Adv
Deepak Bora
  • 0
Newbie

Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (– 1, – 2).

  • 0

This question is Based on Equation of a Straight Line Chapter of M.L Aggarwal book for ICSE BOARD for class 10.
Passing point is given of a perpendicular line. So write down the equation
This is the Question Number 17, Exercise 12.2 of M.L Aggarwal.

Share

1 Answer

  1. Given line: 3x + 8y = 12

    8y = -3x + 12

    y = (-3/8) x + 12

    So, the slope (m1) = -3/8

    Let’s consider the slope of the line perpendicular to the given line as m2

    Then, m1 x m2 = -1

    -3/8 x m2 = -1

    m2 = 8/3

    Now,

    The equation of the line perpendicular to the given line and passing through the point (-1, -2) will be

    y – y1 = m (x – x1)

    y – (-2) = (8/3) (x – (-1))

    y + 2 = (8/3) (x + 1)

    3y + 6 = 8x + 8

    3y = 8x + 2

    Thus, the equation of the required line is 3y = 8x + 2.

    • 0
Leave an answer

Leave an answer

Browse

Choose from here the video type.

Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs".

Captcha Click on image to update the captcha.

Related Questions