First slide

Cartesian plane

Question

$A triangle ABC lying in the first quadrant has two vertices as A (1, 2) and B (3, 1) . If$ $∠ BAC = 90 °, and area of triangle A B C is 5 \sqrt{5} sq units, then the abscissa of the vertex C is:$

Moderate

A

$1 + \sqrt{5}$
B

$2 \sqrt{5} - 1$
C

$2 + \sqrt{5}$
D

$1 + 2 \sqrt{5}$

Solution

$The slope of the line A B is \frac{1 - 2}{3 - 1} = - \frac{1}{2}$
$Hence the slope of A C is 2$
It gives
$\begin{matrix} \frac{b - 2}{a - 1} = 2 \\ b - 2 = 2 a - 2 \\ b = 2 a \end{matrix}$
$Suppose that C is (a, 2 a)$
$The area of the triangle is 5 \sqrt{5} square units$
It implies
$\begin{matrix} \frac{1}{2} (A B) (A C) = 5 \sqrt{5} \\ \frac{1}{2} (\sqrt{5}) A C = 5 \sqrt{5} \\ A C = 10 \end{matrix}$
Hence,
$\begin{matrix} \sqrt{{(a - 1)}^{2} + {(2 a - 2)}^{2}} = 2 \sqrt{5} \\ \sqrt{{(a - 1)}^{2} + 4 {(a - 1)}^{2}} = 10 \\ \sqrt{5} (a - 1) = 10 \\ a - 1 = 2 \sqrt{5} \\ a = 1 + 2 \sqrt{5} \end{matrix}$
$Therefore, abscissa of C is 1 + 2 \sqrt{5}$

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