Review Exercise 7

Review Exercise 7 Solutions

Comprehensive review of Coordinate Geometry, including formulas, line properties, and dynamic conversions.

Question 1 Interactive MCQ Quiz
Choose the correct option. Test your conceptual understanding using the three tabbed self-grading sections below.
Question 2 Distance AB
Find the distance between two points A(2, 3) and B(7, 8) on a coordinate plane.
Solution Step-by-Step
Using the distance formula:
d = √(x2 - x1)2 + (y2 - y1)2
|AB| = √(7 - 2)2 + (8 - 3)2
= √52 + 52
= √25 + 25 = √50 = √25 × 2 = 5√2 units
Question 3 Midpoint
Find the midpoint of the line segment joining the points (4, -2) and (-6, 3).
Solution Step-by-Step
Using the midpoint formula for points (4, -2) and (-6, 3):
M(xm, ym) = M(
x1 + x22
,
y1 + y22
)
M = M(
4 + (-6)2
,
-2 + 32
)
= M(
-22
,
12
) = M(-1, 0.5)
Conclusion: The midpoint of the segment is M(-1, 0.5).
Question 4 Gradient
Calculate the gradient (slope) of the line passing through the points (1, 2) and (4, 6).
Solution Step-by-Step
Using the slope formula:
m =
y2 - y1x2 - x1
m =
6 - 24 - 1
=
43
Conclusion: The gradient of the line is 4/3.
Question 5 Line Equation
Find the equation of the line in the form y = mx + c that passes through the points (3, 7) and (5, 11).
Solution Step-by-Step
First find the slope:
m =
11 - 75 - 3
=
42
= 2
Use point-slope form with point (3, 7) and slope m = 2:
y - 7 = 2(x - 3)
y - 7 = 2x - 6
y = 2x - 6 + 7
Equation: y = 2x + 1
Question 6 Parallel Gradient
If two lines are parallel, and one line has a gradient of 5/2, what is the gradient of the other line?
Solution Step-by-Step
Parallel lines have equal slopes:
m2 = m1 =
52
Conclusion: The gradient of the other line is 5/2.
Question 7 Airplane Distance
An airplane needs to fly from city A at coordinates (12, 5) to city B at coordinates (8, -4). Calculate the straight-line distance between these two cities.
Solution Step-by-Step
Using the distance formula for points A(12, 5) and B(8, -4):
d = √(8 - 12)2 + (-4 - 5)2
= √(-4)2 + (-9)2
= √16 + 81 = √97 units
Conclusion: The straight-line distance is √97 units.
Question 8 Path Midpoint
In a landscaping project, the path starts at (2, 3) and ends at (10, 7). Find the midpoint.
Solution Step-by-Step
Using the midpoint formula for points (2, 3) and (10, 7):
M = M(
2 + 102
,
3 + 72
) = M(
122
,
102
) = M(6, 5)
Conclusion: The midpoint of the landscaping path is M(6, 5).
Question 9 Drone Flight
A drone is flying from point (2, 3) to point (10, 15) on the grid. Calculate the gradient of the line along which the drone is flying and the total distance traveled.
Solution Step-by-Step
(a) Gradient:
m =
15 - 310 - 2
=
128
=
32
 (or 1.5)
(b) Distance Traveled:
d = √(10 - 2)2 + (15 - 3)2
= √82 + 122
= √64 + 144 = √208 = √16 × 13 = 4√13 units
Question 10 Line Forms
For a line with a gradient of -3 and a y-intercept of 2, write the equation of the line in:
(a) Slope-intercept form     (b) Point-slope form (using point (1, 2))
(c) Two-point form (using points (1, 2) and (4, -7))     (d) Intercepts form
(e) Symmetric form     (f) Normal form
Solutions
The general equation is:
y = -3x + 2 ⇒ 3x + y - 2 = 0
(a) Slope-intercept form:
y = -3x + 2
(b) Point-slope form using point (1, 2):
y - 2 = -3(x - 1)
(c) Two-point form using points (1, 2) and (4, -7):
y - 2x - 1
=
-7 - 24 - 1
⇒
y - 2x - 1
= -3
(d) Intercepts form: Divide 3x + y = 2 by 2:
3x2
+
y2
 = 1 ⇒
x2/3
+
y2
= 1
(e) Symmetric form: Dividing 3x + y = 2 by √(3² + 1²) = √10:
3x√10
+
y√10
=
2√10
(f) Normal form: Comparing the symmetric form with x cos(α) + y sin(α) = p:
cos(α) =
3√10
, sin(α) =
1√10
⇒ α = tan-1(1/3) ≈ 18.43°
x cos(18.43°) + y sin(18.43°) = 2/√10
Interactive Sandbox Coordinate Plane Explorer
Input coordinate points below to dynamically calculate distance, midpoints, slopes, and straight line equations on a Cartesian plane.