Question

Vertices of a triangle are $(-1,3),(-2,2)$ and $(3,-1)$. A triangle is formed by translating the sides of the given triangle by one unit inwards. Find the equation of that side of the new triangle which is nearest to the origin.

• A. x+y-(2-√2)=0
• B. x+y-(2+√2)=0
• C. x-y-(2-√2)=0
• D. x-y-(2+√2)=0

Answer 1

The correct option is A x+y-(2-√2)=0

Equation of side $AB$ is $y-2=1(x+2)$
i.e. $x-y+4=0\cdots \left(1\right)$
equation of side $BC$ is $y-2=\frac{-3}{5}(x+2)$
i.e.,$3x+5y-4=0\cdots \left(2\right)$
the perpendicular distance of the three sides from the origin are respectively.
$2\sqrt{2},\frac{4}{\sqrt{34}}$and $\sqrt{2}$.
If $A^{\prime }{B}^{\prime }{C}^{\prime }$ be the new triangle formed by shifting the sides of the triangle $ABC$ inwards by one unit, then the perpendicular distance of the sides of the new triangle from the origin will be (see fig)., respectively $(2\sqrt{2}-1),\frac{4}{\sqrt{34}}+1$ and $(\sqrt{2}-1).$.

we see that side $C^{\prime }{A}^{\prime }$, whose perpendicular distance from the origin is $(\sqrt{2}-1),$ is nearest to the origin
equation of side $C^{\prime }{A}^{\prime }$ can be chosen as $x+y+c=0.$
then , we have $\frac{|c|}{\sqrt{2}}=\sqrt{2}-1$
i.e., $c=\pm (2-\sqrt{2})$.
it can be seen that the side $C^{\prime }{A}^{\prime }$ will lie above the origin. therefore, $c$ must be $-$ve, hence, equation of the required side, i.e., $C^{\prime }{A}^{\prime }$ is $x+y-(2-\sqrt{2})=0.$