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 :
$(p - 2q) x + (q - 2p) y + 1 = 0$
$(p + q) (x + y) - 2 = 0$
$(2pq - 1) (px + qy - 1) = (p^2 + q^2 - 1) (qx + py - 1)$
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The line $3x + 2y = 24$ meets $y$-axis at $A$ and $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, - 1)$ parallel to $x$-axis at $C$. The area of the triangle $ABC$ is ............... $\mathrm{sq. \, units}$
A straight the through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q.$ If $O$ is the origin and the rectangle $OPRQ$ is completed, then the locus of $R$ is:
Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$, then the equation of locus of mid- point $P$ of $BC$ is
If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is
Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :