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प्रश्न
Prove that the perpendicular drawn from the point (4, 1) on the join of (2, −1) and (6, 5) divides it in the ratio 5 : 8.
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उत्तर
Let PD be the perpendicular drawn from P (4, 1) on the line joining the points \[A\left( 2, - 1 \right) \text { and } B\left( 6, 5 \right)\].
Let m be the slope of PD.
\[\therefore m \times \text { Slope of }AB = - 1\]
\[ \Rightarrow m \times \left( \frac{5 + 1}{6 - 2} \right) = - 1\]
\[ \Rightarrow m \times \frac{6}{4} = - 1\]
\[ \Rightarrow m \times \frac{3}{2} = - 1\]
\[ \Rightarrow m = - \frac{2}{3}\]
Thus, the equation of line PD passing through P (4, 1) and having slope \[- \frac{2}{3}\] is
\[y - 1 = - \frac{2}{3}\left( x - 4 \right)\]
\[ \Rightarrow 3y - 3 = - 2x + 8\]
\[ \Rightarrow 2x + 3y - 11 = 0\]
Let D divide the line AB in the ratio k : 1
Then, the coordinates of D are \[\left( \frac{6k + 2}{k + 1}, \frac{5k - 1}{k + 1} \right)\].
Since, D lies on AB whose equation is \[2x + 3y - 11 = 0\]
Therefore, it satisfy the equation.
\[\therefore 2\left( \frac{6k + 2}{k + 1} \right) + 3\left( \frac{5k - 1}{k + 1} \right) - 11 = 0\]
\[ \Rightarrow 12k + 4 + 15k - 3 - 11k - 11 = 0\]
\[ \Rightarrow 16k = 10\]
\[ \Rightarrow k = \frac{5}{8}\]
Hence, the perpendicular drawn from the point (4, 1) on the line joining the points (2, −1) and (6, 5) divides it in the ratio 5 : 8
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