$ABC$ is an isosceles triangle . If the co-ordinates of the base are $(1, 3)$ and $(- 2, 7) $, then co-ordinates of vertex $A$ can be :
$\left( { - \,{\textstyle{1 \over 2}}\,\,,\,\,5} \right)$
$\left( { - \,{\textstyle{1 \over 8}}\,\,,\,\,5} \right)$
$\left( {{\textstyle{5 \over 6}}\,\,,\,\, - \,5} \right)$
$\left( { - \,7\,\,,\,\,{\textstyle{1 \over 8}}} \right)$
The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equation to the side $AB \,\,ane\,\, AC$ are $px + qy = 1 \,\,ane\,\, qx + py = 1$ . The equation of the median through $A$ is :
If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex
Show that the path of a moving point such that its distances from two lines $3 x-2 y=5$ and $3 x+2 y=5$ are equal is a straight line.
Let two points be $\mathrm{A}(1,-1)$ and $\mathrm{B}(0,2) .$ If a point $\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be such that the area of $\Delta \mathrm{PAB}=5\; \mathrm{sq}$ units and it lies on the line, $3 x+y-4 \lambda=0$ then a value of $\lambda$ is