The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ 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 :

If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is

Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positive $\mathrm{x}$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is :

Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle \mathrm{ABC}=\frac{\pi}{6}$ and $\angle \mathrm{BAC}=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $\mathrm{y}=\mathrm{x}+4$, then $\alpha^2+\gamma^2$ is equal to………………..

Two vertices of a triangle are $(5, – 1)$ and $( – 2,3)$. If orthocentre is the origin then coordinates of the third vertex are