The vertices of a triangle are A(-1,3),B(-2,2) and C(3,-1).

Question

The vertices of a triangle are $A(-1,3),B(-2,2)$ and $C(3,-1)$. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Accepted Answer

Equation of $A C: x+y=2$

Equation of $A B: x-y+4=0$

Equation of $B C: 3 x+5 y=4$

The line nearest to origin is parallel to $A C$ and inward. Let its equation is $x+y=C$.

$$\therefore\left|\frac{C-2}{\sqrt{2}}\right|=1$$

$$\therefore \quad C=2-\sqrt{2}$$

$$\therefore$$ required equation line is :

$$x+y-(2-\sqrt{2})=0$$

Solutions · 1