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Deepak Bora
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Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7

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An Important Question of class 10 Based on Equation of a Straight Line Chapter of M.L Aggarwal for ICSE BOARD.
Here given that a straight line is passing through the origin and through the point of intersection of the given lines.
So, you have to find the equation of a straight line.
This is the Question Number 32, Exercise 12.1 of M.L Aggarwal.

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  1. Given line equations,

    5x + 7y = 3 … (i)

    2x – 3 y = 7 … (ii)

    Now, performing multiplication of (i) by 3 and (ii) by 7, we get

    15x + 21y = 9

    14x – 21y = 49

    On adding we get,

    29x = 58

    x = 58/29 = 2

    Substituting the value of x in (i), we get

    5(2) + 7y = 3

    10 + 7y = 3

    7y = 3 – 10

    y = -7/7 = -1

    Hence, the point of intersection of lines is (2, -1)

    Now, the slope of the line joining the points (2, -1) and (0, 0) will be

    m = y2 – y1/ x2 – x1

    = (0 + 1)/ (0 – 2)

    = -1/2

    Equation of the line is given by:

    y – y1 = m (x – x1)

    y – 0 = -1/2 (x – 0)

    2y = -x

    Thus, the required line equation is x + 2y = 0.

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