In the given graph, P and Q are points such that PQ cuts off intercepts of 5 units and 3 units along the x-axis and y-axis respectively. Line RS is…

(a) Given,

PQ cuts off intercepts of 5 units and 3 units along the x-axis and y-axis respectively.

Thus,

Coordinates of P = (5, 0) and coordinates of Q = (0, -3).

Hence, coordinates of P = (5, 0) and coordinates of Q = (0, -3).

(b) By formula,

Slope (m) = $-\frac{y_2 - y_1}{x_2 - x_1}$

⇒ Slope of PQ = m_PQ = $\frac{-3 - 0}{0 - 5} = \frac{-3}{-5} = \frac{3}{5}$.

We know that,

Product of slopes of perpendicular lines = -1.

⇒ Slope of RS × Slope of PQ = -1

⇒ Slope of RS × $\frac{3}{5}$ = -1

⇒ Slope of RS = $-\frac{5}{3}$.

By point-slope formula,

Equation of line : $y - y_{1} = m(x - x_{1})$

Since, slope of RS = $-\frac{5}{3}$ and it passes through the origin.

Equation of RS :

⇒ $y - 0 = -\frac{5}{3}(x - 0)$

⇒ y = $-\frac{5}{3}x$

⇒ 3y = -5x

⇒ 5x + 3y = 0.

Hence, the equation of line RS is 5x + 3y = 0.