Question

The base \[{\rm{BC}}\] of a triangle \[{\rm{ABC}}\] is bisected at the point \[({\rm{p}},{\rm{q}})\] and the equations to the sides \[{\rm{AB}}\] and \[{\rm{AC}}\] are respectively \[x + y + 3 = 0\] and \[qx + py = 1\], then the equation to the median through \[{\rm{A}}\] is
A. \[2x - y = 9\]
B. \[\left( {{p^2} + {q^2} - 1} \right)(px + qy - 1) = (2p - 1)(qx + py - 1)\]
C. \[(pq - 1)(px + qy - 1) = \left( {{p^2} + {q^2} - 1} \right)(qx + py - 1)\]
D. None of these

Answer 1

Hint: First, create the equation for the family of lines going through AB and AC, and then, from that family of lines, one line is the median and passes through point \[\left( {p,{\rm{ }}q} \right)\]and if you satisfy this point \[\left( {p,{\rm{ }}q} \right)\] in the line family, you will receive the equation of the median flowing through A.

Formula Used: Family of lines passing through the point can be calculated using the formula,
\[u + kv = 0\]
Where: \[u = 0\] \[v = 0\] and \[k\]- parameter

Complete step by step solution: We have been provided in the data given in the question that,
A triangle’s \[{\rm{ABC}}\]base \[{\rm{BC}}\] bisects at the point \[({\rm{p}},{\rm{q}})\]
The equations to the sides \[{\rm{AB}}\] and \[{\rm{AC}}\] are respectively
\[x + y + 3 = 0\]---\[{L_1}\]
\[qx + py = 1\]----\[{L_2}\]


 Let us assume that the equation of side \[AB\] is \[{L_1}\]
And the equation of side \[BC\] be assumed as \[{L_2}\]
Now, we have to construct the line family that runs through \[AB\] and \[AC\] we have
\[{L_1} + \lambda {L_2} = 0\]
On substituting the value of \[{L_1}\] and \[{L_2}\] in the above equation, we get
\[px + qy - 1 + \lambda (qx + py - 1) = 0\]------ (1)
Now, one of the families of lines in this equation is the median and passes through point \[\left( {p,{\rm{ }}q} \right)\] hence satisfying the point \[\left( {p,{\rm{ }}q} \right)\] in the preceding equation will produce the value of \[\lambda \]
\[{p^2} + {q^2} - 1 + \lambda (qp + pq - 1) = 0\]
Now, we have to simplify the terms inside the parentheses, we get
\[ \Rightarrow \lambda (2pq - 1) = 1 - {p^2} - {q^2}\]
On solving for \[\lambda \] we get
\[ \Rightarrow \lambda = \dfrac{{1 - {p^2} - {q^2}}}{{2pq - 1}}\]
On substituting the value of \[\lambda \] in the equation (1), we get
\[px + qy - 1 + \dfrac{{1 - {p^2} - {q^2}}}{{2pq - 1}}(qx + py - 1) = 0\]
Now, we have to solve the fraction by multiplying the numerator with terms on LHS of the equation and vice versa, we get
\[ \Rightarrow (2pq - 1)(px + qy - 1) + \left( {1 - {p^2} - {q^2}} \right)(qx + py - 1) = 0\]
Now, let us restructure the obtained equation as,\[ \Rightarrow (2pq - 1)(px + qy - 1) = (qx + py - 1)\left( {{p^2} + {q^2} - 1} \right)\]
Therefore, the equation to the median through \[{\rm{A}}\] is \[(2pq - 1)(px + qy - 1) = (qx + py - 1)\left( {{p^2} + {q^2} - 1} \right)\]

Option ‘C’ is correct

Note: There is another technique to solve the problem by first finding the junction point of lines travelling through sides \[AB,AC\] and then getting the coordinates of \[A\] and a point \[\left( {p,{\rm{ }}q} \right)\]and if we have two points, we can write the equation of a straight line.