Chapter 7: Coordinate Geometry
Exercise 7.1
Q1
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
Distance Formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
(i) (2, 3), (4, 1)
\[ d = \sqrt{(4 - 2)^2 + (1 - 3)^2} \]
\[ = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \]
(ii) (-5, 7), (-1, 3)
\[ d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} \]
\[ = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units} \]
(iii) (a, b), (-a, -b)
\[ d = \sqrt{(-a - a)^2 + (-b - b)^2} \]
\[ = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} \]
\[ = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2} \text{ units} \]
Q2
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
Solution:
Let \( P(0, 0) \) and \( Q(36, 15) \).
\[ PQ = \sqrt{(36 - 0)^2 + (15 - 0)^2} \]
\[ = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} \]
\[ = \sqrt{1521} = 39 \]
Answer: The distance is 39 units. Yes, the distance between town A and town B is 39 km.
Q3
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let \( A(1, 5) \), \( B(2, 3) \), and \( C(-2, -11) \).
Distance AB:
\[ AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.24 \]
Distance BC:
\[ BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} \approx 14.56 \]
Distance AC:
\[ AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265} \approx 16.28 \]
Since \( AB + BC \neq AC \) (2.24 + 14.56 = 16.8 \( \neq \) 16.28), the points are not collinear.
Q4
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let \( A(5, -2) \), \( B(6, 4) \), and \( C(7, -2) \).
\[ AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{37} \]
\[ BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{37} \]
\[ AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \]
Since \( AB = BC \), two sides are equal. Therefore, the points form an isosceles triangle.
Q5
In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
From the figure, the coordinates are: \( A(3, 4) \), \( B(6, 7) \), \( C(9, 4) \), \( D(6, 1) \).
Sides:
\[ AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \]
\[ BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2} \]
\[ CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2} \]
\[ DA = \sqrt{(3-6)^2 + (4-1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \]
All sides are equal.
Diagonals:
\[ AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{6^2} = 6 \]
\[ BD = \sqrt{(6-6)^2 + (1-7)^2} = \sqrt{(-6)^2} = 6 \]
Since sides are equal and diagonals are equal, ABCD is a square. Champa is correct.
Q6
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0).
\( AB = \sqrt{2^2+2^2} = \sqrt{8} \), \( BC = \sqrt{(-2)^2+2^2} = \sqrt{8} \), \( CD = \sqrt{(-2)^2+(-2)^2} = \sqrt{8} \), \( DA = \sqrt{2^2+(-2)^2} = \sqrt{8} \).
Diagonals: \( AC = \sqrt{0^2+4^2} = 4 \), \( BD = \sqrt{(-4)^2+0^2} = 4 \).
All sides equal, diagonals equal. It is a square.
(ii) Points: A(-3, 5), B(3, 1), C(0, 3), D(-1, -4).
Calculating distances reveals no specific pattern of equal sides or diagonals. Checking collinearity: \( AB+BC \neq AC \), but let's check basic lengths.
\( AB = \sqrt{52}, BC = \sqrt{13}, CD = \sqrt{50}, DA = \sqrt{85} \). Also \( AC = \sqrt{13} \). Here \( AB + BC = 2\sqrt{13} + \sqrt{13} \) no match. But \( AC + BC = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = \sqrt{52} = AB \).
Since \( AC + BC = AB \), points A, B, C are collinear. Thus, a quadrilateral is not formed.
(iii) Let A(4, 5), B(7, 6), C(4, 3), D(1, 2).
\( AB = \sqrt{10} \), \( BC = \sqrt{18} \), \( CD = \sqrt{10} \), \( DA = \sqrt{18} \). Opposite sides equal.
Diagonals: \( AC = 2 \), \( BD = \sqrt{52} \). Diagonals unequal.
It is a parallelogram.
Q7
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
A point on the x-axis has coordinates \( P(x, 0) \).
Given \( PA = PB \), where \( A(2, -5) \) and \( B(-2, 9) \).
\[ PA^2 = PB^2 \]
\[ (x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2 \]
\[ (x - 2)^2 + 25 = (x + 2)^2 + 81 \]
\[ x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \]
\[ -4x + 29 = 4x + 85 \]
\[ -8x = 56 \Rightarrow x = -7 \]
Answer: The point is \( (-7, 0) \).
Q8
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:
Given \( PQ = 10 \). So \( PQ^2 = 100 \).
\[ (10 - 2)^2 + (y - (-3))^2 = 100 \]
\[ 8^2 + (y + 3)^2 = 100 \]
\[ 64 + (y + 3)^2 = 100 \]
\[ (y + 3)^2 = 36 \]
\[ y + 3 = \pm 6 \]
Case 1: \( y + 3 = 6 \Rightarrow y = 3 \)
Case 2: \( y + 3 = -6 \Rightarrow y = -9 \)
Answer: \( y = 3 \) or \( y = -9 \).
Q9
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Given \( QP = QR \Rightarrow QP^2 = QR^2 \).
\[ (5 - 0)^2 + (-3 - 1)^2 = (x - 0)^2 + (6 - 1)^2 \]
\[ 25 + 16 = x^2 + 25 \]
\[ x^2 = 16 \Rightarrow x = \pm 4 \]
If \( x = 4 \): \( R(4, 6) \)
\( QR = \sqrt{(4-0)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41} \)
\( PR = \sqrt{(4-5)^2 + (6-(-3))^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82} \)
If \( x = -4 \): \( R(-4, 6) \)
\( QR = \sqrt{(-4-0)^2 + (6-1)^2} = \sqrt{16 + 25} = \sqrt{41} \)
\( PR = \sqrt{(-4-5)^2 + (6-(-3))^2} = \sqrt{(-9)^2 + 9^2} = \sqrt{81 + 81} = 9\sqrt{2} \)
Q10
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution:
Let \( P(x, y) \), \( A(3, 6) \), and \( B(-3, 4) \).
\( PA = PB \Rightarrow PA^2 = PB^2 \).
\[ (x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2 \]
\[ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \]
\[ -6x - 12y + 45 = 6x - 8y + 25 \]
\[ -12x - 4y + 20 = 0 \]
Divide by -4:
\[ 3x + y - 5 = 0 \]
Answer: \( 3x + y - 5 = 0 \)
Exercise 7.2
Q1
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
Solution:
Using Section Formula: \( x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}, y = \frac{m_1y_2 + m_2y_1}{m_1+m_2} \)
Here \( x_1=-1, y_1=7, x_2=4, y_2=-3, m_1=2, m_2=3 \).
\[ x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1 \]
\[ y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \]
Answer: (1, 3)
Q2
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution:
Let A(4, -1) and B(-2, -3). Let P and Q be points of trisection such that AP = PQ = QB.
For Point P: Ratio is 1:2.
\[ x_p = \frac{1(-2) + 2(4)}{1+2} = \frac{-2 + 8}{3} = 2 \]
\[ y_p = \frac{1(-3) + 2(-1)}{1+2} = \frac{-3 - 2}{3} = -\frac{5}{3} \]
Point P is \( (2, -\frac{5}{3}) \).
For Point Q: Ratio is 2:1.
\[ x_q = \frac{2(-2) + 1(4)}{2+1} = \frac{-4 + 4}{3} = 0 \]
\[ y_q = \frac{2(-3) + 1(-1)}{2+1} = \frac{-6 - 1}{3} = -\frac{7}{3} \]
Point Q is \( (0, -\frac{7}{3}) \).
Q3
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
Distance AD = 100 m.
Niharika's position (Green Flag):
Line = 2 (x-coordinate). Distance = \( \frac{1}{4} \times 100 = 25 \) m (y-coordinate).
Coordinates \( G(2, 25) \).
Preet's position (Red Flag):
Line = 8 (x-coordinate). Distance = \( \frac{1}{5} \times 100 = 20 \) m (y-coordinate).
Coordinates \( R(8, 20) \).
Distance between flags:
\[ GR = \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \text{ m} \]
Rashmi's position (Blue Flag):
Midpoint of GR.
\[ x = \frac{2+8}{2} = 5, \quad y = \frac{25+20}{2} = 22.5 \]
Rashmi should post her flag on the 5th line at a distance of 22.5 m.
Q4
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the ratio be \( k:1 \).
Using the section formula for the x-coordinate:
\[ -1 = \frac{k(6) + 1(-3)}{k+1} \]
\[ -1(k+1) = 6k - 3 \]
\[ -k - 1 = 6k - 3 \]
\[ 2 = 7k \Rightarrow k = \frac{2}{7} \]
Answer: The ratio is 2:7.
Q5
Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Solution:
A point on the x-axis has coordinates \( (x, 0) \). Let the ratio be \( k:1 \).
Using the y-coordinate formula:
\[ 0 = \frac{k(5) + 1(-5)}{k+1} \]
\[ 5k - 5 = 0 \Rightarrow 5k = 5 \Rightarrow k = 1 \]
So the ratio is 1:1 (midpoint).
Now find the x-coordinate:
\[ x = \frac{1(-4) + 1(1)}{2} = \frac{-3}{2} \]
Answer: Ratio 1:1, Point \( (-\frac{3}{2}, 0) \).
Q6
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
The diagonals of a parallelogram bisect each other. So, the midpoint of AC is the same as the midpoint of BD.
Let A(1, 2), B(4, y), C(x, 6), D(3, 5).
Midpoint of AC: \( (\frac{1+x}{2}, \frac{2+6}{2}) = (\frac{1+x}{2}, 4) \)
Midpoint of BD: \( (\frac{4+3}{2}, \frac{y+5}{2}) = (\frac{7}{2}, \frac{y+5}{2}) \)
Comparing coordinates:
\[ \frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6 \]
\[ 4 = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3 \]
Answer: \( x = 6, y = 3 \).
Q7
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Solution:
Let coordinates of A be \( (x, y) \). The centre \( C(2, -3) \) is the midpoint of AB.
\[ 2 = \frac{x + 1}{2} \Rightarrow 4 = x + 1 \Rightarrow x = 3 \]
\[ -3 = \frac{y + 4}{2} \Rightarrow -6 = y + 4 \Rightarrow y = -10 \]
Answer: A is \( (3, -10) \).
Q8
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
Solution:
Given \( AP = \frac{3}{7}AB \). This means P divides AB into two parts, AP and PB.
\( AP : AB = 3 : 7 \). Since \( AB = AP + PB \), \( PB = 4 \) parts.
So, ratio \( m_1 : m_2 = 3 : 4 \).
Using section formula:
\[ x = \frac{3(2) + 4(-2)}{3+4} = \frac{6 - 8}{7} = -\frac{2}{7} \]
\[ y = \frac{3(-4) + 4(-2)}{3+4} = \frac{-12 - 8}{7} = -\frac{20}{7} \]
Answer: \( (-\frac{2}{7}, -\frac{20}{7}) \)
Q9
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Let points be \( P_1, P_2, P_3 \). \( P_2 \) is the midpoint of AB.
\[ P_2 = (\frac{-2+2}{2}, \frac{2+8}{2}) = (0, 5) \]
\( P_1 \) is the midpoint of A and \( P_2 \):
\[ P_1 = (\frac{-2+0}{2}, \frac{2+5}{2}) = (-1, \frac{7}{2}) \]
\( P_3 \) is the midpoint of \( P_2 \) and B:
\[ P_3 = (\frac{0+2}{2}, \frac{5+8}{2}) = (1, \frac{13}{2}) \]
Q10
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
Solution:
Let A(3, 0), B(4, 5), C(-1, 4), D(-2, -1).
Area of Rhombus = \( \frac{1}{2} \times d_1 \times d_2 \).
Diagonal \( d_1 = AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2} \).
Diagonal \( d_2 = BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2} \).
\[ \text{Area} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24 \text{ sq units} \]