TS EAMCET · Maths · Straight Lines
\(\mathrm{A}(-4,0)\) and \(\mathrm{B}(4,0)\) are two fixed points. \(\mathrm{C}\) and \(\mathrm{D}\) are two points on \(\mathrm{Y}\)-axis such that \(\mathrm{CD}=4\) and \(\mathrm{C}\) is a point below \(\mathrm{D}\). Then the locus of the point of intersection of the lines \(\mathrm{AC}\) and \(\mathrm{BD}\) is
- A \(x^2-y^2-x y=0\)
- B \(x^2+2 x y-16=0\)
- C \((x+y)^2-16=0\)
- D \(2 x y=16+y^2+x^2\)
Answer & Solution
Correct Answer
(B) \(x^2+2 x y-16=0\)
Step-by-step Solution
Detailed explanation
Let- \(\mathrm{C}\left(0_1 \mathrm{y}_1\right)\) and \(\mathrm{D}\left(0_1 \mathrm{y}_2\right)\) \(\therefore\) Equation of \(\mathrm{AC}\)…
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