Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations.
2x – 3y = 0 …(i)
4x – 5y = 2 …(ii)
Now, we find the point of intersection of eq. (i) and (ii)
Multiply the eq. (i) by 2, we get
4x – 6y = 0 …(iii)
On subtracting eq. (iii) from (ii), we get
4x – 5y – 4x + 6y = 2 – 0
⇒ y = 2
Putting the value of y in eq. (i), we get
2x – 3(2) = 0
⇒ 2x – 6 = 0
⇒ 2x = 6
⇒ x = 3
Hence, the point of intersection P(x1, y1) is (3, 2)
Now, we know that, when two lines are perpendicular, then the product of their slope is equal to -1
m1 × m2 = -1
⇒ Slope of the given line × Slope of the perpendicular line = -1
∴ (-1/2) x Slope of the perpendicular line = -1
⇒ The slope of the perpendicular line = 2
So, the slope of a line which is perpendicular to the given line is 2
Then the equation of the line passing through the point (3, 2) having slope 2 is:
y – y1 = m (x – x1)
⇒ y – 2 = 2(x – 3)
⇒ y – 2 = 2x – 6
⇒ 2x – y – 6 + 2 = 0
⇒ 2x – y – 4 = 0