MHT CET · Maths · Pair of Lines
If the equation \(k x y+5 x+3 y+2=0\) represents a pair of lines, then \(k=\)
- A \(\frac{15}{2}\)
- B \(1, \frac{15}{2}\)
- C 15
- D \(40, \frac{-15}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{15}{2}\)
Step-by-step Solution
Detailed explanation
(C)
Comparing given equation with
\(\mathrm{Ax}^{2}+2 \mathrm{Hxy}+\mathrm{By}^{2}+2 \mathrm{Gx}+2 \mathrm{Fy}+\mathrm{c}=0\), we get
\(A =0, B=0, C=2, F=\frac{3}{2} \cdot G=\frac{5}{2}, H=\frac{k}{2} \)
\( \Delta =A B C+2 F G H-A F^{2}-B G^{2}-C H^{2}=0 \)
\( \Delta =0+2 \frac{(3)}{2} \frac{(5)}{2}\left(\frac{k}{2}\right)-0-0-2\left(\frac{k}{2}\right)^{2}=0 \)
\( =\frac{15}{4} k-\frac{k^{2}}{2}=0 \Rightarrow 15 k-2 k^{2}=0 \Rightarrow k(2 k-15)\) \(=0 \Rightarrow k=0, \frac{15}{2}\)
Comparing given equation with
\(\mathrm{Ax}^{2}+2 \mathrm{Hxy}+\mathrm{By}^{2}+2 \mathrm{Gx}+2 \mathrm{Fy}+\mathrm{c}=0\), we get
\(A =0, B=0, C=2, F=\frac{3}{2} \cdot G=\frac{5}{2}, H=\frac{k}{2} \)
\( \Delta =A B C+2 F G H-A F^{2}-B G^{2}-C H^{2}=0 \)
\( \Delta =0+2 \frac{(3)}{2} \frac{(5)}{2}\left(\frac{k}{2}\right)-0-0-2\left(\frac{k}{2}\right)^{2}=0 \)
\( =\frac{15}{4} k-\frac{k^{2}}{2}=0 \Rightarrow 15 k-2 k^{2}=0 \Rightarrow k(2 k-15)\) \(=0 \Rightarrow k=0, \frac{15}{2}\)
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