Question

The ends of the base of an isosceles triangle are at $\left(2,0\right)$ and $\left(0,1\right)$ and the equation of one side is $x=2$ then the orthocentre of the triangle is

• A. $\left(\frac{3}{2},\frac{3}{2}\right)$
• B. $\left(\frac{5}{4},1\right)$
• C. $\left(\frac{3}{4},1\right)$
• D. $\left(\frac{4}{3},\frac{7}{12}\right)$

Answer 1

The correct option is B $\left(\frac{5}{4},1\right)$

Given the triangle ABC is isosceles

with equation of one side is x = 2

$\therefore$ co-ordinates of the third vertex be A(2,y)

Thus $AC=AB\Rightarrow A{C}^2=A{B}^2$

$\therefore (2-0^2)+(y-1{)}^2=(2.2{)}^2+(y-0{)}^2$

$\Rightarrow 2y=5$

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

So the co-ordinate of A is $(2,\frac{5}{2})$

Now, slope of line $BC=\frac{1-0}{0-2}=\frac{-1}{2}$

Slope perpendicular to BC = 2

Equation through BA and slope 2 is

$y-\frac{5}{2}=2(x-2)\Rightarrow 4x-2y=3$____(1)

Now slope of $AC=\frac{3}{4}$ and $\perp$ to $AC=-\frac{4}{3}$

$E{q}^n$ through B and slope -4/3 is

$y-0=\frac{-4}{3}(x-2)\Rightarrow 4x+3y=8$____(2)

subtracting (1) from (2)

we, get y = 1 and putting y = 1 in (i)

$x=\frac{5}{4}$

Coordinate of orthocentre are $(\frac{5}{4},1)$