From the given figure, find :

(i) the co-ordinates of A, B and C.

(ii) the equation of the line through A and parallel to BC.

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Solution

(i) A = (2, 3), B = (-1, 2), C = (3, 0)

(ii) Slope of BC = $fraction numerator 0 minus 2 over denominator 3 plus 1 end fraction equals negative 1 half$

Slope of required line which is parallel to BC = Slope of BC =

(x1, y1) = (2, 3)

The required equation of the line through A and parallel to BC is given by:

y - y1 = m1(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

x + 2y = 8

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