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KEAM 2015 Math Paper

Question : 75 of 120

Marks: +1, -0

Solution:

The point of intersection of the lines

2 x+y=2

and

x+2 y=2

can be calculated by putting

y=2-2 x

x+2 y=2

\Rightarrow\;\; x+2(2-2x)=2

\Rightarrow\;\; x+4-4x=2

\Rightarrow\;\; -3x=-2

\Rightarrow\;\; x=\;\frac{2}{3}

Also,

\;\; y=2-2\left(\frac{2}{3}\right)=\frac{2}{3}

Hence, point of intersection is

\left(\frac{2}{3}, \frac{2}{3}\right)

Now, the required distance

=\sqrt{\left(\frac{2}{3}-1\right)^2+\left(\frac{2}{3}-2\right)^2}

=\sqrt{\frac{1}{9}+\frac{16}{9}}=\frac{\sqrt{17}}{3}

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