Question

The equations of two equal sides $AB$ and $AC$ of an isosceles $△ABC$ are $x+y=5$ and $7x-y=3$, respectively. Then the equations of the side $BC$, if area of $\left(\Delta ABC\right)=5$ unit${}^2$, can be

• A. $x-3y-1=0$
• B. $x-3y-21=0$
• C. $3x+y-2=0$
• D. $3x+y-12=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $x-3y-1=0$
B $x-3y-21=0$
C $3x+y-12=0$
D $3x+y-2=0$

Side BC will be perpendicular to the bisectors of the angle BAC. Now, equations of the bisectors of lines AB and AC are $\frac{(x+y-5)}{\sqrt{2}}=\pm \frac{(7x-y-3)}{5\sqrt{2}}$
$\Rightarrow x-3y+11=0or3x+y-7=0$
Let altitude through vertex A be AD.
If the equation of AD is $3x+y-7=0$ , then the equation of side BC be $x-3y+\lambda =0$
Let the angle between $AD$ and $AC$ be $\theta$
$tan\theta =|\frac{-3+1}{1+3}|=\frac{1}{2}$

Let $AD=\lambda ,$ then $\Delta ABC=\frac{1}{2}\times \left[BC\right]=\frac{1}{2}\times \lambda =2\lambda |tan\theta |={\lambda }^2|tan\theta |.$
Hence,
${\lambda }^2\times \frac{1}{2}=5\Rightarrow {\lambda }^2=10$
$\Rightarrow (11-\lambda {)}^2=100$
$\Rightarrow 11-\lambda =\pm \Rightarrow \lambda =1,21$.
Hence, equation of BC is $x-3y+1=0$ or $x-3y+21=0$.
Similarly, if equation of BC is $3x+y+k=0$
The equation of AD will be $x-3y+11=0$.
Hence,
$|tan\theta |=\left|\frac{7-\frac{1}{2}}{1+\frac{7}{3}}\right|=2$
$\Rightarrow {\lambda }^2|tan\theta |=2{\lambda }^2=5\Rightarrow {\lambda }^2=\frac{5}{2}$
Also, $\frac{5}{2}=\frac{(3+4+\lambda {)}^2}{10}$
$\Rightarrow 7+\lambda =\pm 5\Rightarrow \lambda =2,-12$
Hence, equation of BC is $3x+y+2=0or3x+y-12=0$. Finally, there are four possible equations of side BC, viz., $x-3y+1=0,x-3y-21=0,3x+y+2=0$ or $3x+y-12=0$.