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(B) Let the vertex of the right angle be $B(a, 10 - 2a)$. Let the other two vertices be $A(4, 1)$ and $C(2, -3)$.Since $\angle B = 90^{\circ}$,the pro

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$3$

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(B) Let the vertex of the right angle be $B(a, 10 - 2a)$. Let the other two vertices be $A(4, 1)$ and $C(2, -3)$.Since $\angle B = 90^{\circ}$,the product of the slopes of $AB$ and $BC$ is $-1$.Slope of $AB = \frac{(10 - 2a) - 1}{a - 4} = \frac{9 - 2a}{a - 4}$.Slope of $BC = \frac{(10 - 2a) - (-3)}{a - 2} = \frac{13 - 2a}{a - 2}$.Since $AB \perp BC$,we have $\left(\frac{9 - 2a}{a - 4}ight) 	imes \left(\frac{13 - 2a}{a - 2}ight) = -1$.$(9 - 2a)(13 - 2a) = -(a - 4)(a - 2)$.$117 - 18a - 26a + 4a^2 = -(a^2 - 6a + 8)$.$4a^2 - 44a + 117 = -a^2 + 6a - 8$.$5a^2 - 50a + 125 = 0$.$a^2 - 10a + 25 = 0 \implies (a - 5)^2 = 0 \implies a = 5$.Thus,the vertex $B$ is $(5, 10 - 2(5)) = (5, 0)$.Length $AB = \sqrt{(5 - 4)^2 + (0 - 1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.Length $BC = \sqrt{(5 - 2)^2 + (0 - (-3))^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.Area of $	riangle ABC = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes \sqrt{2} 	imes 3\sqrt{2} = \frac{1}{2} 	imes 6 = 3$ sq. units.

The vertex of the right angle of a right-angled triangle lies on the straight line $2x + y - 10 = 0$. If the two other vertices are at points $(2, -3)$ and $(4, 1)$,then the area of the triangle in sq. units is:

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