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Question

(i) Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4.

(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).

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Solution

(i) x - 3y = 4

3y = x - 4

y equals 1 third x minus 4 over 3

Slope of this line = 1 third

Slope of a line parallel to this line = 1 third

Required equation of the line passing through (5, -3) is

y - y1 = m(x - x1)

y + 3 = 1 third(x - 5)

3y + 9 = x - 5

x - 3y - 14 = 0

(ii) 2y = -3x + 8

Or y = fraction numerator negative 3 over denominator 2 end fraction x plus 8 over 2

The slope of given line = fraction numerator negative 3 over denominator 2 end fraction

Since the required line is parallel to given straight line.

Slope of required line (m) = fraction numerator negative 3 over denominator 2 end fraction

Now the equation of the required line is given by:

y - y1 = m(x - x1)

y - 1 = fraction numerator negative 3 over denominator 2 end fraction open parentheses x minus 0 close parentheses

2y - 2 = -3x

3x + 2y = 2


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