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

Solution:

Let’s assume the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)

Now, substituting the value y = 0 in the line equation, we have

2x + 5(0) – 4 = 0

2x – 4 = 0

x = 4/2 = 2

Hence, the co-ordinates of the point of intersection is (2, 0)

Also given, line equation: 3x – 7y + 8 = 0

7y = 3x + 8

y = (3/7) x + 8/7

So, the slope (m) = 3/7

We know that the slope of any line parallel to the given line will be the same.

So, the equation of the line having slope 3/7 and passing through the point (2, 0) will be

y – 0 = (3/7) (x – 2)

7y = 3x – 6

3x – 7y – 6 = 0

Thus, the required line equation is 3x – 7y – 6 = 0.