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Question
Find the equation of the straight line which passes through the point P (2, 6) and cuts the coordinate axes at the point A and B respectively so that \[\frac{AP}{BP} = \frac{2}{3}\] .
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, the coordinates of A and B are A (a, 0) and B(0, b).
Given:
\[\therefore 2 = \frac{2 \times 0 + 3 \times a}{2 + 3}, 6 = \frac{2 \times b + 3 \times 0}{2 + 3}\]
\[ \Rightarrow 3a = 10, 2b = 30\]
\[ \Rightarrow a = \frac{10}{3}, b = 15\]
Thus, the equation of the line is
\[\frac{x}{\frac{10}{3}} + \frac{y}{15} = 1\]
\[ \Rightarrow \frac{3x}{10} + \frac{y}{15} = 1\]
\[ \Rightarrow 9x + 2y = 30\]
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