Straight LineHardBloom L3
Question
The vertex of 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:
Options
A.$\sqrt{10}$
B.3
C.$\frac{33}{5}$
D.11
Solution
Sol. [Key Concept: Right-angle vertex on a given line with the other two fixed: locate it from slope-product $- 1$, then area.]
$$M_{1}M_{2} = - 1
$$
$${\frac{9 - 2a}{a - 4} \times \frac{13 - 2a}{a - 2} = - 1 }{117 - 26a - 18a + 4a^{2} = - \left( a^{2} - 6a + 8 \right)}$$
$$5a^{2} - 50a + 125 = 0$$
so $B$ is $(5,0)$
so area $= \frac{1}{2}AB \times AC = \frac{1}{2}\sqrt{2} \times 3\sqrt{2} = 3$
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