A $(2 - 4), B (3, 3)$ and $C (- 1, 5)$ are the vertices of $△ A B C$ . Find the equation of
(i) The median of the triangle $A$ .
(ii) The altitude of the triangle through $B$ .

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Solution

Let mid point $B C$ be $D$ , so, $D = (\frac{3 - 1}{2}, \frac{3 + 5}{2}) = (1, 4)$
Equation of $A D$ $⟹ y - 4 = (x - 1) (\frac{4 - (- 4)}{1 - 2})$
$⟹ y = - 8 x + 12$
Slope of $A C ⟹ \frac{5 - (- 4)}{- 1 - 2} = - 3$
Slope of line is perpendicular to $A C = \frac{- 1}{- 3} = \frac{1}{3}$
So, eqaution of at\ltitude $⟹ y - 3 = \frac{1}{3} (x - 3) ⟹ 3 y = x + 6$

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Similar questions

A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC. Find the equations of :

(i) the median of the triangle through A.

(ii) the altitude of the triangle through B.

(iii) the line through C and parallel to AB.

If A (2, 1), B(-2, 3) and C(4, 5) are the vertices of a $△$ ABC, then find the equation of

(i) the altitude through B.

(ii) the right bisector of side BC.

A (5, 4), B (- 3, - 2)

C (1, - 8)

A (5, 4), B (- 3, - 2)

C (1, - 8)

△ A B C

A D

B M

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