9.Straight Line
hard

If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is

A

$5{a^2}sq$. units

B

$\frac{5}{2}{a^2}sq.$units

C

$\frac{{25{a^2}}}{2}sq.$units

D

None of these

(JEE MAIN-2013)

Solution

(b) Let the co-ordinates of the third vertex be $(2a,\,\,t)$.

$AC = BC \Rightarrow t = \sqrt {4{a^2} + {{(a – t)}^2}} \Rightarrow $$t = \frac{{5a}}{2}$

So the coordinates of third vertex $C$ are $\left( {2a,\frac{{5a}}{2}} \right)$

Therefore area of the triangle

$ = \pm \frac{1}{2}\left| {\,\begin{array}{*{20}{c}}{2a}&{\frac{{5a}}{2}}&1\\{2a}&0&1\\0&a&1\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}a&{\frac{{5a}}{2}}&1\\0&{ – \frac{{5a}}{2}}&0\\0&a&1\end{array}\,} \right| = \frac{{5{a^2}}}{2}sq.$units.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.