Question

The equation of a line is 3x + 4y – 7 = 0. Find:

1. the slope of the line.
2. the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.

Answer 1

3x + 4y − 7 = 0    ...(1)

4y = −3x + 7

${\displaystyle y=\frac{-3}{4}x+\frac{7}{4}}$

i. Slope of the line = m = ${\displaystyle \frac{-3}{4}}$

ii. Slope of the line perpendicular to the given line = ${\displaystyle \frac{-1}{\frac{-3}{4}}=\frac{4}{3}}$

Solving the equations x − y + 2 = 0 and 3x + y − 10 = 0, we get x = 2 and y = 4.

So, the point of intersection of the two given lines is (2, 4).

Given that a line with slope ${\displaystyle \frac{4}{3}}$ passes through point (2, 4).

Thus, the required equation of the line is

y − y₁​ = m(x − x₁​)

${\displaystyle y-4=\frac{4}{3}(x-2)}$

3(y − 4) = 4(x − 2)

3y − 12 = 4x − 8

4x − 3y + 4 = 0