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

Question

A triangle $ABC$ lying in the first quadrant has two vertices as $A(1,2)$ and $B(3,1)$. If $\angle BAC=90^\circ$, and ar$(	riangle ABC)=5\sqrt5$ sq. 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, $$\sin \theta  = {2 \over {\sqrt 5 }}$$, $$\cos \theta  = {1 \over {\sqrt 5 }}$$

$$ \therefore $$ $$a = {x_A} + r\cos \theta $$

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

$$ = 1 + 2\sqrt 5 $$

Solutions · 1