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

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Question

(a) Since the median passes through $A$, the intersection of the given lines. Its equation is given by $(px + qy - 1) + \lambda (qx + py - 1) = 0$, wh

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Since the median passes through $A$, the intersection of the given lines. Its equation is given by $(px + qy - 1) + \lambda (qx + py - 1) = 0$, where $\lambda $ is some real number. Also, since the median passes through the point $(p, q)$, we have $({p^2} + {q^2} - 1) + \lambda (qp + pq - 1) = 0$

==> $\lambda = - \frac{{{p^2} + {q^2} - 1}}{{2pq - 1}}$ and the equation of median through $A$ is $(px + qy - 1) - \frac{{{p^2} + {q^2} - 1}}{{2pq - 1}}(qx + py - 1) = 0$

==> $(2pq - 1)(px + qy - 1) = ({p^2} + {q^2} - 1)(qx + py - 1)$.

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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