Training Geometry Coordinate Geometry
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Coordinate Geometry

20 min Geometry

Coordinate Geometry

Coordinate geometry — also called analytic geometry — uses the coordinate plane to study geometric figures with algebraic tools. By placing shapes on a grid, you can calculate distances, midpoints, slopes, and areas using formulas rather than rulers and protractors.

The distance formula, the midpoint formula, and the slope formula are the three essential tools of coordinate geometry. The distance formula is really just the Pythagorean theorem in disguise, connecting geometry and algebra in a beautiful way.

This lesson covers all three formulas along with applications such as determining whether lines are parallel or perpendicular, classifying triangles, and finding the equation of a circle.

Coordinate geometry (analytic geometry) combines algebra and geometry by placing shapes on the coordinate plane.

Key Formulas
FormulaExpression
Distance between $(x_1,y_1)$ and $(x_2,y_2)$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Midpoint$M = \left(\dfrac{x_1+x_2}{2},\; \dfrac{y_1+y_2}{2}\right)$
Slope$m = \dfrac{y_2-y_1}{x_2-x_1}$
Example 1

Find the distance between $(3, -2)$ and $(-1, 5)$.

$$d = \sqrt{(-1-3)^2 + (5-(-2))^2} = \sqrt{16 + 49} = \sqrt{65}$$

Example 2

Find the midpoint of $(4, -6)$ and $(10, 2)$.

$$M = \left(\frac{4+10}{2},\; \frac{-6+2}{2}\right) = (7, -2)$$

Example 3

Find the slope of the line through $(2, 3)$ and $(5, 9)$.

$$m = \frac{9-3}{5-2} = \frac{6}{3} = 2$$

Example 4

Show that $(1,1)$, $(4,5)$, and $(7,9)$ are collinear.

Slope from $(1,1)$ to $(4,5)$: $m = \frac{4}{3}$.

Slope from $(4,5)$ to $(7,9)$: $m = \frac{4}{3}$.

Equal slopes → collinear. ✓

Example 5

Find the perimeter of a triangle with vertices $A(0,0)$, $B(6,0)$, $C(3,4)$.

$AB = 6$, $BC = \sqrt{9+16} = 5$, $AC = \sqrt{9+16} = 5$.

$P = 6 + 5 + 5 = 16$.

Practice Problems

1. Distance between $(0, 0)$ and $(5, 12)$.
2. Midpoint of $(-3, 7)$ and $(5, -1)$.
3. Slope of line through $(1, 4)$ and $(3, -2)$.
4. Distance between $(-2, 3)$ and $(4, -5)$.
5. Midpoint of $(0, 0)$ and $(8, 6)$.
6. What is the slope of a horizontal line?
7. Two points $(1, 2)$ and $(1, 7)$. Find the slope.
8. Is the triangle with vertices $(0,0)$, $(3,0)$, $(0,4)$ a right triangle? Find the area.
9. Find $x$ if the distance from $(x, 0)$ to $(0, 3)$ is 5.
10. The midpoint of $(a, 3)$ and $(7, 11)$ is $(5, 7)$. Find $a$.
11. Find the perimeter of a square with vertices $(0,0)$, $(4,0)$, $(4,4)$, $(0,4)$.
12. Are lines with slopes $\frac{2}{3}$ and $-\frac{3}{2}$ perpendicular?
13. Find the equation of the line through $(2,5)$ with slope 3.
14. Distance from $(1,1)$ to $(-4,-11)$.
15. Find the center and radius of a circle: $(x-3)^2 + (y+2)^2 = 25$.
Show Answer Key

1. $13$

2. $(1, 3)$

3. $\frac{-6}{2} = -3$

4. $\sqrt{36 + 64} = 10$

5. $(4, 3)$

6. $0$

7. Undefined (vertical line)

8. Yes (legs along axes). $A = \frac{1}{2}(3)(4) = 6$

9. $x^2 + 9 = 25 \implies x = \pm 4$

10. $\frac{a+7}{2} = 5 \implies a = 3$

11. $4 \times 4 = 16$

12. Yes. $\frac{2}{3} \times (-\frac{3}{2}) = -1$ ✓

13. $y - 5 = 3(x - 2) \implies y = 3x - 1$

14. $\sqrt{25 + 144} = 13$

15. Center $(3, -2)$, radius $5$

Surface Area and Volume Placement Test Practice — Geometry