Finding distance of mid-point $M$ of line-segment $PQ$ from origin

Question

In the figure, point $M$ is the midpoint of the line segment $PQ$. Show that $c = \frac12(a + b)$.

Hello, I couldn't figure out how to show it because I don't have vector points. And there are no vector angles. How should I go about it?

Answer 1

$\vec M=\frac12(\vec P+\vec Q)\implies\|\vec M\|=c=\frac12\sqrt{a^2+b^2+2ab\cos\theta}$ where $\theta$ is the angle between $\vec P,\vec Q$. Without knowing $\theta$, we can only find bounds on $c$:$$-1\le\cos\theta\le1\implies(a-b)^2\le a^2+b^2+2ab\cos\theta\le(a+b)^2\\\implies c\in\left[\frac12|a-b|,\frac12(a+b)\right]$$

Answer 2

Length $c$ is a half of their vector sum in a parallelogram resultant (half) computable by trig cosine rule.

$$ c=\frac12 \sqrt{a^2+b^2+ 2ab \cos POQ}$$

When vectors sense is (same, opposite) we have respectively half resultant

$$ c= ( \dfrac{a+b}{2}, \dfrac{a-b}{2})$$

(PM = MQ given lengths, marking error)

Question is misleading, true if they are vectors.