QJEEM20Ma7d33: JEE Main 2020 Mathematics - Class 11 JEEM20Ma7d33 Solution: A triangle ABC lying in the first quadrant has two vertices as…

Question

Verified Solution • 2025–26 Session moderate

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If

\angle BAC = {90^o}

and area

\left( {\Delta ABC} \right) = 5\sqrt 5

s units, then the abscissa of the vertex C is :

Accepted Answer

Distance between A and B

= \sqrt {{{(2)}^2} + {{( - 1)}^2}} = \sqrt 5

Area of triangle =

5\sqrt 5

\Rightarrow {1 \over 2} \times \sqrt 5 \times x = 5\sqrt 5

\Rightarrow x = 10

Slope of AB,

{m_{AB}} = {{1 - 2} \over {3 - 1}} = - {1 \over 2}

AC is perpendicular to AB.

\therefore

{m_{AB}}\,.\,{m_{AC}} = - 1

\Rightarrow {m_{AC}} = 2 = \tan \theta

So,

\sin \theta = {2 \over {\sqrt 5 }}

,

\cos \theta = {1 \over {\sqrt 5 }}

\therefore

a = {x_A} + r\cos \theta

= 1 + 10 \times {1 \over {\sqrt 5 }}

= 1 + 2\sqrt 5

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