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

(i) slope of the line.

(ii) 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.

(i) Given, 3x + 4y - 7 = 0

Converting the equation in the form of y = mx + c,

⇒ 4y = -3x + 7

⇒ y = $-\dfrac{3}{4}x + \dfrac{7}{4}$

Comparing the equation with y = mx + c, we get slope (m₁) = $-\dfrac{3}{4}$.

(ii) Let the slope of the line perpendicular to the given line be m₂.

Then,

$\Rightarrow m$1 \times m2 = -1 \\[1em] \Rightarrow -\dfrac{3}{4} \times m2 = -1 \\[1em] \Rightarrow m2 = \dfrac{4}{3}.

Now to find the point of intersection of

x - y + 2 = 0 … (i)
3x + y - 10 = 0 … (ii)

On adding (i) and (ii), we get

⇒ x - y + 2 + 3x + y - 10 = 0
⇒ 4x - 8 = 0
⇒ 4x = 8
⇒ x = 2.

Putting x = 2 in (i), we get

⇒ 2 - y + 2 = 0
⇒ y = 4.

Hence, the point of intersection of the lines is (2, 4).

The equation of the line with slope $\dfrac{4}{3}$ and passing through (2, 4) can be given by point-slope form,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 4 = \dfrac{4}{3}(x - 2) \\[1em] \Rightarrow 3(y - 4) = 4(x - 2) \\[1em] \Rightarrow 3y - 12 = 4x - 8 \\[1em] \Rightarrow 4x - 3y + 4 = 0.

Hence, the equation of the new line is 4x - 3y + 4 = 0.