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Deepak Bora
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Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.

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This question is Based on Equation of a Straight Line Chapter of M.L Aggarwal book for ICSE BOARD for class 10.
Here you have to find the equation of a line perpendicular from a point on a given line and Also find the co-ordinates of the foot of perpendicular.
This is the Question Number 23, Exercise 12.2 of M.L Aggarwal.

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1 Answer

  1. Given line equation: 4x – 3y – 5 = 0

    3y = 4x – 5

    y = (4/3) x – 5

    Slope of the line (m1) = 4/3

    Let the slope of the line perpendicular to the given line be m2

    Then, m1 x m2 = -1

    (4/3) x m2 = -1

    m2 = -3/4

    Now, the equation of the line having slope m2 and passing through the point (1, -2) will be

    y + 2 = (-3/4) (x – 1)

    4y + 8 = -3x + 3

    3x + 4y + 5 = 0

    Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines

    4x – 3y – 5 = 0 …. (1) and

    3x + 4y + 5 = 0 …. (2)

    On multiplying (1) by 4 and (2) by 3, we get

    16x – 12y – 20 = 0

    9x + 12y + 15 = 0

    Adding we get,

    25x – 5 = 0

    x = 5/25

    x = 1/5

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

    4(1/5) – 3y – 5 = 0

    4/5 – 3y – 5 = 0

    3y = 4/5 – 5 = (4 – 25)/5

    3y = -21/5

    y = -7/5

    Thus, the co-ordinates are (1/5, -7/5)

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