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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)

Q1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) 

Ans: Let P(2, 3) and Q(4, 1)
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)
(ii) (-5, 7), (-1, 3) 

Ans: Let P (-5, 7) and Q (-1, 3)
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)
(iii) (a, b), (-a, -b)

Ans: Let P (a, b) and Q (-a, -b)
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

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?
Ans: Let points be A (0, 0) and B (36, 15)
The distance between the two points is:
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

The distance between towns A and B will be 39 km.

Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Ans: Let points be A(1, 5), B(2, 3) and C(-2, -11)
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

AB + BC ≠ AC
Hence, the given points are not collinear.

Q4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Ans: Since two sides of any isosceles triangle are equal. To check whether the given points are vertices of an isosceles triangle, we will find the distance between all the points.

Let the points be A (5, -2), B (6, 4) and C (7, -2).
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

Here AB = BC
∴ Δ ABC is an isosceles.

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

NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)Ans: Points A (3, 4), B (6, 7), C (9, 4) and D (6, 1)
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

ABCD is a square.
Hence, 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)
Ans: Let the points be: A (-1, -2), B (1, 0), C (-1, 2) and D (-3, 0).
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

Here AC = BD, AB = BC = CD = AD
Hence, the quadrilateral ABCD is a square.
(ii) (-3, 5), (3, 1), (0, 3), (-1, - 4) 
Let points be A (-3, 5), B (3, 1), C (0, 3) and D (-1, -4)
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

It is seen that points A, B and C are collinear.

So, the given points can only form 3 sides i.e, a triangle and not a quadrilateral that has 4 sides.

Therefore, the given points do not form any quadrilateral.
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Let points be A (4, 5), B (7, 6), C (4, 3) and D (1, 2)
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

Here, AB = CD, BC = AD and AC ≠ BD
The quadrilateral ABCD is a parallelogram.

Q7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Ans: Let points be A (2, -5) and B (-2, 9)
Let P (x, 0) be the point on the x-axis.
PA = PB
 Then, PA= PB2

⇒ (x - 2)2 + (0 + 5)2 = (x + 2)2 + (0 - 9)2
⇒ (x - 2)2 -  (x + 2)2 = 81 - 25
⇒ (x - 2 + x + 2)(x - 2 - x - 2) = 56
⇒ (2x)(-4) = 56
⇒ -8x = 56
⇒ x = -7
Hence, the required 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.
Ans: Points P (2, -3), Q (10, y) and PQ = 10 units
The distance between the two points is:
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)
⇒ y - 3 = 0 or y + 9 = 0
y = 3 or - 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.
Ans: PQ = QR
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)
Therefore, point R is (4, 6) or (−4, 6).
When point R is (4, 6),
NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Ans: Points A (3, 6), B (-3, 4) are given and point P(x, y) is equidistant from points A and B.

NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

The document NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry (Exercise 7.1)

1. What is the distance formula in coordinate geometry?
Ans. The distance formula is used to find the distance between two points in a coordinate plane. If the points are \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) between them is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
2. How do you find the midpoint of a line segment in coordinate geometry?
Ans. The midpoint of a line segment connecting two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) can be found using the midpoint formula: \[ M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \] This gives the coordinates of the midpoint \(M\).
3. What are the coordinates of points that lie on the x-axis and y-axis?
Ans. Points that lie on the x-axis have coordinates of the form \((x, 0)\), where \(x\) can be any real number. Points that lie on the y-axis have coordinates of the form \((0, y)\), where \(y\) can also be any real number.
4. How do you determine if three points are collinear in coordinate geometry?
Ans. To determine if three points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) are collinear, you can use the area method. If the area formed by these three points is zero, then they are collinear. The formula for the area is: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] If this area equals 0, the points are collinear.
5. What is the slope of a line in coordinate geometry, and how is it calculated?
Ans. The slope of a line measures its steepness and is calculated as the ratio of the vertical change to the horizontal change between two points on the line. If the points are \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the slope \(m\) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] This formula shows how much \(y\) changes for a unit change in \(x\).
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