Coordinate Method

Use coordinates to prove that the diagonals of a rectangle bisect each other.

Proof. Place the rectangle with one vertex at the origin: $A(0, 0)$, $B(a, 0)$, $C(a, b)$, $D(0, b)$, where $a, b > 0$.

The two diagonals are $AC$ (from $A(0,0)$ to $C(a,b)$) and $BD$ (from $B(a,0)$ to $D(0,b)$).

Midpoint of $AC$:

$M_{AC} = \left(\frac{0 + a}{2},\; \frac{0 + b}{2}\right) = \left(\frac{a}{2},\; \frac{b}{2}\right)$

Midpoint of $BD$:

$M_{BD} = \left(\frac{a + 0}{2},\; \frac{0 + b}{2}\right) = \left(\frac{a}{2},\; \frac{b}{2}\right)$

Since $M_{AC} = M_{BD}$, the diagonals share the same midpoint, which means they bisect each other. $\blacktriangleleft$

Find the locus of all points equidistant from $A(1, 3)$ and $B(5, 1)$.

Solution. Let $P(x, y)$ satisfy $|PA| = |PB|$. Then:

$|PA|^2 = |PB|^2$

$(x - 1)^2 + (y - 3)^2 = (x - 5)^2 + (y - 1)^2$

Expanding both sides:

$x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 - 2y + 1$

$-2x + 1 - 6y + 9 = -10x + 25 - 2y + 1$

$8x - 4y - 16 = 0$

$2x - y - 4 = 0 \quad \text{i.e.,} \quad y = 2x - 4$

Geometric interpretation. This is the perpendicular bisector of $AB$.

Let us verify: the midpoint of $AB$ is $M\left(3, 2\right)$, which satisfies $y = 2(3) - 4 = 2$ $\checkmark$. The slope of $AB$ is $k_{AB} = \frac{1-3}{5-1} = -\frac{1}{2}$, and the slope of the perpendicular bisector is $k = 2$, giving $k_{AB} \cdot k = -\frac{1}{2} \cdot 2 = -1$ $\checkmark$ (perpendicular).

Answer: The locus is the line $2x - y - 4 = 0$. $\blacktriangleleft$

Prove using coordinates that the three medians of triangle $ABC$ are concurrent (meet at a single point).

Proof. Place the triangle with vertices $A(0, 0)$, $B(2b, 0)$, $C(2c, 2d)$ (choosing coordinates to keep the midpoints integer).

The midpoints of the sides are:

• $M_A$ = midpoint of $BC$ = $\left(b + c,\; d\right)$
• $M_B$ = midpoint of $AC$ = $\left(c,\; d\right)$
• $M_C$ = midpoint of $AB$ = $\left(b,\; 0\right)$

The point $G = \left(\dfrac{0 + 2b + 2c}{3},\; \dfrac{0 + 0 + 2d}{3}\right) = \left(\dfrac{2(b+c)}{3},\; \dfrac{2d}{3}\right)$ divides each median in ratio $2:1$ from vertex to midpoint.

Check that $G$ lies on median $AM_A$ (from $A(0,0)$ to $M_A(b+c, d)$):

$G$ divides $AM_A$ in ratio $2:1$, so $G = \left(\frac{2(b+c)}{3}, \frac{2d}{3}\right)$ $\checkmark$.

By symmetry, $G$ lies on all three medians. Therefore the medians are concurrent at $G$. $\blacktriangleleft$

Use coordinates to prove that the midpoints of the sides of any quadrilateral form a parallelogram (Varignon’s theorem).

Hint: Place the vertices of the quadrilateral at $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$. Find the midpoints of the four sides, then show that opposite sides of the midpoint quadrilateral are parallel (equal slopes) and equal in length.

Find the locus of points $P(x, y)$ such that $|PA|^2 + |PB|^2 = 50$, where $A(-3, 0)$ and $B(3, 0)$.

Describe the locus geometrically (what figure does the equation represent?).

Using coordinates, prove that the diagonals of a rhombus are perpendicular.

Hint: Let the rhombus have vertices $A(a, 0)$, $B(0, b)$, $C(-a, 0)$, $D(0, -b)$. Compute the slopes of diagonals $AC$ and $BD$ and verify their product equals $-1$.

A point $P$ moves so that it is always twice as far from $A(6, 0)$ as from the origin $O(0, 0)$: that is, $|PA| = 2|PO|$. Find and identify the locus of $P$.

Hint: Write out $|PA|^2 = 4|PO|^2$, expand, and complete the square.