AP EAMCET · Maths · Three Dimensional Geometry
If a variable straight line passing through the point of intersection of the lines \(x-2 y+3=0\) and \(2 x-y-1=0\) intersects the \(\mathrm{X}, \mathrm{Y}\)-axes at A and B respectively, then the equation of the locus of a point which divides the segment AB in the ratio \(-2: 3\) is
- A \(14 x^2+3 x y-15 y^2=0\)
- B \(x y=14 x+15 y\)
- C \(x^2+x y-y^2=0\)
- D \(14 x+3 x y-15 y=0\)
Answer & Solution
Correct Answer
(D) \(14 x+3 x y-15 y=0\)
Step-by-step Solution
Detailed explanation
Since, equation of a line passing through the two given lines is \(x-2 y+3+K(2 x-y-1)=0\) \(\Rightarrow(2 K+1) x+(-K-2) y-K+3=0\) \(\Rightarrow \frac{(2 K+1)}{(K-3)} x+\frac{(-K-2)}{(K-3)} y=1\) So, \(A\left(\frac{K-3}{2 K+1}, 0\right), B\left(0, \frac{3-K}{K+2}\right)\) Now,…
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