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Question

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.

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Solution

(i) We know the median through A will pass through the mid-point of BC. Let AD be the median through A.

Co-ordinates of the mid-point of BC, i.e., D are

open parentheses fraction numerator 2 minus 2 over denominator 2 end fraction comma fraction numerator 2 plus 4 over denominator 2 end fraction close parentheses equals open parentheses 0 comma 3 close parentheses

Slope of AD = fraction numerator 3 plus 5 over denominator 0 minus 1 end fraction equals negative 8

Equation of the median AD is

y - 3 = -8(x - 0)

8x + y = 3

(ii) Let BE be the altitude of the triangle through B.

Slope of AC = fraction numerator 4 plus 5 over denominator negative 2 minus 1 end fraction equals fraction numerator 9 over denominator negative 3 end fraction equals negative 3

Slope of BE = fraction numerator negative 1 over denominator s l o p e space o f space A C end fraction equals 1 third

Equation of altitude BE is

y - 2 = 1 third(x - 2)

3y - 6 = x - 2

3y = x + 4

(iii) Slope of AB = fraction numerator 2 plus 5 over denominator 2 minus 1 end fraction equals 7

Slope of the line parallel to AB = Slope of AB = 7

So, the equation of the line passing through C and parallel to AB is

y - 4 = 7(x + 2)

y - 4 = 7x + 14

y = 7x + 18


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