The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{green}{d.)x+2y-5=0}\\$
$\quad\\~\\~\\$
Let the slope of the required line be $m$.
I used the slope formula between two lines
$\angle B=\angle C\\~\\\left|{\dfrac{(-1-m)}{(1-m)}}\right|=\left|{\dfrac{(7-m)}{(1+7m)}}\right|\\
\implies m=-3,\dfrac{1}{3}$
But the book is giving right answer as option $d.)$.
By finding $m$ i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using $Area=5$ .
I have studied maths upto $12th$ grade.
Update : by using Geogebra I found that $x-3y+1=0$ fits perfectly
Best Answer
Let the lines: $x+y=5$ & $7x-y=3$ represent AC & AB. Then the acute angle $\theta$ between the equal sides AB & AC is given as $$\tan \theta=\left|\frac{7-(-1)}{1+7(-1)}\right|=\left|\frac{8}{-6}\right|=\frac{4}{3} \implies \sin \theta=\frac{4}{5}$$ The length of two sides AB & AC of isosceles triangle are equal hence, the area of triangle is given as $$\frac{1}{2}(AB)(AC)\sin\theta=5 \implies \frac{1}{2}(AB)^2\left(\frac{4}{5}\right)=5 \implies AB=\frac{5}{\sqrt{2}}$$ The equations of lines bisecting the angle between the equal sides: $x+y=5$ & $7x-y=3$ are given as $$\frac{x+y-5}{\sqrt{1^2+1^2}}=\pm \frac{7x-y-3}{\sqrt{7^2+(-1)^2}}\implies 5x+5y-25=\pm(7x-y-3)$$$$ \implies x-3y+11=0 \quad \text{&} \quad 3x+y-7=0 $$ From above, equations it's clear that $3x+y-7=0$ is acute angle bisector having negative slope & passing through vertex A. We can also check it out by a rough sketch.
Let the equation of third (unknown) line i.e. side BC be $x-3y+c=0$ normal to the angle bisector: $3x+y-7=0$. Now, solving: $x-3y+c=0$ & any given line say $7x-y=3$, we get the intersection point $\left(\frac{c+9}{20}, \frac{7c+3}{20} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(1, 4 \right)$. Now, calculating the length $AB=\frac{5}{\sqrt{2}}$ of equal sides of isosceles triangle using point-to-point distance-formula as $$AB=\sqrt{\left(1-\frac{c+9}{20}\right)^2+\left(4-\frac{7c+3}{20}\right)^2}=\frac{5}{\sqrt{2}}$$$$ (c-11)^2=100 \implies c-11=\pm 10 \implies c=21 \quad \text{&} \quad c=1$$ Thus by setting the values of $c$, we get two equations of parallel lines representing the third unknown side BC as: $\color{#0ae}{x-3y+1=0}$ & $ \color{#0ae} {x-3y+21=0}$ lying on either side of vertex A of given isosceles triangle ABC. Thus, we see that the option (b) $\color{#0b4}{x-3y+1=0}$ is correct.
According to the option (d) in your book the unknown side is: $x+2y-5=0$. Note that this line has slope $\frac{-1}{2}$ i.e. negative but see in the figure you drew the slope of unknown side must be only positive. Thus option provided in your book is absolutely wrong (may be due to some printing mistake).