Question

Two sides of isosceles triangle are given by the equations 7x-y+3=0and x+y-3=0 and the third side passes through thepoints (1,10).determine the equation of the third side

Answer 1

here is your answer.................

Answer 2

Answer:

The equations are

$3x+y-7=0$

$x-3y+31=0$

Step-by-step explanation:

Given Two sides of isosceles triangle are given by the equations 7x-y+3=0and x+y-3=0 and the third side passes through the points (1,10).

We have to find the equation of the third side.

Slope intercept form is

y=mx+c

Slope of 7x-y+3=0 is 7

Slope of x+y-3=0 is -1

Let slope of third line i.e BC is m

Since ΔABC is isosceles therefore angle on the bases are equal.

⇒ $\tan \angle B=\tan \angle C$

As, $\tan \theta=|{\frac{m_2-m_1}{1+m_1.m_2}}|$

⇒ $|{\frac{7-m}{1+7m}}|=|{\frac{m-(-1)}{1+(-1)m}}|$

⇒ $\frac{7-m}{1+7m}=\pm(\frac{m+1}{1-m})$

$7-8m+m^2=7m^2+8m+1$ and $7-8m+m^2=-(7m^2+8m+1)$

$6m^2+16m-6=0$ and $8m^2+8=0$

$(m+3)(3m-1)=0$ and $m^2=-1$

$m=-3,\thinspace m=\frac{1}{3}$

$m^2=-1$ not possible

Let m=-3, then equation of third side is

$y-10=-3(x-1)$

⇒ $3x+y-7=0$

Let $m=\frac{1}{3}$, then equation of third side is

$y-10=\frac{1}{3}(x-1)$

⇒ $x-3y+31=0$