Exercise 7.1

Exercise 7.1 Solutions

Coordinate plane quadrants, distance formula, midpoint formula, and geometric shape verification.

Question 1 Point Locations
Describe the location in the plane of the point P(x, y) for which:
(i) x > 0     (ii) x > 0 and y > 0     (iii) x = 0     (iv) y = 0     (v) x > 0 and y ≤ 0
(vi) y = 0, x = 0     (vii) x = y     (viii) x ≥ 3     (ix) y > 0     (x) x and y have opposite signs
Solutions
(i) x > 0: The set of all points in the open right half-plane of the Cartesian plane (Quadrants I and IV, excluding the y-axis).
(ii) x > 0 and y > 0: The set of all points in the First Quadrant (Quadrant I).
(iii) x = 0: The set of all points on the y-axis.
(iv) y = 0: The set of all points on the x-axis.
(v) x > 0 and y ≤ 0: The set of all points in the Fourth Quadrant and on the positive x-axis (excluding the origin).
(vi) y = 0, x = 0: The point at the origin: O(0, 0).
(vii) x = y: The straight line bisecting the First and Third quadrants, passing through the origin at a 45° angle.
(viii) x ≥ 3: The set of all points lying on and to the right of the vertical line x = 3.
(ix) y > 0: The set of all points in the open upper half-plane (above the x-axis, Quadrants I and II).
(x) x and y have opposite signs: The set of all points lying in the Second Quadrant (x < 0, y > 0) and the Fourth Quadrant (x > 0, y < 0).
Question 2 Distance Formula
Find the distance between the points:
(i) A(6, 7), B(0, -2)     (ii) C(-5, -2), D(3, 2)
(iii) L(0, 3), M(-2, -4)     (iv) P(-8, -7), Q(0, 0)
(i) A(6, 7), B(0, -2)
Solution Step-by-Step
The distance formula is:
d = √(x2 - x1)2 + (y2 - y1)2
Substitute x1 = 6, y1 = 7 and x2 = 0, y2 = -2:
|AB| = √(0 - 6)2 + (-2 - 7)2
= √(-6)2 + (-9)2
= √36 + 81
= √117 = √9 × 13 = 3√13 units
(ii) C(-5, -2), D(3, 2)
Solution Step-by-Step
Substitute values into the distance formula:
|CD| = √[3 - (-5)]2 + [2 - (-2)]2
= √(3 + 5)2 + (2 + 2)2
= √82 + 42
= √64 + 16
= √80 = √16 × 5 = 4√5 units
(iii) L(0, 3), M(-2, -4)
Solution Step-by-Step
Substitute values:
|LM| = √(-2 - 0)2 + (-4 - 3)2
= √(-2)2 + (-7)2
= √4 + 49 = √53 units
(iv) P(-8, -7), Q(0, 0)
Solution Step-by-Step
Substitute values:
|PQ| = √[0 - (-8)]2 + [0 - (-7)]2
= √82 + 72
= √64 + 49 = √113 units
Question 3 Distance & Midpoint
Find in each of the following: (i) The distance between the two given points, (ii) Midpoint of the line segment joining the two points.
(a) A(3, 1), B(-2, -4)     (b) A(-8, 3), B(2, -1)     (c) A(-√5, -1/3), B(-3√5, 5)
(a) A(3, 1), B(-2, -4)
Distance Calculation
|AB| = √(-2 - 3)2 + (-4 - 1)2
= √(-5)2 + (-5)2
= √25 + 25 = √50 = 5√2 units
Midpoint Calculation
The midpoint formula is:
M(xm, ym) = M(
x1 + x22
,
y1 + y22
)
Substitute values:
M = M(
3 + (-2)2
,
1 + (-4)2
) = M(0.5, -1.5)
(b) A(-8, 3), B(2, -1)
Distance Calculation
|AB| = √[2 - (-8)]2 + (-1 - 3)2
= √102 + (-4)2
= √100 + 16 = √116 = 2√29 units
Midpoint Calculation
Substitute values:
M = M(
-8 + 22
,
3 + (-1)2
) = M(
-62
,
22
) = M(-3, 1)
(c) A(-√5, -1/3), B(-3√5, 5)
Distance Calculation
|AB| = √[-3√5 - (-√5)]2 + [5 - (-1/3)]2
= √(-2√5)2 + (5 + 1/3)2
= √20 + (16/3)2
= √20 + 256/9 = √(180 + 256)/9 = √436/9 =
2√1093
units
Midpoint Calculation
Substitute values:
M = M(
-√5 + (-3√5)2
,
-1/3 + 52
)
= M(
-4√52
,
14/32
) = M(-2√5,
73
)
Question 4 Origin Distance
Which of the following points are at a distance of 15 units from the origin?
(i) (√176, 7)     (ii) (10, -10)     (iii) (1, 15)
Solution Step-by-Step
The distance from the origin O(0,0) to a point A(x, y) is:
|OA| = √x2 + y2
(i) (√176, 7):
|OA| = √(√176)2 + 72 = √176 + 49 = √225 = 15 units
Conclusion: Yes, this point is at a distance of 15 units from the origin.
(ii) (10, -10):
|OA| = √102 + (-10)2 = √100 + 100 = √200 = 10√2 ≈ 14.14 units
Conclusion: No, this point is not at a distance of 15 units.
(iii) (1, 15):
|OA| = √12 + 152 = √1 + 225 = √226 ≈ 15.03 units
Conclusion: No, this point is not at a distance of 15 units.
Question 5 Shape Vertices
Show that:
(i) The points A(0, 2), B(√3, 1) and C(0, -2) are vertices of a right triangle.
(ii) The points A(3, 1), B(-2, -3) and C(2, 2) are vertices of an isosceles triangle.
(iii) The points A(5, 2), B(-2, 3), C(-3, -4) and D(4, -5) are vertices of a parallelogram.
(i) Right Triangle Verification
Solution Step-by-Step
We find the square of the lengths of the sides:
|AB|2 = (√3 - 0)2 + (1 - 2)2 = 3 + 1 = 4
|BC|2 = (0 - √3)2 + (-2 - 1)2 = 3 + 9 = 12
|AC|2 = (0 - 0)2 + (-2 - 2)2 = 0 + 16 = 16
Since |AB|2 + |BC|2 = 4 + 12 = 16 = |AC|2, Pythagoras' theorem is satisfied.
Conclusion: Points A, B, and C form a right-angled triangle with the right angle at vertex B.
(ii) Isosceles Triangle Verification
Solution Step-by-Step
Find lengths of all three sides:
|AB| = √(-2 - 3)2 + (-3 - 1)2 = √25 + 16 = √41
|BC| = √[2 - (-2)]2 + [2 - (-3)]2 = √42 + 52 = √16 + 25 = √41
|AC| = √(2 - 3)2 + (2 - 1)2 = √(-1)2 + 12 = √2
Since |AB| = |BC| = √41 (two sides are equal in length), the triangle is isosceles.
Conclusion: Vertices A, B, and C form an isosceles triangle.
(iii) Parallelogram Verification
Solution Step-by-Step
Find lengths of opposite sides:
|AB| = √(-2 - 5)2 + (3 - 2)2 = √(-7)2 + 12 = √50
|CD| = √[4 - (-3)]2 + [-5 - (-4)]2 = √72 + (-1)2 = √50
|BC| = √[-3 - (-2)]2 + (-4 - 3)2 = √(-1)2 + (-7)2 = √50
|AD| = √(4 - 5)2 + (-5 - 2)2 = √(-1)2 + (-7)2 = √50
Since |AB| = |CD| and |BC| = |AD|, opposite sides are equal. Let us verify if the diagonals bisect each other:
Midpoint of diagonal AC = M1(
5 + (-3)2
,
2 + (-4)2
) = M1(1, -1)
Midpoint of diagonal BD = M2(
-2 + 42
,
3 + (-5)2
) = M2(1, -1)
Since the midpoints of the diagonals coincide (M1 = M2 = (1, -1)), the diagonals bisect each other.
Conclusion: The points form a parallelogram (specifically, a rhombus since all four sides are equal).
Question 6 Right Triangle at A
Find h such that the points A(√3, -1), B(0, 2) and C(h, -2) are vertices of a right triangle with right angle at the vertex A.
Solution Step-by-Step
If the right angle is at vertex A, then BC is the hypotenuse. By Pythagoras' theorem:
|BC|2 = |AB|2 + |AC|2
Find the square of each side length:
|AB|2 = (0 - √3)2 + [2 - (-1)]2 = 3 + 9 = 12
|BC|2 = (h - 0)2 + (-2 - 2)2 = h2 + 16
|AC|2 = (h - √3)2 + [-2 - (-1)]2 = h2 - 2√3h + 3 + 1 = h2 - 2√3h + 4
Substitute these into the relation:
h2 + 16 = 12 + (h2 - 2√3h + 4)
h2 + 16 = h2 - 2√3h + 16
Simplify the equation by subtracting h2 and 16 from both sides:
0 = -2√3h
h = 0
Question 7 Collinear Points
Find h such that A(-1, h), B(3, 2) and C(7, 3) are collinear.
Solution Step-by-Step
Points are collinear if they lie on the same straight line, meaning the slope of line segment AB equals the slope of line segment BC.
Slope(AB) = Slope(BC)
Compute both slopes:
Slope(AB) =
2 - h3 - (-1)
=
2 - h4
Slope(BC) =
3 - 27 - 3
=
14
Equate the slopes:
2 - h4
 =
14
2 - h = 1
h = 1
Question 8 Circle Diameter
The points A(-5, -2) and B(5, -4) are ends of a diameter of a circle. Find the centre and radius of the circle.
Solution Step-by-Step
(i) Finding the Centre: The centre of a circle is the midpoint of its diameter AB.
Centre C(xc, yc) = M(
-5 + 52
,
-2 + (-4)2
)
= M(0,
-62
) = M(0, -3)
(ii) Finding the Radius: The radius r is the distance from the centre C(0, -3) to endpoint A(-5, -2).
r = √[-5 - 0]2 + [-2 - (-3)]2
= √(-5)2 + (1)2
= √25 + 1 = √26 units
Question 9 Right Triangle Variable
Find h such that the points A(h, 1), B(2, 7) and C(-6, -7) are vertices of a right triangle with right angle at the vertex A.
Solution Step-by-Step
Since the right angle is at vertex A, the hypotenuse is BC. By Pythagoras' theorem:
|BC|2 = |AB|2 + |AC|2
Calculate the squared lengths:
|BC|2 = (-6 - 2)2 + (-7 - 7)2 = (-8)2 + (-14)2 = 64 + 196 = 260
|AB|2 = (2 - h)2 + (7 - 1)2 = (4 - 4h + h2) + 36 = h2 - 4h + 40
|AC|2 = (-6 - h)2 + (-7 - 1)2 = (36 + 12h + h2) + 64 = h2 + 12h + 100
Substitute the values:
260 = (h2 - 4h + 40) + (h2 + 12h + 100)
260 = 2h2 + 8h + 140
Rearrange into a quadratic equation:
2h2 + 8h - 120 = 0
h2 + 4h - 60 = 0
Factorize the quadratic equation:
h2 + 10h - 6h - 60 = 0
h(h + 10) - 6(h + 10) = 0
(h + 10)(h - 6) = 0
Therefore:
h = -10   or   h = 6
The value of h is either -10 or 6.
Question 10 Midpoints Parallelogram
A quadrilateral has the points A(9, 3), B(-7, 7), C(-3, -7) and D(5, -5) as its vertices. Find the midpoints of its sides. Show that the figure formed by joining the midpoints consecutively is a parallelogram.
Solution Step-by-Step
Step 1: Find the midpoints of all four sides.
Let P, Q, R, and S be the midpoints of sides AB, BC, CD, and DA respectively.
P (midpoint of AB) = M(
9 + (-7)2
,
3 + 72
) = P(1, 5)
Q (midpoint of BC) = M(
-7 + (-3)2
,
7 + (-7)2
) = Q(-5, 0)
R (midpoint of CD) = M(
-3 + 52
,
-7 + (-5)2
) = R(1, -6)
S (midpoint of DA) = M(
5 + 92
,
-5 + 32
) = S(7, -1)
Step 2: Show quadrilateral PQRS is a parallelogram. We can show that the diagonals PR and QS bisect each other by computing their midpoints:
Midpoint of diagonal PR = M(
1 + 12
,
5 + (-6)2
) = (1, -0.5)
Midpoint of diagonal QS = M(
-5 + 72
,
0 + (-1)2
) = (1, -0.5)
Since the midpoints of the diagonals coincide at (1, -0.5), they bisect each other.
Conclusion: The quadrilateral PQRS formed by joining the midpoints is a parallelogram.
Interactive Sandbox Shape Solver & Plotter
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