The vertex of a right angle of a right angled | vedclass

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Question

$M_1M_2 = -1 $

$\frac{{9 - 2a}}{{a - 4}}\,\,	imes\,\,\frac{{13 - 2a}}{{a - 2}} = -1$

$117 - 26a - 18a + 4a^2 = -(a^2 - 6a + 8)$

$5a^2 - 50a

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$M_1M_2 = -1 $

$\frac{{9 - 2a}}{{a - 4}}\,\,	imes\,\,\frac{{13 - 2a}}{{a - 2}} = -1$

$117 - 26a - 18a + 4a^2 = -(a^2 - 6a + 8)$

$5a^2 - 50a + 125 = 0$

$a = 5$

so $B$ is $(5, 0)$

so area $= \frac {1}{2} AB &times; AC = \frac{1}{2} \sqrt{2} &times; 3 \sqrt{2} = 3 $

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

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