First slide

Cartesian plane

Question

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

Easy

A

$1 + \sqrt{5}$
B

$2 + \sqrt{5}$
C

$1 + 2 \sqrt{5}$
D

$2 \sqrt{5} - 1$

Solution

Let the coordinates of vertex C be $(α, β)$ . It is given that $∠ B A C = 90^{\circ}$
$\Rightarrow$ Slope of AB x Slope of AC = -1
$\Rightarrow \frac{1 - 2}{3 - 1} \times \frac{β - 2}{α - 1} = - 1$
$\Rightarrow 2 α = β$ …(i)
Area of $Δ (A B C) = 5 \sqrt{5}$
$\begin{array}{l} \Rightarrow \frac{1}{2} A B \times A C = 5 \sqrt{5} \\ \Rightarrow \sqrt{(3 - 1)^{2} + (1 - 2)^{2}} \times \sqrt{(α - 1)^{2} + (β - 2)^{2}} = 10 \sqrt{5} \\ \Rightarrow \sqrt{5} \sqrt{(α - 1)^{2} + (2 α - 2)^{2}} = 10 \sqrt{5} \end{array}$
$\begin{array}{l} \Rightarrow \sqrt{5} | α - 1 | = 10 \\ \Rightarrow | α - 1 | = 2 \sqrt{5} \Rightarrow α - 1 = \pm 2 \sqrt{5} \Rightarrow α = 1 \pm 2 \sqrt{5} \end{array}$

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