Coordinate Geometry | Jamb Mathematics

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Hey love, it's time to get ready for your big date with Coordinate Geometry—trust me, it’s a subject worth falling for! Let’s make sure you and those equations have the perfect chemistry, from slopes that never let you down to circles that keep you wrapped in their embrace. So grab your notebook, and let’s turn this study session into a love story with the most beautiful graphs you've ever seen! 💕📐

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Coordinate Geometry? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Coordinate Geometry together and move one step closer to achieving your exam success! Blissful learning.

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Question 1

Find the midpoint and gradient of the line segment joining the points

A(2,3)

and

B(8,7)

Solution* Step 1: Midpoint formula The midpoint

M

of a line segment with endpoints

(x_1, y_1)

and

(x_2, y_2)

is given by:

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

Substituting

A(2,3)

and

B(8,7)

M = \left(\frac{2+8}{2}, \frac{3+7}{2}\right) = (5, 5)

Step 2: Gradient formula The gradient

m

of a line through

(x_1, y_1)

and

(x_2, y_2)

is:

m = \frac{y_2 - y_1}{x_2 - x_1}

Substituting values:

m = \frac{7 - 3}{8 - 2} = \frac{4}{6} = \frac{2}{3}

Final Answer: Midpoint =

(5,5)

, Gradient =

\frac{2}{3}

Question 2

Find the midpoint and gradient of the line segment connecting

C(-4,6)

and

D(10,-2)

Solution Step 1: Midpoint

M = \left(\frac{-4+10}{2}, \frac{6+(-2)}{2}\right) = \left(\frac{6}{2}, \frac{4}{2}\right) = (3,2)

Step 2: Gradient

m = \frac{-2 - 6}{10 - (-4)} = \frac{-8}{14} = \frac{-4}{7}

Final Answer: Midpoint =

(3,2)

, Gradient =

\frac{-4}{7}

Question 3

Find the midpoint and gradient for the points

E(1, -5)

and

F(7,3)

Solution

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Step 1: Midpoint

M = \left(\frac{1+7}{2}, \frac{-5+3}{2}\right) = \left(\frac{8}{2}, \frac{-2}{2}\right) = (4, -1)

Step 2: Gradient

m = \frac{3 - (-5)}{7 - 1} = \frac{3+5}{6} = \frac{8}{6} = \frac{4}{3}

Final Answer: Midpoint =

(4,-1)

, Gradient =

\frac{4}{3}

Question 4

Find the midpoint and gradient for

G(-6,4)

and

H(2,10)

Solution

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Step 1: Midpoint

M = \left(\frac{-6+2}{2}, \frac{4+10}{2}\right) = \left(\frac{-4}{2}, \frac{14}{2}\right) = (-2, 7)

Step 2: Gradient

m = \frac{10 - 4}{2 - (-6)} = \frac{6}{8} = \frac{3}{4}

Final Answer: Midpoint =

(-2,7)

, Gradient =

\frac{3}{4}

Question 5

Find the midpoint and gradient for the points

I(5,8)

and

J(-3,2)

Solution

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Step 1: Midpoint

M = \left(\frac{5+(-3)}{2}, \frac{8+2}{2}\right) = \left(\frac{2}{2}, \frac{10}{2}\right) = (1, 5)

Step 2: Gradient

m = \frac{2 - 8}{-3 - 5} = \frac{-6}{-8} = \frac{3}{4}

Final Answer: Midpoint =

(1,5)

, Gradient =

\frac{3}{4}

Question 6

Find the midpoint and gradient for the points

K(-8,-6)

and

L(4,10)

Solution

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Step 1: Midpoint

M = \left(\frac{-8+4}{2}, \frac{-6+10}{2}\right) = \left(\frac{-4}{2}, \frac{4}{2}\right) = (-2, 2)

Step 2: Gradient

m = \frac{10 - (-6)}{4 - (-8)} = \frac{10+6}{4+8} = \frac{16}{12} = \frac{4}{3}

Final Answer: Midpoint =

(-2,2)

, Gradient =

\frac{4}{3}

Question 7

Find the midpoint and gradient of the line segment joining

M(-5,-3)

and

N(9,7)

Solution

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Step 1: Midpoint

M = \left(\frac{-5+9}{2}, \frac{-3+7}{2}\right) = (2, 2)

Step 2: Gradient

m = \frac{7 - (-3)}{9 - (-5)} = \frac{7+3}{9+5} = \frac{10}{14} = \frac{5}{7}

Final Answer: Midpoint =

(2,2)

, Gradient =

\frac{5}{7}

Question 8

Find the midpoint and gradient for

P(0,0)

and

Q(6,12)

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Solution

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Step 1: Midpoint

M = \left(\frac{0+6}{2}, \frac{0+12}{2}\right) = (3,6)

Step 2: Gradient

m = \frac{12 - 0}{6 - 0} = \frac{12}{6} = 2

Final Answer: Midpoint =

(3,6)

, Gradient =

2

Question 9

Find the midpoint and gradient for

R(-7,5)

and

S(7,-9)

Solution

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Step 1: Midpoint

M = \left(\frac{-7+7}{2}, \frac{5+(-9)}{2}\right) = (0, -2)

Step 2: Gradient

m = \frac{-9 - 5}{7 - (-7)} = \frac{-14}{14} = -1

Final Answer: Midpoint =

(0,-2)

, Gradient =

-1

Question 10

Find the midpoint and gradient for

T(-2,-4)

and

U(6,8)

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Solution Step 1: Midpoint

M = \left(\frac{-2+6}{2}, \frac{-4+8}{2}\right) = (2,2)

Step 2: Gradient

m = \frac{8 - (-4)}{6 - (-2)} = \frac{8+4}{6+2} = \frac{12}{8} = \frac{3}{2}

Final Answer: Midpoint =

(2,2)

, Gradient =

\frac{3}{2}

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Calculation problems involving distance between two points

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Formula for Distance Between Two Points

The distance

d

between two points

(x_1, y_1)

and

(x_2, y_2)

is given by:

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

Question 1

Find the distance between the points

A(2,3)

and

B(8,7)

Solution

Using the distance formula:

d = \sqrt{(8 - 2)^2 + (7 - 3)^2}

d = \sqrt{6^2 + 4^2}

d = \sqrt{36 + 16}

d = \sqrt{52}

d = 2\sqrt{13} \approx 7.21

Final Answer:

2\sqrt{13}

7.21

(approx).

Question 2

Find the distance between

C(-4,6)

and

D(10,-2)

Solution

d = \sqrt{(10 - (-4))^2 + (-2 - 6)^2}

d = \sqrt{(10 + 4)^2 + (-8)^2}

d = \sqrt{14^2 + 8^2}

d = \sqrt{196 + 64}

d = \sqrt{260}

d = 2\sqrt{65} \approx 16.12

Final Answer:

2\sqrt{65}

16.12

(approx).

Question 3

Find the distance between

E(1,-5)

and

F(7,3)

Solution

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d = \sqrt{(7 - 1)^2 + (3 - (-5))^2}

d = \sqrt{6^2 + 8^2}

d = \sqrt{36 + 64}

d = \sqrt{100}

d = 10

Final Answer:

10

Question 4

Find the distance between

G(-6,4)

and

H(2,10)

Solution

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d = \sqrt{(2 - (-6))^2 + (10 - 4)^2}

d = \sqrt{(2+6)^2 + 6^2}

d = \sqrt{8^2 + 6^2}

d = \sqrt{64 + 36}

d = \sqrt{100}

d = 10

Final Answer:

10

Question 5

Find the distance between

I(5,8)

and

J(-3,2)

Solution

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d = \sqrt{(-3 - 5)^2 + (2 - 8)^2}

d = \sqrt{(-8)^2 + (-6)^2}

d = \sqrt{64 + 36}

d = \sqrt{100}

d = 10

Final Answer:

10

Question 6

Find the distance between

K(-8,-6)

and

L(4,10)

Solution

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d = \sqrt{(4 - (-8))^2 + (10 - (-6))^2}

d = \sqrt{(4+8)^2 + (10+6)^2}

d = \sqrt{12^2 + 16^2}

d = \sqrt{144 + 256}

d = \sqrt{400}

d = 20

Final Answer:

20

Question 7

Find the distance between

M(-5,-3)

and

N(9,7)

Solution

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d = \sqrt{(9 - (-5))^2 + (7 - (-3))^2}

d = \sqrt{(9+5)^2 + (7+3)^2}

d = \sqrt{14^2 + 10^2}

d = \sqrt{196 + 100}

d = \sqrt{296}

d = 2\sqrt{74} \approx 17.2

Final Answer:

2\sqrt{74}

17.2

(approx).

Question 8

Find the distance between

P(0,0)

and

Q(6,12)

Solution

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d = \sqrt{(6 - 0)^2 + (12 - 0)^2}

d = \sqrt{6^2 + 12^2}

d = \sqrt{36 + 144}

d = \sqrt{180}

d = 6\sqrt{5} \approx 13.42

Final Answer:

6\sqrt{5}

13.42

(approx).

Question 9

Find the distance between

R(-7,5)

and

S(7,-9)

Solution

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d = \sqrt{(7 - (-7))^2 + (-9 - 5)^2}

d = \sqrt{(7+7)^2 + (-9-5)^2}

d = \sqrt{14^2 + (-14)^2}

d = \sqrt{196 + 196}

d = \sqrt{392}

d = 2\sqrt{98} = 14\sqrt{2} \approx 19.8

Final Answer:

14\sqrt{2}

19.8

(approx).

Question 10

Find the distance between

T(-2,-4)

and

U(6,8)

Solution

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d = \sqrt{(6 - (-2))^2 + (8 - (-4))^2}

d = \sqrt{(6+2)^2 + (8+4)^2}

d = \sqrt{8^2 + 12^2}

d = \sqrt{64 + 144}

d = \sqrt{208}

d = 4\sqrt{13} \approx 14.42

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Calculation problem involving conditions for parallelism and perpendicularity;

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Concepts:

Parallelism Condition
Two lines are parallel if their slopes are equal. If the equations of two lines are:
$y = m_1x + c_1$ and $y = m_2x + c_2$ ,
then the lines are parallel if $m_1 = m_2$ .
Perpendicularity Condition
Two lines are perpendicular if the product of their slopes is $-1$ .
That is, if $m_1 \times m_2 = -1$ , then the lines are perpendicular.

Question 1

Determine whether the lines passing through

A(2,3), B(6,7)

and

C(1,5), D(5,9)

are parallel.

Solution*

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Step 1: Find the slopes of the two lines
Slope of

AB

m_1 = \frac{7 - 3}{6 - 2} = \frac{4}{4} = 1

Slope of

CD

m_2 = \frac{9 - 5}{5 - 1} = \frac{4}{4} = 1

Since

m_1 = m_2

, the lines are parallel.

Final Answer: The lines are parallel.

Question 2

Determine whether the lines passing through

E(-4,6), F(10,-2)

and

G(3,5), H(9,2)

are perpendicular.

Solution

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Step 1: Find the slopes
Slope of

EF

m_1 = \frac{-2 - 6}{10 - (-4)} = \frac{-8}{14} = \frac{-4}{7}

Slope of

GH

m_2 = \frac{2 - 5}{9 - 3} = \frac{-3}{6} = \frac{-1}{2}

Step 2: Check perpendicularity

m_1 \times m_2 = \left(\frac{-4}{7}\right) \times \left(\frac{-1}{2}\right) = \frac{4}{14} = \frac{2}{7} \neq -1

Since

m_1 \times m_2 \neq -1

, the lines are not perpendicular.

Final Answer: The lines are not perpendicular.

Question 3

Find the missing coordinate

y

such that the line through

P(4, y)

and

Q(10,8)

is parallel to the line through

R(2,1)

and

S(6,5)

Solution

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Step 1: Find the slope of $RS$

m = \frac{5 - 1}{6 - 2} = \frac{4}{4} = 1

Step 2: Set the slope of $PQ$ equal to 1

\frac{8 - y}{10 - 4} = 1

Step 3: Solve for $y$

8 - y = 6

y = 2

Final Answer:

y = 2

Question 4

Find the missing coordinate

x

so that the line through

A(x, 3)

and

B(6,7)

is perpendicular to the line through

C(1,2)

and

D(5,10)

Solution

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Step 1: Find the slope of $CD$

m_1 = \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2

Step 2: Set the slope of $AB$ as $m_2$ and use the perpendicularity condition

m_1 \times m_2 = -1

2 \times \frac{7 - 3}{6 - x} = -1

2 \times \frac{4}{6 - x} = -1

\frac{8}{6 - x} = -1

Step 3: Solve for $x$

8 = -(6 - x)

8 = -6 + x

x = 14

Final Answer:

x = 14

Question 5

Verify whether the lines

y = 3x + 4

and

y = 3x - 2

are parallel.

Solution

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Both lines have the same slope

m = 3

, hence they are parallel.

Final Answer: The lines are parallel.

Question 6

Check if the lines

y = -\frac{1}{2}x + 7

and

y = 2x - 4

are perpendicular.

Solution

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Slope of first line:

m_1 = -\frac{1}{2}

Slope of second line:

m_2 = 2

Product:

m_1 \times m_2 = \left(-\frac{1}{2}\right) \times 2 = -1

Since the product is

-1

, the lines are perpendicular.

Final Answer: The lines are perpendicular.

Calculation problem involving the equation of a line in the two-point form, point-slope form, slope intercept form and the general form.

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Formulas to Remember

Two-Point Form (Given two points $(x_1, y_1)$ and $(x_2, y_2)$ ): $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$
Point-Slope Form (Given one point $(x_1, y_1)$ and slope $m$ ): $y - y_1 = m (x - x_1)$
Slope-Intercept Form (If slope is $m$ and intercept is $c$ ): $y = mx + c$
General Form: $Ax + By + C = 0$

Question 1

Find the equation of the line passing through

(2,3)

and

(6,7)

in:

Two-point form
Point-slope form
Slope-intercept form
General form

Solution

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Step 1: Find the slope
Using the slope formula:

m = \frac{7 - 3}{6 - 2} = \frac{4}{4} = 1

Step 2: Two-Point Form
Using the formula:

y - 3 = \frac{4}{4} (x - 2)

y - 3 = 1(x - 2)

y - 3 = x - 2

Step 3: Point-Slope Form
Using point

(2,3)

y - 3 = 1(x - 2)

Step 4: Slope-Intercept Form
Expanding:

y = x - 2 + 3

y = x + 1

Step 5: General Form
Rearrange:

x - y + 1 = 0

Final Answer:

Two-Point Form: $y - 3 = 1(x - 2)$
Point-Slope Form: $y - 3 = x - 2$
Slope-Intercept Form: $y = x + 1$
General Form: $x - y + 1 = 0$

Question 2

Find the equation of the line passing through

(4, -1)

and

(8,3)

Solution

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Step 1: Find the slope

m = \frac{3 - (-1)}{8 - 4} = \frac{4}{4} = 1

Step 2: Two-Point Form

y + 1 = 1(x - 4)

Step 3: Point-Slope Form
Using

(4,-1)

y + 1 = x - 4

Step 4: Slope-Intercept Form

y = x - 5

Step 5: General Form

x - y - 5 = 0

Final Answer:

Two-Point Form: $y + 1 = x - 4$
Point-Slope Form: $y + 1 = x - 4$
Slope-Intercept Form: $y = x - 5$
General Form: $x - y - 5 = 0$

Question 3

Find the equation of the line passing through

(3,5)

and

(7,9)

Solution

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Step 1: Find the slope

m = \frac{9 - 5}{7 - 3} = \frac{4}{4} = 1

Step 2: Two-Point Form

y - 5 = 1(x - 3)

Step 3: Point-Slope Form
Using

(3,5)

y - 5 = x - 3

Step 4: Slope-Intercept Form

y = x + 2

Step 5: General Form

x - y + 2 = 0

Final Answer:

Two-Point Form: $y - 5 = x - 3$
Point-Slope Form: $y - 5 = x - 3$
Slope-Intercept Form: $y = x + 2$
General Form: $x - y + 2 = 0$

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Thank you for taking the time to read my blog post! Your interest and engagement mean so much to me, and I hope the content provided valuable insights and sparked your curiosity. Your journey as a student is inspiring, and it’s my goal to contribute to your growth and success.

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This is all we can take on "Jamb Mathematics - Lesson Notes on coordinate geometry for UTME Candidate"

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Coordinate Geometry | Jamb Mathematics

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Calculation problems involving distance between two points

Formula for Distance Between Two Points

Question 1

Solution

Question 2

Solution

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Calculation problem involving conditions for parallelism and perpendicularity;

Concepts:

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Calculation problem involving the equation of a line in the two-point form, point-slope form, slope intercept form and the general form.

Formulas to Remember

Question 1

Question 2

Question 3

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