Question

A line AB meets the x-axis at point A and y-axis at point B. The point P(−4, −2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:

1. the co-ordinates of A and B.
2. equation of line through P and perpendicular to AB.

Answer 1

∵ Line AB intersects x-axis at A and y-axis at B.

i. Let co-ordinates of A be (x, 0) and of B be (0, y)

Point P(−4, –2) intersects AB in the ratio 1 : 2 internally

∴ ${\displaystyle -4=\frac{1\times 0+2\times x}{1+2}}$  ...${\displaystyle (x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2})}$

${\displaystyle \Rightarrow -4=\frac{0+2x}{3}}$

${\displaystyle \Rightarrow }$ 2x = –12

∴ ${\displaystyle x=\frac{-12}{2}=-6}$ and ${\displaystyle -2=\frac{1\times y+2\times 0}{1+2}}$  ...${\displaystyle [\because y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}]}$

${\displaystyle \Rightarrow -2=\frac{y+0}{3}}$

${\displaystyle \Rightarrow }$ y = −6

∴ Co-ordinates of A will be (−6, 0) and Co-ordinates of B will be (0, −6)

ii. Now, slope of AB (m₁)

= ${\displaystyle \frac{y_2-y_1}{x_2-x_2}}$

= ${\displaystyle \frac{-6-0}{0-(-6)}}$

= ${\displaystyle \frac{-6}{6}}$

= −1

∴ Slope of the line perpendicular to AB (m₂) = 1  ...(∵ m₁m₂ = −1)

∴ Equation of line perpendicular to AB and drawn through P(−4, −2) will be

y − y₁ = m(x − x₁)

${\displaystyle \Rightarrow }$ y + 2 = 1(x + 4)

${\displaystyle \Rightarrow }$ y + 2 = x + 4

${\displaystyle \Rightarrow }$ y = x + 4 − 2

${\displaystyle \Rightarrow }$ y = x + 2

Hence, required equation of the line is y = x + 2