Question

Find the equation of the straight line which divides the join of the points (2, 3) and (−5, 8) in the ratio 3 : 4 and is also perpendicular to it.

Answer 1

Let the required line divide the line joining the points \[A \left( 2, 3 \right) \text { and } B \left( - 5, 8 \right)\] at P (x₁, y₁).
Here, AP : PB = 3 : 4

\[\therefore P \left( x_1 , y_1 \right) = \left( \frac{4 \times 2 - 5 \times 3}{3 + 4}, \frac{4 \times 3 + 3 \times 8}{3 + 4} \right) = \left( - 1, \frac{36}{7} \right)\]

Now, slope of AB = \[\frac{8 - 3}{- 5 - 2} = - \frac{5}{7}\]

Let m be the slope of the required line.
Since, the required line is perpendicular to the line joining the points \[A \left( 2, 3 \right) \text { and } B \left( - 5, 8 \right)\]

\[\therefore m \times \text {Slope of the line joining the points }A\left( 2, 3 \right) \text { and } B\left( - 5, 8 \right) = - 1\]

\[ \Rightarrow m \times \left( \frac{- 5}{7} \right) = - 1\]

\[ \Rightarrow m = \frac{7}{5}\]

Substituting

\[m = \frac{7}{5}, x_1 = - 1\text {  and }y_1 = \frac{36}{7}\] in \[y - y_1 = m\left( x - x_1 \right)\] we get,

\[y - \frac{36}{7} = \frac{7}{5}\left( x + 1 \right)\]

\[ \Rightarrow 35y - 180 = 49x + 49\]

\[ \Rightarrow 49x - 35y + 229 = 0\]

Hence, the equation of the required line is \[49x - 35y + 229 = 0\]