Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis

Solution:

The equation of the given line is x/4 + y/6 = 1

This equation can also be written as 3x + 2 y - 12 = 0

y = - 3/2x + 6 , which is of the form y = mx + c

Hence, the slope of the given line is - 3/2

Slope of the line perpendicular to the given line is - 1/(- 3/2) = 2/3

Let the given line intersect the y-axis at (0, y) .

On substituting x = 0 in the equation of the given line, we obtain

y/6 = 1 ⇒ y = 6

The given line intersects the y-axis at (0, 6). The equation of the line that has a slope of 2/3 and passes through point (0, 6) is

(y - 6) = 2/3 (x - 0)

3y - 18 = 2x

2x - 3y + 18 = 0

Thus, the required equation of the line is 2x - 3y + 18 = 0

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 7

Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis

Summary:

The equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis is 2x - 3y + 18 = 0