Question

The equation of two equal sides $AB$ and $AC$ of isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$, respectively. Then the equation of the side $BC$ if are of $△ABC=5{\text{unit}}^2$, is/are

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

Answer 1

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

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

Equation of the Angle bisector of the $AB$ and $AC$ will be perpendicular bisector of the $BC$
$\therefore$ eqution of the angle bisector is
$\frac{x+y-5}{\sqrt{2}}=\pm \frac{7x-y-3}{5\sqrt{2}}\Rightarrow \text{obtuse angle bisector}:x-3y+11=0;\text{acute angle bisector}:3x+y-7=0$
So equation of $BC$ is $3x+y+k=0;x-3y+l=0$
Coordinates of $A\equiv (1,4)$
Angle between lines is $\tan 2\theta =|\frac{7+1}{1-7}|=\pm \frac{4}{3}$
Acute angle between the bisector line and $AB$ is
for, $\tan 2\theta =\frac{4}{3}$
$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\Rightarrow \tan \theta =\frac{1}{2}$
for, $\tan 2\theta =-\frac{4}{3}$
$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\Rightarrow \tan \theta =2$
Let the length of the perpendicular is $x$ then
Area of $△ABC=2\times \frac{1}{2}x\times x\tan \theta$
For $\tan \theta =2\Rightarrow x=\sqrt{\frac{5}{2}}$
For $\tan \theta =\frac{1}{2}\Rightarrow x=\sqrt{10}$
Distance $x$ of the $BC$ from the $(1,4)$
$=\sqrt{10}\text{or}\sqrt{\frac{5}{2}}$
For $x=\sqrt{10}\frac{|1-3\times 4+l|}{\sqrt{10}}=\sqrt{10}l=21,1$
For $x=\sqrt{\frac{5}{2}}\frac{|3\times 1+4+k|}{\sqrt{10}}=\sqrt{\frac{5}{2}}k=-12,-2$