Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4.

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Solution

The given equations are:

x + 2y = 7 ....(1)

x - y = 4 ....(2)

Subtracting (2) from (1), we get,

3y = 3

y = 1

From (2), x = 4 + y = 4 + 1 = 5

The required line passes through (0, 0) and (5, 1).

$S l o p e space o f space t h e space l i n e equals fraction numerator 1 minus 0 over denominator 5 minus 0 end fraction equals 1 fifth$

Required equation of the line is given by:

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Q. If a line passes through the intersection point of the graphs of the lines

x + 2 y = 7

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Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4.

The given equations are: x + 2y = 7 ....(1) x - y = 4 ....(2) Subtracting (2) from (1), we get, 3y = 3 y = 1 From (2), x = 4 + y = 4 + 1 = 5 The required line passes through (0, 0) and (5, 1). Required equation of the line is given by: