If A $(- 1, 8)$ , B $(4, - 2)$ and C $(- 5, - 3)$ are the vertices of a triangle, then the equation of the altitude through $(- 1, 8)$ is given as $a x + b = y$ . Find $a + b$ .
26

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Solution

The correct option is A 26
In the given figure, AD is perpendicular to BC.

Slope of a line passing through (a, b) and (c, d) $= \frac{d - b}{c - a}$
Slope of BC $= \frac{- 3 - (- 2)}{- 5 - 4} = \frac{1}{9}$

Product of slopes of perpendicular lines is -1.
Let the slope of AD be m.
AD $⊥$ BC
So, $\frac{m}{9} = - 1 \to m = - 9$

Equation of a line passing through (a, b) and slope 'm' = $\frac{y - b}{x - a} = m$

Equation of AD $= \frac{y - 8}{x - (- 1) = 9 \to y - 8 = 9 x + 9 \to y = 9 x + 17}$

Comparing $y = 9 x + 17$ with $y = a x + b$ , we get $a = 9$ and $b = 17 a + b = 9 + 17 = 26$

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If A(−1,8), B(4,−2) and C(−5,−3) are the vertices of a triangle, then the equation of the altitude through (−1,8) is given as ax+b=y. Find a+b.26

If A $(- 1, 8)$ , B $(4, - 2)$ and C $(- 5, - 3)$ are the vertices of a triangle, then the equation of the altitude through $(- 1, 8)$ is given as $a x + b = y$ . Find $a + b$ .
26