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

Answer 1

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

Given vertices

$A(2,0),B(0,1)$

Equation of side AB

$y=\frac{1-0}{0-2}(x-2)$

$y=\frac{-1}{2}(x-2)$

$2y=-x+2$

$x+2y=2$

The altitude of triangle will passing through mid point of A$(2,0)$ and B$(0,1)$ and perpendicular to AB

Hence equation of altitude be with slope $2$ and point $(1,\frac{1}{2})$

$y-\frac{1}{2}=2(x-1)$

$2y-1=4x-4$

$4x-2y=3$

And given eq of side AC is $x=2$

Intersection point of AC and altitude give vertex C

$8-2y=3\Rightarrow y=\frac{5}{2}$

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

Slope of line BC $m=\frac{\frac{5}{2}-1}{2}$

Slope of line BC $m=\frac{3}{4}$