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

(i) the slope of the line

(ii) the equation of the line perpendicular to the given line and passing through the intersection of the lines x - y + 2 and 3x + y - 10 = 0.

(i) Given,

Equation : 7 - 3x - 4y = 0

⇒ 4y = 7 - 3x

⇒ 4y = -3x + 7

⇒ y = $-\dfrac{3}{4}x + \dfrac{7}{4}$

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

m = $-\dfrac{3}{4}$.

Hence, the slope of line = $-\dfrac{3}{4}$.

(ii) We know that,

Product of slope of perpendicular lines = -1.

Let slope of line perpendicular to 7 - 3x - 4y = 0 be m₁.

∴ m × m₁ = -1

⇒ $-\dfrac{3}{4} \times m_1 = -1$

⇒ $m_1 = \dfrac{4}{3}$.

Solving equations x - y + 2 and 3x + y - 10 = 0 simultaneously.

⇒ x - y + 2 = 0

⇒ y = x + 2 ……..(1)

⇒ 3x + y - 10 = 0

⇒ y = 10 - 3x ……..(2)

From equation (1) and (2), we get :

⇒ x + 2 = 10 - 3x

⇒ x + 3x = 10 - 2

⇒ 4x = 8

⇒ x = $\dfrac{8}{4}$

⇒ x = 2.

Substituting value of x in equation (1), we get :

⇒ y = 2 + 2 = 4.

∴ Point of intersection of equations x - y + 2 and 3x + y - 10 = 0 are (2, 4).

By point slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get :

⇒ y - 4 = $\dfrac{4}{3}(x - 2)$

⇒ 3(y - 4) = 4(x - 2)

⇒ 3y - 12 = 4x - 8

⇒ 4x - 3y - 8 + 12 = 0

⇒ 4x - 3y + 4 = 0.

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