KCET · Maths · Straight Lines
A variable line \(\frac{x}{a}+\frac{y}{b}=1\) is such that \(a+b=4\). The locus of the mid point of the portion of the line intercepted between the axes is
- A \(x+y=4\)
- B \(x+y=8\)
- C \(x+y=1\)
- D \(x+y=2\)
Answer & Solution
Correct Answer
(D) \(x+y=2\)
Step-by-step Solution
Detailed explanation
Let the coordinate of mid point of \(A B\) is \(\left(x_{1}, y_{1}\right)\).
\(
\begin{aligned}
&\Rightarrow \quad \mathrm{x}_{1}=\frac{\mathrm{a}+0}{2}, \mathrm{y}_{1}=\frac{0+\mathrm{b}}{2} \\
&\therefore \quad \mathrm{a}=2 \mathrm{x}_{1}, \mathrm{~b}=2 \mathrm{y}_{1} \\
&\text { Given, } \quad \mathrm{a}_{1}+2 \mathrm{~b}_{1}=4 \\
&\therefore \quad \mathrm{x}_{1}+\mathrm{y}_{1}=2
\end{aligned}
\)
Hence, the locus of the mid point is
\(
x+y=2
\)
\(
\begin{aligned}
&\Rightarrow \quad \mathrm{x}_{1}=\frac{\mathrm{a}+0}{2}, \mathrm{y}_{1}=\frac{0+\mathrm{b}}{2} \\
&\therefore \quad \mathrm{a}=2 \mathrm{x}_{1}, \mathrm{~b}=2 \mathrm{y}_{1} \\
&\text { Given, } \quad \mathrm{a}_{1}+2 \mathrm{~b}_{1}=4 \\
&\therefore \quad \mathrm{x}_{1}+\mathrm{y}_{1}=2
\end{aligned}
\)
Hence, the locus of the mid point is
\(
x+y=2
\)
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