Question

Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2, and

1. parallel to the line x + 2y − 5 = 0 and
2. perpendicular to the X-axis.

Answer 1

To find the point common to both lines, let's solve the system of linear equations:

(1) 4x + 3y = 1

(2) 5x + 4y = 2

Multiply the (1) equation by 4 and the (2) by 3:

(3) 16x + 12y = 4

(4) 15x + 12y = 6

Subtracting equation (4) from equation (3):

(16x − 15x) + (12y − 12y) = 4 − 6

∴ x = −2

Substitute x = −2 into equation (1):

4(−2) + 3y = 1

−8 + 3y = 1

3y = 9

`y = 9/3`

∴ y = 3

The intersection point is (−2, 3).

A line parallel to x + 2y − 5 = 0 will have the same slope and can be written in the form x + 2y + k = 0.

Substitute the intersection point (−2, 3) into this equation:

(−2) + 2(3) + k = 0

−2 + 6 + k = 0

4 + k = 0

∴ k = −4

The equation is x + 2y − 4 = 0.

A line perpendicular to the X-axis is a vertical line, which has the general form x = c,

Since the line passes through the point (−2, 3), the value of x must be constant at −2,

Hence, the equation is x = −2, which can also be written as x + 2 = 0,

Thus, the equations of the lines are x + 2y − 4 = 0 and x + 2 = 0.