NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.1

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.1 are provided to help students understand the concepts of coordinate geometry and the distance formula. These solutions focus on finding the distance between two points in a coordinate plane and are designed to help students master the basics of coordinate geometry.

Exercise 7.1 primarily deals with:

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.1

Coordinate Geometry

Medium

Q.

Find the distance between following pairs of points:
(i)   (2, 3), (4, 1)                     
(ii)  (–5, 7), (–1, 3)                
(iii) (a, b), (–a, –b)                        

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Coordinate Geometry

Medium

Q.

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.

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Coordinate Geometry

Medium

Q.

Determine if the points (1, 5), (2, 3) and (– 2, –11) are collinear.

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Coordinate Geometry

Medium

Q.

Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

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Coordinate Geometry

Medium

Q.

In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. 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.

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Coordinate Geometry

Medium

Q.

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)

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Coordinate Geometry

Medium

Q.

Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

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Coordinate Geometry

Medium

Q.

Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

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Coordinate Geometry

Medium

Q.

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.

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Coordinate Geometry

Medium

Q.

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and  (–3, 4).

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Coordinate Geometry

Medium

Q.

$\text{Find the coordinates of the point which divides the join of (}-\text{1, 7) and (4, }-\text{3) in the ratio 2}:\text{3.}$

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Coordinate Geometry

Medium

Q.

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{To conduct Sports Day activities, in your rectangular }\\ \text{shaped school ground ABCD, lines have been drawn}\\ \text{with chalk powder at a distance of 1m each}\text{. 100 flower}\\ \text{pots have been placed at a distance of 1m from each}\\ \text{other along AD, as shown in the following figure}\text{. Niharika}\\ \text{runs }\frac{1}{4}\text{th the distance AD on the 2nd line and posts a}\\ \text{green flag}\text{. Preet runs }\frac{\text{1}}{5}\text{th the distance AD on the eighth}\\ \text{line and posts a red flag}\text{. What is the distance between}\\ \text{both the flags? If Rashmi has to post a blue flag exactly}\\ \text{halfway between the line segment joining the two flags, }\\ \text{where should she post her flag?}\end{array}$

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Coordinate Geometry

Medium

Q.

Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).

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Coordinate Geometry

Medium

Q.

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.

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• The distance formula to find the distance between two points

  $(x_1, y_1)$(x1​,y1​) and

  $(x_2, y_2)$(x2​,y2​).

• Understanding the relationship between the coordinates of points on a Cartesian plane.

• Applying the distance formula in real-life problems involving distance calculation between two points.

The solutions are explained in a step-by-step manner, which makes it easy for students to grasp the concepts of coordinate geometry and apply them to solve problems confidently in their Class 10 exams.

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)

Solution:

The distance between two points

$(x_1, y_1)$

(x1​,y1​) and

$(x_2, y_2)$

(x2​,y2​) is given by the formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

d=(x2​−x1​)2+(y2​−y1​)2​

(i) Distance between (2, 3) and (4, 1):

$$d = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$

d=(4−2)2+(1−3)2​=22+(−2)2​=4+4​=8​=22​≈2.83

(ii) Distance between (−5, 7) and (−1, 3):

$$d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66$$

d=(−1−(−5))2+(3−7)2​=(4)2+(−4)2​=16+16​=32​=42​≈5.66

(iii) Distance between (a, b) and (−a, −b):

$$d = \sqrt{(-a - a)^2 + (-b - b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2}$$

d=(−a−a)2+(−b−b)2​=(−2a)2+(−2b)2​=4a2+4b2​=2a2+b2​

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:

Using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

d=(x2​−x1​)2+(y2​−y1​)2​

For points (0, 0) and (36, 15):

$$d = \sqrt{(36 - 0)^2 + (15 - 0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39$$

d=(36−0)2+(15−0)2​=362+152​=1296+225​=1521​=39

Thus, the distance between the two points is 39 units. This can also represent the distance between two towns, A and B.

Q3. Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.

Solution:

Three points are collinear if the area of the triangle formed by them is zero or if the slopes between them are equal.

Using the distance formula:

• Distance between A(1, 5) and B(2, 3):

   

  $$AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$AB=(2−1)2+(3−5)2​=12+(−2)2​=1+4​=5​

• Distance between B(2, 3) and C(−2, −11):

   

  $$BC = \sqrt{(−2 - 2)^2 + (−11 - 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212}$$BC=(−2−2)2+(−11−3)2​=(−4)2+(−14)2​=16+196​=212​

• Distance between A(1, 5) and C(−2, −11):

   

  $$AC = \sqrt{(−2 - 1)^2 + (−11 - 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$$AC=(−2−1)2+(−11−5)2​=(−3)2+(−16)2​=9+256​=265​

Since

$AB + BC \neq AC$

AB+BC=AC and the distances are not proportional, the points (1, 5), (2, 3), and (−2, −11) are not collinear.

Q4. Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.

Solution:

To check if the points form an isosceles triangle, calculate the distances between all pairs of points.

• Distance between A(5, −2) and B(6, 4):

   

  $$AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$$AB=(6−5)2+(4−(−2))2​=12+62​=1+36​=37​

• Distance between B(6, 4) and C(7, −2):

   

  $$BC = \sqrt{(7 - 6)^2 + (−2 - 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$$BC=(7−6)2+(−2−4)2​=12+(−6)2​=1+36​=37​

• Distance between A(5, −2) and C(7, −2):

   

  $$AC = \sqrt{(7 - 5)^2 + (−2 - (−2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$$AC=(7−5)2+(−2−(−2))2​=22+02​=4​=2

Since

$AB = BC$

AB=BC, the triangle is isosceles.

Q5. In a classroom, 4 friends are seated at the points A, B, C, and D. 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 the distance formula, find which of them is correct.

Solution:

Let the points be:

• A(3, 4), B(6, 7), C(9, 4), D(6, 1)

Using the distance formula:

• Distance AB:

   

  $$AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$$AB=(6−3)2+(7−4)2​=32+32​=18​=32​

• Distance BC:

   

  $$BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$$BC=(9−6)2+(4−7)2​=32+(−3)2​=18​=32​

• Distance CD:

   

  $$CD = \sqrt{(9 - 6)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$$CD=(9−6)2+(4−1)2​=32+32​=18​=32​

• Distance DA:

   

  $$DA = \sqrt{(6 - 3)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$$DA=(6−3)2+(1−4)2​=32+(−3)2​=18​=32​

Since all four sides are equal and the diagonals (AC and BD) are also 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)

Solution:

(i) Quadrilateral formed by points (−1, −2), (1, 0), (−1, 2), (−3, 0):

• The quadrilateral has equal sides and opposite sides are parallel, so it forms a square.

(ii) Quadrilateral formed by points (−3, 5), (3, 1), (0, 3), (−1, −4):

• The distances between points are not equal, and the diagonals are also not equal. Therefore, this forms a parallelogram.

(iii) Quadrilateral formed by points (4, 5), (7, 6), (4, 3), (1, 2):

• This quadrilateral has different side lengths and does not form a square, rectangle, or rhombus. Hence, it is a general quadrilateral.

Q7. Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).

Solution:

Let the point on the x-axis be

$P(x, 0)$

P(x,0).

We use the distance formula to find that the distances from P to (2, −5) and P to (−2, 9) are equal:

$$\text{Distance between } P(x, 0) \text{ and } (2, -5) = \text{Distance between } P(x, 0) \text{ and } (-2, 9)$$

Distance between P(x,0) and (2,−5)=Distance between P(x,0) and (−2,9)

After solving, we find

$x = -7$

x=−7.
Thus, the point on the x-axis equidistant from the given points is

$(-7, 0)$

(−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:

Using the distance formula between P(2, −3) and Q(10, y):

$$\sqrt{(10 - 2)^2 + (y - (-3))^2} = 10$$

(10−2)2+(y−(−3))2​=10

Solving this gives

$y = 3$

y=3 or

$y = -9$

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:

Using the distance formula for equidistant points:

$$\text{Distance from Q to P} = \text{Distance from Q to R}$$

Distance from Q to P=Distance from Q to R

Solving the equations, we find

$x = 4$

x=4 or

$x = -4$

x=−4.
The distances QR and PR are both

$5$

5 units.

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:

Using the distance formula, the point (x, y) satisfies:

$$\sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}$$

(x−3)2+(y−6)2​=(x+3)2+(y−4)2​

After squaring both sides and simplifying, we get the relation:

$$3x + y = 5$$

3x+y=5

FAQs: Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.1

Q1. What is the focus of Exercise 7.1?
Answer:
Exercise 7.1 focuses on the distance formula and solving problems involving finding the distance between two points in the coordinate plane.

Q2. What is the distance formula in coordinate geometry?
Answer:
The distance formula is used to calculate the distance between two points

$(x_1, y_1)$

(x1​,y1​) and

$(x_2, y_2)$

(x2​,y2​) on the coordinate plane:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

d=(x2​−x1​)2+(y2​−y1​)2​

where

$d$

d is the distance between the two points.

Q3. How do I use the distance formula to solve problems?
Answer:

1. Identify the coordinates of the two points.

2. Substitute the values of

   $x_1, y_1, x_2, y_2$x1​,y1​,x2​,y2​ into the distance formula.

3. Simplify the expression to find the distance.

Q4. Why is the distance formula important in coordinate geometry?
Answer:
The distance formula is essential in coordinate geometry because it allows us to find the distance between any two points on the Cartesian plane, which is a key concept for many geometry problems, including calculating the length of line segments, midpoints, and solving real-life geometric problems.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
NCERT Solutions provide step-by-step solutions and clear explanations for applying the distance formula to find the distance between points. By practicing these solutions, students can build their confidence in solving geometry-related problems and prepare well for Class 10 exams.