A triangle ABC lying in the first quadrant ha | vedclass

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(C) Let the coordinates of vertex $C$ be $(h, K)$.Since $\angle BAC = 90^{\circ}$,the product of the slopes of $AB$ and $AC$ is $-1$.Slope of $AB = \f

Vedclass · Accepted Answer

$1 + 2\sqrt{5}$

Vedclass · Answer

(C) Let the coordinates of vertex $C$ be $(h, K)$.Since $\angle BAC = 90^{\circ}$,the product of the slopes of $AB$ and $AC$ is $-1$.Slope of $AB = \frac{1 - 2}{3 - 1} = -\frac{1}{2}$.Slope of $AC = \frac{K - 2}{h - 1}$.Thus,$\left(\frac{K - 2}{h - 1}ight) 	imes \left(-\frac{1}{2}ight) = -1$ $\Rightarrow K - 2 = 2(h - 1)$ $\Rightarrow K = 2h$.Length of $AB = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{5}$.Area of $\Delta ABC = \frac{1}{2} 	imes AB 	imes AC = 5\sqrt{5}$.$\frac{1}{2} 	imes \sqrt{5} 	imes \sqrt{(h - 1)^2 + (K - 2)^2} = 5\sqrt{5}$.$\sqrt{(h - 1)^2 + (2h - 2)^2} = 10$.$\sqrt{(h - 1)^2 + 4(h - 1)^2} = 10$.$\sqrt{5(h - 1)^2} = 10 \Rightarrow \sqrt{5}|h - 1| = 10$.$|h - 1| = \frac{10}{\sqrt{5}} = 2\sqrt{5}$.Since the triangle lies in the first quadrant,$h > 0$. $h - 1 = 2\sqrt{5} \Rightarrow h = 1 + 2\sqrt{5}$ or $h - 1 = -2\sqrt{5} \Rightarrow h = 1 - 2\sqrt{5}$ (rejected as $h Thus,the abscissa of $C$ is $1 + 2\sqrt{5}$.

$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is

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