Question

ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD ; BC and AD. Show that\(\vec{PA}+\vec{PB}+\vec {PC}+\vec{PD}=4\vec{PQ},\) where P is any point.

Answer 1

Let E, F, G and H be the midpoints of sides AB, BC, CD and DA respectively of quadrilateral ABCD. Let the position vectors of these vertices and midpoints be as shown in the figure.

As E is the midpoint of AB, using midpoint formula,

we have

We know that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other. ⇒ Q is the midpoint of EG and HF.

Once again using midpoint formula, we get \(\vec q=\cfrac{\vec e+\vec g}2\)

But, we found \(\vec e=\cfrac{\vec a+\vec b}2\) and \(\vec g=\cfrac{\vec c+\vec d}2\)

Now, consider the vector \(\vec {PA}+\vec{PB}+\vec{PC}+\vec{PD}\)

Let the position vector of point P be \(\vec p\).

Recall the vector \(\vec{PA}\) is given by

\(\vec {PA}\) = position vector of A - position vector of P

⇒ \(\vec {PA}\)= \(\vec a-\vec p\)

Similarly, \(\vec {PB}=\vec b-\vec p,\vec{PC}=\vec c-\vec p\) and \(\vec{PD}=\vec d-\vec p.\)

Observe, \(\vec q-\vec p\) = position vector of Q - position vector of P