Question

The variable line $\frac{x}{a}+\frac{y}{b}=1$ is such that $a+b=10$. The locus of the mid-point of the portion of the line intercepted between the axes is:

• A. $x+y=10$
• B. $10x+5y=1$
• C. $x+y=5$
• D. $5x+10y=1$

Answer 1

Answer: (C)

$x+y=5$

Since $a+b=10$ hence the equation of the line can be re-written as
$\frac{x}{a}+\frac{y}{10-a}=1$ ...(i)
Now the line intersects the axes at

$A=(a,0)$ and $B=(0,10-a)$

The midpoint of AB is
$G=\frac{a+0}{2},\frac{0+10-a}{2}$
$=(\frac{a}{2},\frac{10-a}{2})$
Hence
$(x,y)=(\frac{a}{2},\frac{10-a}{2})$

Therefore

$x=\frac{a}{2}$ ...(i) and $y=\frac{10-a}{2}$...(ii)
$y=5-\frac{a}{2}$
$y=5-x$ ...(from i)

Hence

$x+y=5$

Thus the required locus is $x+y=5$.