Question

Find the equation of a straight line passing through the intersection of 2x + 5y − 4 = 0 with the X-axis and parallel to the line 3x − 7y + 8 = 0.

Answer 1

⇒ Substitute y = 0 into the equation 2x + 5y − 4 = 0:

2x + 5(0) − 4 = 0

2x = 4

∴ x = 2

The point of intersection is (2, 0).

⇒ The line is parallel to 3x − 7y + 8 = 0, rewriting it into the slope-intercept form, y = mx + c to find the slope (m₁):

−7y = −3x − 8

`y = 3/7 x + 8/7`

∴ The slope (m₁) of the given line is `3/7`.

⇒ Since the line is parallel, its slope (m₂) is the same:

`m_2 = 3/7`

⇒ Using the point-slope formula with `m = 3/7` and the point (2, 0):

y − y₁ = m(x − x₁)

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

`y = 3/7 x - 6/7`   ...[Multiplying entire equation by 7]

7y = 3x − 6

⇒ Rearranging the above equation in the standard form (Ax + By + C = 0),

3x − 7y − 6 = 0

Hence, the equation of the straight line is 3x − 7y − 6 = 0.