Question

Equation of two equal sides of a triangle are the lines $7x+3y-20=0$ and $3x+7y-20=0$ and the third side passes through the point $(-3,3)$, then equation of the third side can be-

• A. $x+2y=0$
• B. $x-y+6=0$
• C. $x-3=0$
• D. $y=-3$

Answer 1

The correct option is B $x-y+6=0$

Let the equation

$7x+3y-20=0$ ---- (i)

$y=-\frac{7}{3}x+\frac{20}{3}$

Hence the slope $m_1=-\frac{7}{3}$

and, $3x+7y-20=0$ ---- (ii)

$y=-\frac{3}{7}x+\frac{20}{7}$

Hence the slope $m_2=-\frac{3}{7}$

and let the third line be $(y-y_1)=m(x-x_1)$ ---- (iii)

where $(x_1,y_1)$ and $m$ is points of passing and slope respectively.

Let the angle between the line (i) and line (ii) on line (iii) be $\theta$

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

angle between (i) and (iii)

$\tan \theta =\left|\frac{m-m_1}{1+m{m}_1}\right|$

$\tan \theta =\left|\frac{m-\left(-\frac{7}{3}\right)}{1+m\left(-\frac{7}{3}\right)}\right|$ ----- (iv)

and between (ii) and (iii)

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

$\tan \theta =\left|\frac{m-\left(-\frac{3}{7}\right)}{1+m\left(-\frac{3}{7}\right)}\right|$ ---- (v)

Now equating (iv) and (v), because sides are equal.

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

$\frac{m-\left(-\frac{3}{7}\right)}{1+m\left(-\frac{3}{7}\right)}=\frac{m-\left(-\frac{7}{3}\right)}{1+m\left(-\frac{7}{3}\right)}$

$\frac{7m+3}{7-3m}=\frac{3m+7}{3-7m}$

$\left(7m+3\right)\left(3-7m\right)=\left(3m+7\right)\left(7-3m\right)$

$m=1$

Now from (iii), equation of third side

$y-3=1(x-(-3\left)\right)$

$y-3=x+3$

$x-y+6=0$