Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations.
x – 7y + 5 = 0 …(i)
3x + y – 7 = 0 …(ii)
Now, we find the point of intersection of eq. (i) and (ii)
Multiply the eq. (i) by 3, we get
3x – 21y + 15 = 0 …(iii)
On subtracting eq. (iii) from (ii), we get
3x + y – 7 – 3x + 21y – 15 = 0
⇒ 22y - 22 = 0
⇒ 22y = 22
⇒ y = 1
Putting the value of y in eq. (i), we get
x – 7(1) + 5 = 0
⇒ x – 7 + 5 = 0
⇒ x – 2 = 0
⇒ x = 2
Hence, the point of intersection P(x1, y1) is (2, 1)

The equation of line parallel to x – axis is of the form
y = b where b is some constant
Given that this equation of the line passing through the point of intersection (2, 1)
Hence, point (2, 1) will satisfy the equation of a line.
Putting y = 1 in the equation y = b, we get
y = b
⇒ 1 = b
or b = 1
Now, the required equation of a line is y = 1
