A and B are two points on the x-axis and y-axis respectively.
(a) Write down the co-ordinates of A and B.
(b) P is a point on AB such that AP : PB = 3 : 1. Using section formula find the coordinates of point P.
(c) Find the equation of a line passing through P and perpendicular to AB.
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(a) From figure,
A = (4, 0) and B = (0, 4).
(b) Let coordinates of P be (x, y).
By section formula,
(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$
Substituting values we get :
$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{3 \times 0 + 1 \times 4}{3 + 1}, \frac{3 \times 4 + 1 \times 0}{3 + 1}\Big) \\[1em] = \Big(\frac{0 + 4}{4}, \frac{12 + 0}{4}\Big) \\[1em] = \Big(\frac{4}{4}, \frac{12}{4}\Big) \\[1em] = (1, 3).\end{aligned}$$
Hence, coordinates of P = (1, 3).
(c) By formula,
Slope = $\frac{y_2 - y_1}{x_2 - x_1}$
Substituting values we get :
Slope of AB = $\frac{4 - 0}{0 - 4} = \frac{4}{-4}$ = -1.
We know that,
Product of slope of perpendicular lines = -1.
∴ Slope of AB × Slope of line perpendicular to AB = -1
⇒ -1 × Slope of line perpendicular to AB = -1
⇒ Slope of line perpendicular to AB = $\frac{-1}{-1}$ = 1.
Line passing through P and perpendicular to AB :
⇒ y - y1 = m(x - x1)
⇒ y - 3 = 1(x - 1)
⇒ y - 3 = x - 1
⇒ y = x - 1 + 3
⇒ y = x + 2.
Hence, required equation is y = x + 2.