Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.

Given line equations are,

7x + 6y = 71 …….(1)

and

5x - 8y = -23 ………(2)

Multiplying (1) by 4 and (2) by 3 we get,

⇒ 28x + 24y = 284 ……… (3)

⇒ 15x – 24y = -69 ………(4)

On adding (3) and (4), we get

⇒ 28x + 15x + 24y - 24y = 284 + (-69)

⇒ 43x = 215

⇒ x = $\dfrac{215}{43}$

⇒ x = 5.

From (2), we get

⇒ 8y = 5x + 23

⇒ 8y = 5(5) + 23

⇒ 8y = 25 + 23

⇒ 8y = 48

⇒ y = $\dfrac{48}{8}$

⇒ y = 6.

Hence, the required line passes through the point (5, 6).

Given, 4x – 2y = 1

⇒ 2y = 4x – 1

⇒ y = 2x – $\dfrac{1}{2}$.

Comparing above equation with y = mx + c we get,

Slope (m) = 2

Let slope of required line be m₁.

As, the lines are perpendicular to each other so product of their slopes = -1.

⇒ m × m₁ = -1

⇒ 2 × m₁ = -1

⇒ m₁ = $-\dfrac{1}{2}$.

Thus, equation of the line is,

⇒ y – y₁ = m(x – x₁)

⇒ y – 6 = $-\dfrac{1}{2}$(x – 5)

⇒ 2(y – 6) = -1(x - 5)

⇒ 2y - 12 = -x + 5

⇒ 2y + x = 5 + 12

⇒ x + 2y = 17.

Hence, equation of required line is x + 2y = 17.