Question 1(i)
Find the distance between (2, 3) and (4, 1).
Steps only
d = √[(4 − 2)² + (1 − 3)²] = √8 = 2√2
Answer
2√2 units.
CLASS 10 · MATHEMATICS · CHAPTER 7
Coordinate Geometry
Use the distance, section and midpoint formulas through concise working or open a student-friendly explanation for any question.
24 worked answersAll 2 exercise sets
What you will learn
- Find distances between points.
- Classify shapes using their coordinates.
- Divide a line segment in a given ratio.
24 worked answers · short steps first, explanations when you need them
Exercise 7.1
Textbook pages 105–106
Question 1(ii)
Find the distance between (−5, 7) and (−1, 3).
Steps only
d = √[4² + (−4)²] = √32 = 4√2
Answer
4√2 units.
Question 1(iii)
Find the distance between (a, b) and (−a, −b).
Steps only
d = √[(−2a)² + (−2b)²] = 2√(a² + b²)
Answer
2√(a² + b²) units.
Question 2
Find the distance between (0, 0) and (36, 15), and hence the distance between towns A and B.
Steps only
d = √(36² + 15²) = √1521 = 39
Answer
39 km.
Question 3
Determine whether (1, 5), (2, 3) and (−2, −11) are collinear.
Steps only
Slope AB = −2; slope BC = 14/4 = 7/2
The slopes are unequal.
Answer
The three points are not collinear.
Question 4
Check whether (5, −2), (6, 4) and (7, −2) form an isosceles triangle.
Steps only
AB = √37, BC = √37, AC = 2
Answer
Yes. AB = BC, so the triangle is isosceles.
Question 5
For the classroom points A(3, 4), B(6, 7), C(9, 4) and D(6, 1), decide whether ABCD is a square.
Steps only
AB = BC = CD = DA = 3√2
AC = BD = 6
A(3, 4) · B(6, 7) · C(9, 4) · D(6, 1)
Answer
ABCD is a square, so Champa is correct.
Question 6(i)
Classify the quadrilateral with vertices (−1, −2), (1, 0), (−1, 2), (−3, 0), in order.
Steps only
All four sides = 2√2; both diagonals = 4
Answer
It is a square.
Question 6(ii)
Classify the figure formed by (−3, 5), (3, 1), (0, 3), (−1, −4), in order.
Steps only
Slope from (−3,5) to (3,1) = −2/3
Slope from (−3,5) to (0,3) = −2/3
Answer
No quadrilateral is formed; three points are collinear.
Question 6(iii)
Classify the quadrilateral with vertices (4, 5), (7, 6), (4, 3), (1, 2), in order.
Steps only
AB = CD = √10; BC = DA = 3√2
Answer
It is a parallelogram.
Question 7
Find the point on the x-axis that is equidistant from (2, −5) and (−2, 9).
Steps only
Let P = (x, 0). (x − 2)² + 25 = (x + 2)² + 81
−8x = 56, so x = −7
Answer
(−7, 0).
Question 8
Find y if the distance between P(2, −3) and Q(10, y) is 10 units.
Steps only
8² + (y + 3)² = 10²
(y + 3)² = 36, so y + 3 = ±6
Answer
y = 3 or y = −9.
Question 9
Q(0, 1) is equidistant from P(5, −3) and R(x, 6). Find x, QR and PR.
Steps only
QP² = 5² + (−4)² = 41
QR² = x² + 5² = 41, so x = ±4
Answer
x = 4 or −4; QR = √41. PR = √82 for x = 4, and 9√2 for x = −4.
Question 10
Find the relation between x and y if (x, y) is equidistant from (3, 6) and (−3, 4).
Steps only
(x − 3)² + (y − 6)² = (x + 3)² + (y − 4)²
3x + y = 5
Answer
3x + y = 5.
Exercise 7.2
Textbook pages 111
Question 1
Find the point dividing the join of (−1, 7) and (4, −3) in the ratio 2:3.
Steps only
P = ((2×4 + 3×(−1))/5, (2×(−3) + 3×7)/5) = (1, 3)
Answer
(1, 3).
Question 2
Find the trisection points of the segment joining (4, −1) and (−2, −3).
Steps only
Ratio 1:2 gives (2, −5/3)
Ratio 2:1 gives (0, −7/3)
Answer
(2, −5/3) and (0, −7/3).
Question 3
On the school-ground grid, find the distance between the green flag at (2, 25) and red flag at (8, 20), then locate the blue flag halfway between them.
Steps only
GR = √[(8 − 2)² + (20 − 25)²] = √61 m
Midpoint = ((2 + 8)/2, (25 + 20)/2) = (5, 22.5)
G(2, 25) · R(8, 20) · B(5, 45/2)
Answer
The flags are √61 m apart; place the blue flag at (5, 22.5).
Question 4
Find the ratio in which (−1, 6) divides the segment joining (−3, 10) and (6, −8).
Steps only
−1 = (6m − 3n)/(m + n)
7m = 2n, so m:n = 2:7
Answer
2:7.
Question 5
Find the ratio in which the x-axis divides the segment joining A(1, −5) and B(−4, 5), and find the point of division.
Steps only
0 = (5m − 5n)/(m + n), so m:n = 1:1
P = ((1 − 4)/2, 0) = (−3/2, 0)
Answer
Ratio 1:1; point (−3/2, 0).
Question 6
If (1, 2), (4, y), (x, 6) and (3, 5) are consecutive vertices of a parallelogram, find x and y.
Steps only
(1 + x)/2 = (4 + 3)/2, so x = 6
(2 + 6)/2 = (y + 5)/2, so y = 3
Answer
x = 6 and y = 3.
Question 7
AB is a diameter of a circle with centre (2, −3). If B = (1, 4), find A.
Steps only
((x + 1)/2, (y + 4)/2) = (2, −3)
x = 3, y = −10
Answer
A = (3, −10).
Question 8
For A(−2, −2) and B(2, −4), find P on AB such that AP = 3/7 of AB.
Steps only
AP:PB = 3:4
P = ((3×2 + 4×(−2))/7, (3×(−4) + 4×(−2))/7)
Answer
P = (−2/7, −20/7).
Question 9
Find the points dividing the segment from A(−2, 2) to B(2, 8) into four equal parts.
Steps only
One-quarter point: (−1, 7/2)
Midpoint: (0, 5)
Three-quarter point: (1, 13/2)
Answer
(−1, 7/2), (0, 5), (1, 13/2).
Question 10
Find the area of the rhombus with vertices (3, 0), (4, 5), (−1, 4) and (−2, −1), in order.
Steps only
Diagonal AC = √[(−4)² + 4²] = 4√2
Diagonal BD = √[(−6)² + (−6)²] = 6√2
Area = 1/2 × 4√2 × 6√2 = 24
Answer
24 square units.