The base BC of a triangle ABC is bisected at | vedclass

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(C) Let the equation of the median through $A$ be $(px + qy - 1) + \lambda(qx + py - 1) = 0$. Since this line passes through the midpoint $D(p, q)$ of

Vedclass · Accepted Answer

$(2pq - 1) (px + qy - 1) = (p^2 + q^2 - 1) (qx + py - 1)$

Vedclass · Answer

(C) Let the equation of the median through $A$ be $(px + qy - 1) + \lambda(qx + py - 1) = 0$. Since this line passes through the midpoint $D(p, q)$ of $BC$,it must satisfy the equation: $(p(p) + q(q) - 1) + \lambda(q(p) + p(q) - 1) = 0$ $(p^2 + q^2 - 1) + \lambda(2pq - 1) = 0$ $\lambda = -\frac{p^2 + q^2 - 1}{2pq - 1}$ Substituting $\lambda$ back into the equation of the line: $(px + qy - 1) - \frac{p^2 + q^2 - 1}{2pq - 1}(qx + py - 1) = 0$ $(2pq - 1)(px + qy - 1) = (p^2 + q^2 - 1)(qx + py - 1)$

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations of the sides $AB$ and $AC$ are $px + qy = 1$ and $qx + py = 1$. The equation of the median through $A$ is:

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