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9.Straight Line
medium
ચલિત રેખા $\frac{x}{a} + \frac{y}{b} = 1, a + b = 10$ માટે, યામ અક્ષો વચ્ચે આ રેખાના અંત: ખંડના મધ્યબિંદુના બિંદુપથનું સમીકરણ
A
$10x + 5y = 1$
B
$x + y = 10$
C
$x + y = 5$
D
$5x + 10y = 1$
Solution
Since $a+b=10$ hence the equation of the line can be re-written as
$\frac{x}{a}+\frac{y}{10-a}=1 \ldots \text { (i) }$
Now the line intersects the axes at
$A=(a, 0) \text { and } B=(0,10-a)$
The midpoint of $A B$ is
$G =\frac{ a +0}{2}, \frac{0+10- a }{2}$
$=\left(\frac{ a }{2}, \frac{10- a }{2}\right)$
Hence
$(x, y)=\left(\frac{a}{2}, \frac{10-a}{2}\right)$
Therefore
$x=\frac{a}{2} \ldots \text { (i) and } y=\frac{10-a}{2} \ldots \text { (ii) }$
$y=5-\frac{a}{2}$
$y=5-x \ldots(\text { from i) }$
Hence
$x+y=5$
Thus the required locus is $x+y=5$.
Standard 11
Mathematics
