Coordinate Geometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Coordinate Geometry — Quick Facts

Distance Formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Midpoint Formula: $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Gradient (Slope) of a Line: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$$

Equation of a Straight Line:

• Slope-intercept form: $y = mx + c$
• Point-slope form: $y - y_1 = m(x - x_1)$
• General form: $Ax + By + C = 0$

Parallel Lines: $m_1 = m_2$ Perpendicular Lines: $m_1 \times m_2 = -1$

⚡ JAMB Exam Tip: When finding the equation of a line through two points, first find the gradient $m = (y_2 - y_1)/(x_2 - x_1)$, then substitute one point into $y - y_1 = m(x - x_1)$.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Coordinate Geometry — Study Guide

Gradient Interpretation:

• $m > 0$: Line slopes upward (left to right)
• $m < 0$: Line slopes downward
• $m = 0$: Horizontal line
• $m$ undefined: Vertical line

Angle Between Two Lines: $$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$

Worked Example:

Find the equation of the line passing through (2, 5) and (-3, 7).

$m = \frac{7 - 5}{-3 - 2} = \frac{2}{-5} = -\frac{2}{5}$

Using point-slope form with (2, 5): $y - 5 = -\frac{2}{5}(x - 2)$ $5y - 25 = -2x + 4$ $2x + 5y - 29 = 0$

Perpendicular Lines Example:

Find the equation of the line perpendicular to $3x - 2y + 5 = 0$ passing through (4, 1).

Original line: $3x - 2y + 5 = 0$ $m_1 = \frac{3}{2}$

Perpendicular: $m_2 = -\frac{2}{3}$

Using point (4, 1): $y - 1 = -\frac{2}{3}(x - 4)$ $3y - 3 = -2x + 8$ $2x + 3y - 11 = 0$

Section Formula:

For point P dividing AB in ratio $m:n$ (internally): $$P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$$

⚡ Common Student Mistake: Confusing internal and external section formulas. For internal division, $m$ and $n$ are both positive. For external division, use $m:n$ but one coordinate formula has a minus sign.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Coordinate Geometry — Comprehensive Notes

Area of a Triangle:

Given vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_2, y_3)$: $$\text{Area} = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$$

Collinearity Test: Three points are collinear if the area of the triangle formed is zero, or if: $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_1}{x_3 - x_1}$$

Equation of a Line in Intercept Form:

If a line cuts the x-axis at $(a, 0)$ and y-axis at $(0, b)$: $$\frac{x}{a} + \frac{y}{b} = 1$$

Perpendicular Distance:

From point $(x_1, y_1)$ to line $Ax + By + C = 0$: $$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Worked Example:

Find the distance from (3, 2) to the line $4x - 3y + 7 = 0$.

$d = \frac{|4(3) - 3(2) + 7|}{\sqrt{16 + 9}} = \frac{|12 - 6 + 7|}{\sqrt{25}} = \frac{13}{5} = 2.6$

Locus Problems:

Example 1: Find the locus of points equidistant from (2, 3) and (4, 7).

Using distance formula, set distances equal: $(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-7)^2$ Expanding and simplifying: $x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 14y + 49$ $4x + 8y - 52 = 0$ $x + 2y - 13 = 0$

Example 2: Locus of points 3 units from the x-axis.

$y = \pm 3$

Circle in Coordinate Geometry:

The general equation of a circle with centre $(h, k)$ and radius $r$: $$(x - h)^2 + (y - k)^2 = r^2$$

Expanding: $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$

Intersection of Line and Circle:

Substitute the line equation into the circle equation, then solve the resulting quadratic. The discriminant determines:

• $D > 0$: Two intersection points (secant)
• $D = 0$: One intersection point (tangent)
• $D < 0$: No intersection points

JAMB Pattern Analysis (2015-2024):

• 2015: Distance between two points
• 2017: Gradient and equation of a line through two points
• 2019: Perpendicular distance from point to line
• 2021: Area of triangle using coordinates
• 2023: Equation of circle given centre and radius
• 2024: Perpendicular bisector as a locus

⚡ Exam Strategy: For “find the equation of” problems, always identify the gradient first. For locus problems, use the distance formula method: express the condition algebraically, then simplify.