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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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

Similar Questions

The portion of the line $4x + 5y = 20$ in the first quadrant is trisected by the lines $L_1$ and $L_2$ passing through the origin. The tangent of an angle between the lines $L_1$ and $L_2$ is:

$A$ line,with a slope greater than $1$,passes through the point $A(4, 3)$ and intersects the line $x - y - 2 = 0$ at the point $B$. If the length of the line segment $AB$ is $\frac{\sqrt{29}}{3}$,then $B$ also lies on the line:

The area of the triangle formed by the coordinate axes and the line $px + qy = r$ (where $p, q$ and $r$ are positive real numbers) is $\frac{r^2}{2pq}$. If this area is $\frac{1}{54}$ sq. units, then which of the following is true?

The system of equations $4x + 6y = 5$ and $8x + 12y = 10$ has

$A$ straight line passing through the origin $O$ meets the parallel lines $4x + 2y = 9$ and $2x + y + 6 = 0$ at the points $P$ and $Q$ respectively. Then the point $O$ divides the line segment $PQ$ in the ratio:

Vedclass Test Series

Exam Paper Generator

Online Exam Module