First slide

Equation of line in Straight lines

Question

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

Moderate

A

$5 a^{2} sq . units$
B

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

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

none of these

Solution

Given vertices are $A (2 a, 0)$ and $B (0, a)$
Let the coordinates of the third vertex C are $(2 a, t)$ . Now, $AC = BC .$ Hence,
$t = \sqrt{4 a^{2} + {(a - t)}^{2}}$ or $t = \frac{5 a}{2}$
So, the coordinates of the third vertex C are $(2 a, 5 a / 2) .$
Therefore, area of the triangle is
$\frac{1}{2} ||\begin{matrix} 2 a & 5 a / 2 & 1 \\ 2 a & 0 & 1 \\ 0 & a & 1 \end{matrix}|| = \frac{5 a^{2}}{2} sq . units$

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