Question

The extremities of the base of an isosceles triangle $ABC$ are the points $A(2,0)$ and $B(0,1)$. If the equation of the side $AC$ is $x=2$ then the slope of the side $BC$ is -

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{3}{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is A $\frac{3}{4}$

Point C lies on $x=2$

$\Rightarrow C=(2,p)$

ABC is isosceles $\Rightarrow AC=BC$

$\sqrt{(2-2{)}^2+p^2}=\sqrt{(0-2{)}^2+(p-1{)}^2}$

$p^2=4+(p-1{)}^2$

$\Rightarrow p^2=4+p^2-2p+1$

$\Rightarrow 2p=5$

$\Rightarrow p=\frac{5}{2}$

$\therefore C=(2,\frac{5}{2})$

Slope of BC:

$\Rightarrow \frac{(\frac{5}{2}-1)}{(2-0)}$

$\Rightarrow \frac{\frac{3}{2}}{2}$

$\Rightarrow \frac{3}{4}$

$\therefore$ Slope of BC$=\frac{3}{4}$.