MHT CET · Maths · Straight Lines
The equations of the lines which make intercepts on the axes whose sum is 8 and
product is 15 are
- A \(3 x-5 y+15=0,5 x+3 y+15=0\)
- B \(5 x-3 y+15=0,3 x+5 y+15=0\)
- C \(3 x+5 y-15=0 \quad, 3 y+5 x-15=0\)
- D \(3 x+5 y+15=0,5 x+3 y-15=0\)
Answer & Solution
Correct Answer
(C) \(3 x+5 y-15=0 \quad, 3 y+5 x-15=0\)
Step-by-step Solution
Detailed explanation
Let ' \(\mathrm{a}\) ' and ' \(\mathrm{b}\) ' be the intercepts made by line.
We have \(a+b=8\) and \(a b=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\) ...(1)
When \(a=3, b=5\) from \((1)\), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\) ...(2)
When \(a=5, b=3\) from \((1)\), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\) ...(3)
We have \(a+b=8\) and \(a b=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\) ...(1)
When \(a=3, b=5\) from \((1)\), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\) ...(2)
When \(a=5, b=3\) from \((1)\), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\) ...(3)
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