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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let the line $\frac{x}{4}+\frac{y}{2}=1$ meets the x-axis and y-axis at A and B, respectively. M is the midpoint
of side AB, and M' is the image of the point M across the line x + y = 1. Let the point P lie on
the line x + y = 1 such that the $\Delta$ABP is an isosceles triangle with AP = BP. Then the
distance between M' and P is
Solution
A = $(4,0)$, B = $(0,2)$Midpoint of AB:
$M=\left(\frac{4+0}{2},\frac{0+2}{2}\right)=(2,1)$
Image of $M(2,1)$ across line $x+y=1$:
$M'=(0,-1)$
Since $AP=BP$, point $P$ lies on perpendicular bisector of AB.
Perpendicular bisector of AB passes through $M(2,1)$ and has slope $2$:
$y-1=2(x-2)$
$y=2x-3$
Also $P$ lies on:
$x+y=1$
So,
$x+2x-3=1$
$3x=4$
$x=\frac{4}{3}$
$y=1-\frac{4}{3}=-\frac{1}{3}$
So,
$P=\left(\frac{4}{3},-\frac{1}{3}\right)$
Distance between $M'(0,-1)$ and $P$:
$d=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(-\frac{1}{3}+1\right)^2}$
$d=\sqrt{\frac{16}{9}+\frac{4}{9}}$
$d=\sqrt{\frac{20}{9}}$
$d=\frac{2\sqrt5}{3}$
Final Answer: $\boxed{\frac{2\sqrt5}{3}}$
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