Vertex of a right angle of a right angled triangle lies on straight line 2x + y - 10 = 0 and two oth

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9.Straight Line

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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

A

$\sqrt{10}$

B

$3$

C

$\frac{33}{5}$

D

$11$

Solution

$M_1M_2 = -1 $

$\frac{{9 – 2a}}{{a – 4}}\,\,\times\,\,\frac{{13 – 2a}}{{a – 2}} = -1$

$117 – 26a – 18a + 4a^2 = -(a^2 – 6a + 8)$

$5a^2 – 50a + 125 = 0$

$a = 5$

so $B$ is $(5, 0)$

so area $= \frac {1}{2} AB × AC = \frac{1}{2} \sqrt{2} × 3 \sqrt{2} = 3 $

Standard 11

Mathematics

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