Question

The equation of a straight line can be written in the
form 3x + 2y - 8 = 0.
(a) Rearrange this equation to make y the subject.
(b) Write down the gradient of the line.
(c) Write down the coordinates of the point where
the line crosses the y-axis.

Answer 1

Given line : 3x + 2y − 8 = 0

2y = − 3x + 8

y = (-3/2)x + 4

Here, slope (m₁) = -3/2

Now, the co-ordinates of the mid-point of the line segment joining the points (5,-2) and (2, 2) will be

((5 + 2)/7, (-2 +2)/7) = (7/2,0)

Let's consider the slope of the perpendicular to the given line be m2

Then,

m₁ x m₂ = -1

(-3/2) × m₂ = -1

m₂ = -2/3

So, the equation of the line with slope m₂

and passing through (7/2,0) will be

y - 0 = (2/3) (x-7/2)

3y = 2x - 7

2x - 3y - 7 = 0

Thus, the required line equation is 2x - 3y - 7 = 0