The coordinate geometry questions are given here, along with answers, to help students understand the concept easily. The chapter coordinate geometry has been included in Class 9 and 10. The Class 9 coordinate geometry chapter includes a basic introduction to coordinate geometry, how to locate the points in a coordinate plane and the equality of two points on a coordinate system. In Class 10, the coordinate geometry chapter deals with finding the distance between two points, section formula and area of a triangle whose vertices are given in the form of coordinate points, etc. In this article, you will get some important questions on coordinate geometry, as per the latest NCERT curriculum.
What is Coordinate Geometry?
Coordinate geometry is one of the important branches of Mathematics in which the position of a point in a plane is described using coordinates. Hence, the plane is called the Cartesian system or Cartesian plane.
Learn: Coordinate geometry
| The coordinates of a point in all the quadrants are of the form-
First quadrant: (+, +) Second quadrant: (-, +) Third quadrant: (-, -) Fourth quadrant: (+, -) Here, + denotes a positive real number and – denotes a negative real number. If x ≠ y, then (x, y) ≠ (y, x), and (x, y) = (y, x), if x = y. |
Coordinate Geometry Questions and Answers
1. What is the name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane?
Solution:
As we know, to locate the position of an object or a point in a plane (Cartesian or coordinate plane), we require two perpendicular lines. One of them is horizontal, and the other is vertical. The horizontal line is the x-axis, and the vertical line is the y-axis.
2. Without plotting the points, indicate the quadrant in which they will lie, if
(i) ordinate is 3 and abscissa is –5
(ii) abscissa is –3 and ordinate is – 5
(iii) ordinate is 3 and abscissa is 5
Solution:
(i) ordinate is 3 and abscissa is –5
Here, the x-coordinate is -5, and the y-coordinate is 3.
The point = (-5, 3), i.e. (-, +)
Hence, the point lies in quadrant II.
(ii) abscissa is –3 and ordinate is – 5
Here, the x-coordinate is -3, and the y-coordinate is -5.
The point = (-3, -5), i.e. (-, -)
Hence, the point lies in quadrant III.
(iii) ordinate is 3 and abscissa is 5
Here, the x-coordinate is 5, and the y-coordinate is 3.
The point = (5, 3), i.e. (+, +)
Hence, the point lies in quadrant I.
3. Find the coordinates of the point
(i) which lies on both the x and y axes.
(ii) whose ordinate is -6, and which lies on the y-axis.
(iii) whose abscissa is 4, and which lies on the x-axis.
Solution:
(i) The point which lies on both x and y-axes is the origin whose coordinates are (0, 0).
(ii) Given that the ordinate is –6, the point lies on the y-axis.
So, the x-coordinate will be zero.
Therefore, the point is (0, -6).
(iii) Given that the abscissa is 4, the point lies on the x-axis.
So, the y-coordinate will be zero.
Therefore, the point is (4, 0).
4. A point lies on the x-axis at a distance of 8 units from the y-axis. What are its coordinates? What will be the coordinates if it lies on the y-axis at a distance of –8 units from the x-axis?
Solution:
Given that the point lies on the x-axis at a distance of 8 units from the y-axis.
That means the point lies in the positive direction of the x-axis, and its y-coordinate is 0.
So, its coordinates are (8, 0).
If the point lies on the y-axis at a distance of –8 units from the x-axis, its x-coordinate must be 0, and the point lies in the negative direction of the y-axis.
So its coordinates are (0, -8).
| The distance between P(x1, y1) and Q(x2, y2) is: \(\begin{array}{l}d =\sqrt{(x_2-x_1)^2 + (y_2 – y_1)^2}\end{array} \)
The distance of a point P(x, y) from the origin is \(\begin{array}{l}\sqrt{x^2 + y^2}\end{array} \) |
|---|
5. Find the distance between two points, A(–1, 2) and B(3, 2).
Solution:
Let the given points be:
A(-1, 2) = (x1, y1)
B(3, 2) = (x2, y2)
Using the distance formula,
d = √[(x2 – x1)2 + (y2 – y1)2]
Distance between A and B is:
AB = √[(3 + 1)2 + (2 – 2)2]
= √16
= 4
Therefore, the distance between two points, A(–1, 2) and B(3, 2) is 4 units.
6. Find the value of a, if the distance between the points P(–3, –14) and Q(a, –5) is 9 units.
Solution:
Let the given points be:
P(-3, -14) = (x1, y1)
Q(a, -5) = (x2, y2)
Using the distance formula,
d = √[(x2 – x1)2 + (y2 – y1)2]
PQ = √[(a + 3)2 + (-5 + 14)2]
√[(a + 3)2 + 81] = 9 {from the given}
Squaring on both sides,
(a + 3)2 + 81 = 81
(a + 3)2 = 0
a + 3 = 0
a = -3
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7. Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.
Solution:
Let P(x, y) be the required point.
From the given,
(4, -3) = (x1, y1)
(8, 5) = (x2, y2)
Using the section formula, we get;
(x, y) =
So,
x = [3(8) + 1(4)]/(3 + 1) = (24 + 4)/4 = 28/4 = 7
y = [3(5) + 1(-3)]/(3 + 1) = (15 – 3)/4 = 12/4 = 3
Therefore, (7, 3) is the required point.
8. If the mid-point of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y) and x + y – 10 = 0, find the value of k.
Solution:
Given that the mid-point of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y).
Using the mid-point formula,
x = (3 + k)/2
y = (4 + 6)/2
y = 10/2 = 5
From the given,
x + y – 10 = 0
Substituting y = 5 in this equation, we get;
[(3 + k)/2] + 5 – 10 = 0(3 + k)/2 = 5
3 + k = 10
k = 10 – 3 = 7
Therefore, k = 7.
9. Find the area of a triangle whose vertices are (1, –1), (–4, 6) and (–3, –5).
Solution:
Let the given points be:
A(1,-1) = (x1, y1)
B(-4, 6) = (x2, y2)
C(-3, -5) = (x3, y3)
The area of the triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is the numerical value of the expression 1/2 |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|.
Now, the area of triangle ABC = (1/2)|1(6 + 5) – 4(-5 + 1) -3(-1 – 6)|
= (1/2) |11 – 4(-4) – 3(-7)|
= (1/2) |11 + 16 + 21|
= (1/2) × 48
= 24
Therefore, the area of the triangle is 24 square units.
10. Find the value of m if the points (5, 1), (–2, –3) and (8, 2m ) are collinear.
Solution:
Let the given points be:
A(5, 1) = (x1, y1)
B(-2, -3) = (x2, y2)
C(8, 2m) = (x3, y3)
We know that the area of the triangle formed by collinear points is 0.
So, the area of triangle ABC = 0
⇒ (1/2) |5(-3 – 2m) – 2(2m – 1) + 8(1 + 3)| = 0
⇒ |5(-3 – 2m) – 2(2m – 1) + 8(4)| = 0
⇒ |-15 – 10m – 4m + 2 + 32| = 0
⇒ |-14m + 19| = 0
⇒ -14m + 19 = 0
⇒ 14m = 19
⇒ m = 19/14
Video Lesson on Coordinate Geometry Toughest Problems
Practice Questions
Solve the following coordinate geometry problems.
- Taking 0.5 cm as 1 unit, plot the following points on the graph paper :
- A (1, 3), B (– 3, – 1), C (1, – 4), D (– 2, 3), E (0, – 8), F (1, 0)
- Find a point that is equidistant from points A (–5, 4) and B (–1, 6)? How many such points are there?
- If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
- The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC.
- 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).
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