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Deepak Bora
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Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and (i) parallel to the line x + 2y – 5 = 0 (ii) perpendicular to the x-axis.

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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 intersection of the lines are given though which a line is passing. Solve this question
This is the Question Number 38, Exercise 12.2 of M.L Aggarwal.

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1 Answer

  1. Given, line equations:

    4x + 3y = 1 … (1)

    5x + 4y = 2 … (2)

    On solving the above equation to find the point of intersection, we have

    Multiplying (1) by 4 and (2) by 3

    16 x + 12y = 4

    15x + 12y = 6

    On subtracting, we get

    x = -2

    Putting the value of x in (1), we have

    4(-2) + 3y = 1

    -8 + 3y = 1

    3y = 1 + 8 = 9

    y = 9/3 = 3

    Hence, the point of intersection is (-2, 3).

    (i) Given line, x + 2y – 5 = 0

    2y = -x + 5

    y = -(1/2) x + 5/2

    Slope (m) = -1/2

    A line parallel to this line will have the same slope m = -1/2

    So, the equation of line having slope m and passing through (-2, 3) will be

    y – 3 = (-1/2) (x + 2)

    2y – 6 = -x – 2

    x + 2y = 4

    (ii) As any line perpendicular to x-axis will be parallel to y-axis.

    So, the equation of line will be

    x = -2 ⇒ x + 2 = 0

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