If the ends of the base of an isosceles triangle are at (2,0) and (0,1) and the equation of one side is x=2 then the orthocenter of the triangle is

If the ends of the base of an isosceles triangle are at $(2, 0)$ and $(0, 1)$ and the equation of one side is $x = 2$ then the orthocenter of the triangle is

A
$(3 / 2, 3 / 2)$
B
$(5 / 4, 1)$
C
$(3 / 4, 1)$
D
$(4 / 3, 7 / 12)$

Solution:

Form the figure,

$2^{2} + {(y_{1} - 1)}^{2} = y_{1}^{2}$

$4 + y_{1}^{2} + 1 - 2 y_{1} = y_{1}^{2}$

$5 = 2 y_{1} o r y_{1} = 5 / 2$

Equation of the line form $(2, 5 / 2)$ to the given base is

$y - 5 / 2 = 2 (x - 2)$ or $2 y - 5 = 4 (x - 2)$

At $y = 1$ , $- 3 / 4 = x - 2$

Or $x = 5 / 4$

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