Very Short Answer Questions: Coordinate Geometry

# Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Coordinate Geometry

Q1. Find a point on the y-axis equidistant from (− 5, 2) and (9, − 2).

Let the required point on the y-axis be P (0, y)

∴ PA = PB

⇒ y2 − y2 − 4y − 4y = 81 + 4 − 4 − 25
⇒ − 8y = 85 − 29
⇒ − 8y = 56

⇒

∴ The required point is (0, −7).

Q2. Find a point on x-axis at a distance of 4 units from point A (2, 1).

Let the required point on x-axis be P (x, 0).

∴ PA = 4

⇒ x2 − 4x + 4 + 1 = 42 = 16
⇒ x2 − 4x + 1 + 4 − 16 = 0
⇒ x2 − 4x − 11 = 0

⇒

Q3. Find the distance of the point (3, − 4) from the origin.

The coordinates of origin (0, 0).
∴ Distance of (3, − 4) from the origin

Q4. For what value of x is the distance between the points A (− 3, 2) and B (x, 10) 10 units?

The distance between A (− 3, 2) and B (x, 10)

⇒ (x + 3)2 + (8)2 = 102
⇒ (x + 3)2 = 102 − 82
⇒ (x + 3)2 = (10 − 8) (10 + 8) = 36
⇒
For +ve sign, x =6 − 3 = 3
For −ve sign, x = − 6 − 3 = − 9

Q5. Find a point on the x-axis which is equidistant from points A (5, 2) and B (1, − 2).

The given points are: A (5, 2) and B (1, − 2) Let the required point on the x-axis be C (x, 0).
Since, C is equidistant from A and B.
∴ AC = BC

∴ The required point is (0, 3).

Q6. Establish the relation between x and y when P (x, y) is equidistant from the points A (− 1, 2) and B (2, − 1).

As P is equidistant from A and B
∴ PA = PB

which is the required relation.

Q7. Find a relation between x and y such that the point P (x, y) is equidistant from the points A (−5, 3) and B (7, 2)

Since, P (x, y) is equidistant from A (−5, 3) and B (7, 2)
∴ AP = BP

Q8. In the given figure, ABC is a triangle. D and E are the mid points of the sides BC and AC respectively. Find the length of DE. Prove that

Coordinates of the mid point of BC are:

Coordinates of the mid point of AC are:

Q9. Find the distance between the points

Distance between   is given by

Q10. If the mid point of the line joining the points P (6, b − 2) and Q (− 2, 4) is (2, − 3), find the value of b.

Here, P (6, b − 2) and Q (− 2, 4) are the given points.
∴ Mid point of PQ is given by:

The document Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

 1. What is coordinate geometry?
Ans. Coordinate geometry is a branch of mathematics that deals with the study of geometry using the principles of algebra. It involves the use of coordinates to represent points, lines, and shapes on a plane.
 2. How are coordinates represented in coordinate geometry?
Ans. Coordinates are represented by a pair of numbers (x, y), where x represents the horizontal position and y represents the vertical position of a point on a plane. These numbers are known as the x-coordinate and y-coordinate, respectively.
 3. What is the distance formula in coordinate geometry?
Ans. The distance formula in coordinate geometry is used to find the distance between two points (x1, y1) and (x2, y2) on a plane. It is given by the formula: Distance = √((x2 - x1)^2 + (y2 - y1)^2)
 4. How do you find the midpoint of a line segment in coordinate geometry?
Ans. To find the midpoint of a line segment with endpoints (x1, y1) and (x2, y2), you can use the midpoint formula. It is given by: Midpoint = ((x1 + x2)/2, (y1 + y2)/2) The midpoint is the point that lies exactly halfway between the two endpoints of the line segment.
 5. How do you determine if three points are collinear in coordinate geometry?
Ans. In coordinate geometry, three points are collinear if the slopes of the lines connecting any two of the points are equal. If the slopes of the lines between (x1, y1) and (x2, y2), and between (x1, y1) and (x3, y3) are equal, then the three points are collinear.

## Mathematics (Maths) Class 10

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