(x₁,y₁)
A = ½|...|
√(x₂-x₁)²

✨ Complete Coordinate Geometry ✨

All Problems Solved with Step-by-Step Explanations

Triangles
Collinear Points
Quadrilaterals
Advanced Problems
Paint Problems

Area of Triangle Problems

Problem 1(i)

Find area of triangle with points:
A(1, -1), B(-4, 6), C(-3, -5)

Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
1 Assign coordinates:
(x₁,y₁) = (1,-1), (x₂,y₂) = (-4,6), (x₃,y₃) = (-3,-5)
2 Plug into formula:
½ |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|
3 Simplify:
½ |1(11) + (-4)(-4) + (-3)(-7)| = ½ |11 + 16 + 21|
4 Final calculation:
½ |48| = 24 square units

Problem 1(ii)

Find area of triangle with points:
A(-10, -4), B(-8, -1), C(-3, -5)

1 Assign coordinates:
(x₁,y₁) = (-10,-4), (x₂,y₂) = (-8,-1), (x₃,y₃) = (-3,-5)
2 Apply formula:
½ |-10(-1-(-5)) + (-8)(-5-(-4)) + (-3)(-4-(-1))|
3 Calculate:
½ |-10(4) + (-8)(-1) + (-3)(-3)| = ½ |-40 + 8 + 9|
4 Final result:
½ |-23| = 11.5 square units

Problem 3(i)

Find value of 'p' for vertices:
(0,0), (p,8), (6,2) with area = 20

1 Apply area formula:
½ |0(8-2) + p(2-0) + 6(0-8)| = 20
2 Simplify:
½ |0 + 2p - 48| = 20 → |2p - 48| = 40
3 Solve absolute value:
Case 1: 2p - 48 = 40 → p = 44
Case 2: 2p - 48 = -40 → p = 4

Possible values: p = 4 or 44

Problem 3(ii)

Find value of 'p' for vertices:
(p,p), (5,6), (5,-2) with area = 32

1 Apply area formula:
½ |p(6-(-2)) + 5(-2-p) + 5(p-6)| = 32
2 Simplify:
½ |8p - 10 - 5p + 5p - 30| = 32 → |8p - 40| = 64
3 Solve absolute value:
Case 1: 8p - 40 = 64 → p = 13
Case 2: 8p - 40 = -64 → p = -3

Possible values: p = -3 or 13

Collinear Points Problems

Problem 2(i)

Are these points collinear?
(-½,3), (-5,6), (-8,8)

Points are collinear if area of triangle = 0
1 Assign coordinates:
(x₁,y₁) = (-½,3), (x₂,y₂) = (-5,6), (x₃,y₃) = (-8,8)
2 Calculate area:
½ |(-½)(6-8) + (-5)(8-3) + (-8)(3-6)|
3 Simplify:
½ |(-½)(-2) + (-5)(5) + (-8)(-3)| = ½ |1 - 25 + 24| = 0

Since area = 0, points are collinear

Problem 2(ii)

Are these points collinear?
(a,b+c), (b,c+a), (c,a+b)

1 Assign coordinates:
(x₁,y₁) = (a,b+c), (x₂,y₂) = (b,c+a), (x₃,y₃) = (c,a+b)
2 Calculate area:
½ |a(c+a - (a+b)) + b(a+b - (b+c)) + c(b+c - (c+a))|
3 Simplify:
½ |a(c-b) + b(a-c) + c(b-a)|
= ½ |ac - ab + ab - bc + bc - ac| = ½ |0| = 0

Since area = 0, points are collinear

Problem 4(i)

Find 'a' for collinear points:
(2,3), (4,a), (6,-3)

1 For collinearity, area must be 0:
½ |2(a-(-3)) + 4(-3-3) + 6(3-a)| = 0
2 Simplify:
|2a + 6 - 24 + 18 - 6a| = 0 → |-4a| = 0
3 Solve:
-4a = 0 → a = 0

Value of a = 0

Problem 4(ii)

Find 'a' for collinear points:
(a,2-2a), (-a+1,2a), (-4-a,6-2a)

1 For collinearity, area must be 0:
½ |a(2a - (6-2a)) + (-a+1)(6-2a - (2-2a)) + (-4-a)(2-2a - 2a)| = 0
2 Simplify step by step:
First term: a(4a - 6) = 4a² - 6a
Second term: (-a+1)(4) = -4a + 4
Third term: (-4-a)(2-4a) = -8 + 16a -2a + 4a²
3 Combine:
|4a² - 6a - 4a + 4 - 8 + 16a - 2a + 4a²| = 0
|8a² + 4a - 4| = 0 → |4(2a² + a - 1)| = 0
4 Solve quadratic:
2a² + a - 1 = 0 → a = [-1 ± √(1+8)]/4
a = (-1 ± 3)/4 → a = ½ or a = -1

Possible values: a = -1 or ½

Problem 7

Points A(-3,9), B(a,b), C(4,-5) are collinear and a + b = 1.
Find a and b.

1 Collinearity condition:
½ |-3(b - (-5)) + a(-5 - 9) + 4(9 - b)| = 0
2 Simplify:
|-3b -15 -14a +36 -4b| = 0 → |-7b -14a +21| = 0
3 Divide by -7:
b + 2a = 3
4 Given a + b = 1, solve system:
b = 1 - a
Substitute: (1-a) + 2a = 3 → a = 2
Then b = 1 - 2 = -1

Solution: a = 2, b = -1

Quadrilateral Area Problems

Problem 5(i)

Find area of quadrilateral with vertices:
(-9,-2), (-8,-4), (2,2), (1,-3)

Divide quadrilateral into two triangles and sum their areas
1 Divide into:
Triangle ABC: (-9,-2), (-8,-4), (2,2)
Triangle ACD: (-9,-2), (2,2), (1,-3)
2 Area of ABC:
½ |-9(-4-2) + -8(2-(-2)) + 2(-2-(-4))|
= ½ |54 - 32 + 4| = 13
3 Area of ACD:
½ |-9(2-(-3)) + 2(-3-(-2)) + 1(-2-2)|
= ½ |-45 -2 -4| = 25.5

Total area = 13 + 25.5 = 38.5 square units

Problem 5(ii)

Find area of quadrilateral with vertices:
(-9,0), (-8,6), (-1,-2), (-6,-3)

1 Divide into:
Triangle ABC: (-9,0), (-8,6), (-1,-2)
Triangle ACD: (-9,0), (-1,-2), (-6,-3)
2 Area of ABC:
½ |-9(6-(-2)) + -8(-2-0) + -1(0-6)|
= ½ |-72 + 16 + 6| = 25
3 Area of ACD:
½ |-9(-2-(-3)) + -1(-3-0) + -6(0-(-2))|
= ½ |-9 + 3 -12| = 9

Total area = 25 + 9 = 34 square units

Problem 6

Find value of k if quadrilateral area is 28:
(-4,-2), (-3,k), (3,-2), (2,3)

1 Divide into:
Triangle ABC: (-4,-2), (-3,k), (3,-2)
Triangle ACD: (-4,-2), (3,-2), (2,3)
2 Area of ABC:
½ |-4(k-(-2)) + -3(-2-(-2)) + 3(-2-k)|
= ½ |-4k -8 +0 -6 -3k| = ½ |-7k -14|
3 Area of ACD:
½ |-4(-2-3) + 3(3-(-2)) + 2(-2-(-2))|
= ½ |20 +15 +0| = 17.5
4 Total area:
½ |-7k -14| + 17.5 = 28 → |-7k -14| = 21
5 Solve:
Case 1: -7k -14 = 21 → k = -5
Case 2: -7k -14 = -21 → k = 1

Possible values: k = -5 or 1

Problem 11(i)

Find area of triangle AGF (assuming coordinates from figure)

1 Without figure, we can't determine exact coordinates.
General method:
2 If given coordinates A(x₁,y₁), G(x₂,y₂), F(x₃,y₃):
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
3 Example: If A(0,0), G(4,0), F(0,3):
Area = ½ |0(0-3) + 4(3-0) + 0(0-0)| = ½ |12| = 6

Without figure, exact area cannot be determined. Follow the formula with given coordinates.

Advanced Problems

Problem 8

P(11,7), Q(13.5,4), R(9.5,4) are midpoints of sides AB, BC, AC of ΔABC.
Find coordinates of A, B, C and areas.

1 Let A = (x₁,y₁), B = (x₂,y₂), C = (x₃,y₃)
Using midpoint formulas:
2 AB midpoint P:
(x₁+x₂)/2 = 11 → x₁ + x₂ = 22
(y₁+y₂)/2 = 7 → y₁ + y₂ = 14
3 BC midpoint Q:
(x₂+x₃)/2 = 13.5 → x₂ + x₃ = 27
(y₂+y₃)/2 = 4 → y₂ + y₃ = 8
4 AC midpoint R:
(x₁+x₃)/2 = 9.5 → x₁ + x₃ = 19
(y₁+y₃)/2 = 4 → y₁ + y₃ = 8
5 Solve for x-coordinates:
x₃ = 19 - x₁
x₂ = 22 - x₁
Substitute into x₂ + x₃ = 27:
(22 - x₁) + (19 - x₁) = 27 → 41 - 2x₁ = 27 → x₁ = 7
Then x₂ = 15, x₃ = 12
6 Solve for y-coordinates:
y₃ = 8 - y₁
y₂ = 14 - y₁
Substitute into y₂ + y₃ = 8:
(14 - y₁) + (8 - y₁) = 8 → 22 - 2y₁ = 8 → y₁ = 7
Then y₂ = 7, y₃ = 1
7 Final coordinates:
A(7,7), B(15,7), C(12,1)
8 Area of ΔABC:
½ |7(7-1) + 15(1-7) + 12(7-7)| = ½ |42 - 90| = 24
9 Area of ΔPQR:
½ |11(4-4) + 13.5(4-7) + 9.5(7-4)| = ½ |0 - 40.5 + 28.5| = 6

Vertices: A(7,7), B(15,7), C(12,1)
Area ΔABC = 24, Area ΔPQR = 6 (1/4 of ABC)

Problem 9

Find area of patio around quadrilateral swimming pool (figure needed)

1 Without the figure, general approach:
2 Method:
1. Find area of quadrilateral pool using coordinates
2. Find area of outer quadrilateral (pool + patio)
3. Subtract pool area from total area to get patio area
3 Example:
If pool vertices: (0,0), (4,0), (4,3), (0,3) → Area = 12
Outer vertices: (-1,-1), (5,-1), (5,4), (-1,4) → Area = 30
Patio area = 30 - 12 = 18

Without figure, exact area cannot be determined. Follow this method with given coordinates.

Paint Calculation Problems

Problem 10

Triangular glass with vertices at A(-5,-4), B(1,6), C(7,-4).
One paint bucket covers 6 sq.ft. How many buckets needed?

1 First find area of triangle ABC
2 Apply area formula:
½ |-5(6-(-4)) + 1(-4-(-4)) + 7(-4-6)|
3 Calculate:
½ |-5(10) + 1(0) + 7(-10)| = ½ |-50 + 0 -70| = ½ |-120| = 60
4 Calculate paint buckets:
Area = 60 sq.ft, Coverage = 6 sq.ft/bucket
Buckets needed = 60 / 6 = 10

Number of paint buckets required = 10

Problem 11(ii)

Find area of triangle FED (assuming coordinates from figure)

1 Without figure, we can't determine exact coordinates.
General method:
2 If given coordinates F(x₁,y₁), E(x₂,y₂), D(x₃,y₃):
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
3 Example: If F(0,0), E(4,0), D(0,3):
Area = ½ |0(0-3) + 4(3-0) + 0(0-0)| = ½ |12| = 6

Without figure, exact area cannot be determined. Follow the formula with given coordinates.

Problem 11(iii)

Find area of quadrilateral BCEG (assuming coordinates from figure)

1 Without figure, we can't determine exact coordinates.
General method:
2 Divide quadrilateral into two triangles:
Triangle BCE and Triangle BEG (or other combination)
3 Calculate each triangle's area using the formula:
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
4 Example: If B(0,0), C(4,0), E(4,3), G(0,3):
Area BCE = ½ |0(0-3) + 4(3-0) + 4(0-0)| = 6
Area BEG = ½ |0(3-3) + 4(3-0) + 0(0-3)| = 6
Total area = 6 + 6 = 12

Without figure, exact area cannot be determined. Follow this method with given coordinates.