Question

The distance between the point $(1,2)$ and the point of intersection of the line $2x+y=2$ and $x+2y=2$ is

• A. $\frac{\sqrt{17}}{3}$
• B. $\frac{\sqrt{16}}{3}$
• C. $\frac{\sqrt{17}}{5}$
• D. $\frac{\sqrt{19}}{3}$
• E. $\frac{\sqrt{19}}{5}$

Answer 1

The correct option is A $\frac{\sqrt{17}}{3}$

The point of intersection of the lines $2x+y=2$ and $x+2y=2$ can be calculated by putting $y=2-2x$ in $x+2y=2$
$\Rightarrow x+2(2-2x)=2$
$\Rightarrow x+4-4x=2$
$\Rightarrow -3x=-2$
$\Rightarrow x=\frac{2}{3}$
Also, $y=2-2\left(\frac{2}{3}\right)=\frac{2}{3}$
Hence, point of intersection is $(\frac{2}{3},\frac{2}{3})$
Now, the required distance $=\sqrt{{(\frac{2}{3}-1)}^2+{(\frac{2}{3}-2)}^2}$
$=\sqrt{\frac{1}{9}+\frac{16}{9}}=\frac{\sqrt{17}}{3}$