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Question
Find the equation of the line, which passes through P (1, −7) and meets the axes at A and Brespectively so that 4 AP − 3 BP = 0.
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Solution
The equation of the line with intercepts a and b is
\[\frac{x}{a} + \frac{y}{b} = 1\].
Since the line meets the coordinate axes at A and B, so the coordinates are A (a, 0) and B (0, b).
Given:
\[4AP - 3BP = 0\]
\[ \Rightarrow AP : BP = 3 : 4\]
Here,
\[P \equiv \left( 1, - 7 \right)\]
\[\therefore 1 = \frac{3 \times 0 + 4 \times a}{3 + 4}, - 7 = \frac{3 \times b + 4 \times 0}{3 + 4}\]
\[ \Rightarrow 4a = 7, 3b = - 49\]
\[ \Rightarrow a = \frac{7}{4}, b = - \frac{49}{3}\]
Thus, the equation of the line is \[\frac{x}{\frac{7}{4}} + \frac{y}{- \frac{49}{3}} = 1\]
\[\Rightarrow \frac{4x}{7} - \frac{3y}{49} = 1\]
\[ \Rightarrow 28x - 3y = 49\]
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