Question

Find the equation of the line passing through the point of intersection of the lines $$5x - 8y + 23 = 0$$ and $$7x + 6y - 71 = 0$$ and perpendicular to the line $$4x - 2y = 3$$.

Answer 1

First, find the point of intersection of the two lines:

$$5x - 8y = -23 \quad ...(1)$$ 

$$7x + 6y = 71 \quad ...(2)$$

Multiply (1) by 3 and (2) by 4:

$$15x - 24y = -69$$

$$28x + 24y = 284$$

Adding these two equations:

$$43x = 215$$

$$x = 5$$

Substitute $$x = 5$$ into (1): 

$$5(5) - 8y = -23$$

$$25 - 8y = -23$$

$$8y = 48$$

$$\implies y = 6$$ 

So the point of intersection is $$(5, 6)$$.

Now, find the slope of the given line $$4x - 2y = 3$$: 

$$2y = 4x - 3$$

$$\implies y = 2x - \frac{3}{2}$$ 

Slope $$m_1 = 2$$

Slope of the perpendicular line:

$$m = -\frac{1}{m_1} = -\frac{1}{2}$$

Equation of the line passing through $$(5, 6)$$ with slope $$-\frac{1}{2}$$:

$$y - 6 = -\frac{1}{2}(x - 5)$$ 

$$2(y - 6) = -(x - 5)$$

$$2y - 12 = -x + 5$$

$$x + 2y - 17 = 0$$