The vertices of $ riangle PQR$ are $P(2, 1)$, $Q(-2, 3)$ and $R(4, 5)$. Find the equation of the median through the vertex R.

**OR**

If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axis are $a$ and $b$, then show that $\dfrac{1}{p^2} =

Question

The vertices of $	riangle PQR$ are $P(2, 1)$, $Q(-2, 3)$ and $R(4, 5)$. Find the equation of the median through the vertex R.

**OR**

If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axis are $a$ and $b$, then show that $\dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$.

Accepted Answer

>

> [!TLDR]
> The median from $R$ goes to the midpoint of $PQ$, which is $(0,2)$; the line through $R(4,5)$ and $(0,2)$ is $3x-4y+8=0$.
>
> The median through vertex $R$ joins $R$ to the midpoint $M$ of the opposite side $PQ$.
>
> Given $P(2,1)$ and $Q(-2,3)$:
> $$M = \left(\dfrac{2+(-2)}{2}, \dfrac{1+3}{2}\right) = (0, 2)$$
>
> The median is the line through $R(4,5)$ and $M(0,2)$. Its slope is:
> $$m = \dfrac{5-2}{4-0} = \dfrac34$$
>
> Using point-slope form through $M(0,2)$:
> $$y-2 = \dfrac34(x-0) \implies 4y-8=3x \implies 3x-4y+8=0$$
>
> **Check:** at $R(4,5)$: $3(4)-4(5)+8=12-20+8=0$. ✓
> …

Quadrant	x-sign	y-sign	Example
I	$+$	$+$	$(2, 3)$
II	$-$	$+$	$(-1, 4)$
III	$-$	$-$	$(-3, -2)$
IV	$+$	$-$	$(5, -1)$

Concept understanding — Coordinate Geometry

Coordinate Geometry: Where Algebra Meets Geometry

The Big Idea

The Coordinate System

A Concrete Example

The Four Quadrants

Why This Matters

The Precise Statement

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