=7$ and slope of $AC\left(m_2\right)=-1$
Let slope of $BC=m$ $\because \angle ABC=\angle ACB$
$[\theta$ is angle between two lines]
So, slope is
$\left|\frac{m-m_1}{1+m{m}_1}\right|=\left|\frac{m-m_2}{1+{mm}_2}\right|$
Put $m_1=7,m_2=-1$
$\left|\frac{m-7}{1+7m}\right|=\left|\frac{m+1}{1-m}\right|$
$\Rightarrow \frac{m-7}{1+7m}=+\frac{m+1}{1-m}$ or $\frac{m-7}{1+7m}=-\frac{m+1}{1-m}$ $\Rightarrow m-m^2-7+7m=m+1+{7m}^2+7m$
$\Rightarrow -8m^2-8=0\Rightarrow m^2=-1$
Which is not possible
So, $\frac{m-7}{1+7m}=-\frac{m+1}{1-m}$
$\Rightarrow (m-7)(m-1)=(m+1)(1+7m)$ $\Rightarrow m^2-8m+7=7m^2+8m+1$
$\Rightarrow 6m^2+16m-6=0$
$\Rightarrow 3m^2+8m-3=0$