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

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(A) The median passes through vertex $A$,which is the intersection of the lines $px + qy - 1 = 0$ and $qx + py - 1 = 0$. Any line passing through the

Vedclass · Accepted Answer

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

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(A) The median passes through vertex $A$,which is the intersection of the lines $px + qy - 1 = 0$ and $qx + py - 1 = 0$. Any line passing through the intersection of these two lines is given by the family of lines equation: $(px + qy - 1) + \lambda(qx + py - 1) = 0$,where $\lambda$ is a constant. Since the median also passes through the midpoint of the base $BC$,which is the point $(p, q)$,we substitute this point into 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 family of lines equation: $(px + qy - 1) - \frac{p^2 + q^2 - 1}{2pq - 1}(qx + py - 1) = 0$ $(2pq - 1)(px + qy - 1) = (p^2 + q^2 - 1)(qx + py - 1)$.

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

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