Base BC of a triangle ABC is bisected at point (p, q) and equation to side AB ,,ane,, AC are px + qy

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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 :

A

$(p - 2q) x + (q - 2p) y + 1 = 0$

B

$(p + q) (x + y) - 2 = 0$

C

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

D

none

Solution

Equation through $A$ is

$(px + qy -1) + (qx + py – 1) = 0 …..(1)$

if this represent $AB$ then, point $D (p, q)$ satisfies $(1)$

$\Rightarrow[pp + qq – 1] + \lambda [pq + pq – 1] = 0$

$\lambda = \frac{{ – (2pq – 1)}}{{{p^2} + {q^2} – 1}}$

Standard 11

Mathematics

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