Three sides AB, BC and CA of a triangle ABC are $5 x - 3 y + 2 = 0, x - 3 y - 2 = 0$ and $x + y - 6 = 0$ respectively. Find the equation of the altitude through the vertex A.

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Solution

The sides AB, Bc and CA of a triangle ABC are as follows :

$5 x - 3 y + 2 = 0 . . . (i)$

$x - 3 y - 2 = 0 . . . (i i)$

$x + y - 6 = 0 . . . (i i i)$

Solving (i) and (iii):

$x - 2, y = 4$

Thus, AB and CA intersect at A $(2, 4)$

Let AD be the altitude

Thus, AB and CA intersect at A $(2, 4)$

Let AD be the altutude

$A D ⊥ B C$

$∴$ Slope of AD $\times$ Slope of BC = -1

Here, slope of BC = slope of the line (ii)

$= \frac{1}{3}$

$∴$ Slope of $A D \times \frac{1}{3} = - 1$

$\Rightarrow$ Slope of AD = - 3

Hence, the equation of the altitude AD passing through A $(2, 4)$ and having slope -3 is

$y - 4 = - 3 (x - 2)$

$\Rightarrow 3 x + y = 10$

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Three sides AB, BC and CA of a triangle ABC are 5x−3y+2=0, x−3y−2=0 and x+y−6=0 respectively. Find the equation of the altitude through the vertex A.

Three sides AB, BC and CA of a triangle ABC are $5 x - 3 y + 2 = 0, x - 3 y - 2 = 0$ and $x + y - 6 = 0$ respectively. Find the equation of the altitude through the vertex A.