# CBSE Class 10 Maths Notes & Important Questions Chapter 7 Coordinate Geometry

## CBSE Class 10 Maths Notes Chapter 7 Coordinate Geometry

Chapter 7 of CBSE Class 10 Maths is “Coordinate Geometry.” This chapter mainly deals with the study of geometrical shapes using the concept of coordinates in a two-dimensional plane. Some of the main topics covered in this chapter are:

1. Introduction to Coordinate Geometry
2. Cartesian System
3. Plotting a Point in a Plane
4. Distance Formula
5. Section Formula
6. Area of a Triangle

Here’s a brief overview of each of these topics:

1. Introduction to Coordinate Geometry: The chapter starts with an introduction to the concept of coordinate geometry and its relevance in mathematics.
2. Cartesian System: This section deals with the Cartesian coordinate system, which is a system of coordinates that allows us to locate any point on a plane.
3. Plotting a Point in a Plane: In this section, we learn how to plot a point on a plane using the Cartesian coordinate system.
4. Distance Formula: This section explains the distance formula, which is used to find the distance between two points in a plane.
5. Section Formula: This formula is used to find the coordinates of a point that divides a line segment in a given ratio.
6. Area of a Triangle: This section explains how to find the area of a triangle using coordinates.

Overall, this chapter is important because it lays the foundation for the study of coordinate geometry and helps students develop their spatial reasoning skills.

## CBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Here are some important questions from CBSE Class 10 Maths Chapter 7 – Coordinate Geometry:

1. What is the distance between the points (-2, 3) and (4, -5)?
2. Find the coordinates of the midpoint of the line segment joining (3, 4) and (7, 8).
3. What is the slope of the line passing through the points (3, -2) and (5, 6)?
4. Find the area of the triangle whose vertices are (-1, 2), (3, 4), and (2, -2).
5. Find the equation of the line passing through the points (1, 2) and (3, 4) in slope-intercept form.
6. Find the value of k if the points (k, 1), (4, -3), and (7, -6) are collinear.
7. Find the distance between the x-intercept and y-intercept of the line 2x + 3y = 12.
8. Find the coordinates of the point which divides the line segment joining (-3, -4) and (5, 6) in the ratio 2:3.
9. Find the equation of the line passing through the point (2, -5) and parallel to the line 3x + 4y = 8.
10. Find the equation of the line passing through the point (-2, 5) and perpendicular to the line 4x – 3y = 6.

These questions cover the major topics in the chapter, such as distance formula, midpoint formula, slope of a line, area of a triangle, collinear points, intercepts, and equations of lines. It is important to practice these questions to gain a thorough understanding of the concepts and their applications in coordinate geometry.

## CBSE Class 10 Maths Important Questions Answers Chapter 7 Coordinate Geometry

Here are the answers to some important questions from CBSE Class 10 Maths Chapter 7 – Coordinate Geometry:

1. Distance between the points (-2, 3) and (4, -5) = √[(4 – (-2))^2 + (-5 – 3)^2] = √(6^2 + (-8)^2) = √(36 + 64) = √100 = 10.
2. Midpoint of the line segment joining (3, 4) and (7, 8) = [(3+7)/2, (4+8)/2] = (5, 6).
3. Slope of the line passing through the points (3, -2) and (5, 6) = (6 – (-2))/(5 – 3) = 8/2 = 4.
4. Area of the triangle whose vertices are (-1, 2), (3, 4), and (2, -2) = 1/2 |(-1(4 – (-2)) + 3(-2 – 2) + 2(2 – 4))| = 1/2 |(-12)| = 6 square units.
5. Slope of the line passing through the points (1, 2) and (3, 4) = (4 – 2)/(3 – 1) = 2/2 = 1. Equation of the line in slope-intercept form = y – y1 = m(x – x1), where m is the slope and (x1, y1) is any point on the line. Using (1, 2) as the point, we get y – 2 = 1(x – 1), which simplifies to y = x + 1.
6. The slope of the line passing through the points (k, 1), (4, -3), and (7, -6) must be the same. Slope between (k, 1) and (4, -3) = (-3 – 1)/(4 – k) = -4/(k – 4). Slope between (4, -3) and (7, -6) = (-6 – (-3))/(7 – 4) = -1. Equating both slopes, we get -4/(k – 4) = -1, which gives k = 0.
7. x-intercept is obtained by putting y = 0 in the equation 2x + 3y = 12, which gives x = 6. y-intercept is obtained by putting x = 0, which gives y = 4. Therefore, distance between the intercepts = √((6 – 0)^2 + (0 – 4)^2) = √(36 + 16) = √52 = 2√13 units.
8. Let the coordinates of the required point be (x, y). Then, (x – (-3))/5 = 2/3, which gives x = 1 and (y – (-4))/6 = 3/5, which gives y = 2. Therefore, the required point is (1, 2).
9. Slope of the line 3x + 4y = 8 in slope-intercept form = -3/4. Since the given line is parallel to this line, it must have the same slope. Equation of the line passing through (2, -5) with slope -3/4 in point-slope form = y – (-5) = (-3/4)(x – 2),

## CBSE Class 10 Maths Important Questions Answers MCQs Chapter 7 Coordinate Geometry

The coordinates of a point on the y-axis are:
a) (a, b)
b) (0, b)
c) (a, 0)
d) (0, 0)

The area of the triangle formed by the points (-2, 3), (4, 1) and (3, -5) is:
a) 30 sq units
b) 15 sq units
c) 25 sq units
d) 20 sq units

The distance between the points (-2, 4) and (6, -3) is:
a) 9 units
b) 13 units
c) 10 units
d) 11 units

The coordinates of the midpoint of the line segment joining the points (2, 3) and (-4, 5) are:
a) (-1, 4)
b) (3, 2)
c) (-2, 4)
d) (-1, 3)

If the point (x, -3) lies on the x-axis, then the value of x is:
a) 0
b) 3
c) -3
d) any real number

If the points (4, 6), (p, 3) and (8, q) are collinear, then the values of p and q are:
a) p=6, q=10
b) p=10, q=6
c) p=6, q=6
d) p=10, q=10

The equation of the y-axis is:
a) x=0
b) y=0
c) x=y
d) x=-y

The slope of the line passing through the points (3, 5) and (2, 7) is:
a) -2
b) 2
c) 1/2
d) -1/2

If the points (a, b) and (c, d) are equidistant from the point (p, q), then the value of p is:
a) (a+c)/2
b) (b+d)/2
c) (a+c+d)/3
d) (b+d+q)/3

If the equation of a line is y = 3x – 2, then the slope of the line perpendicular to it is:
a) 1/3
b) -3
c) -1/3
d) 3

## CBSE Class 10 Maths Notes Chapter 7 Coordinate Geometry

• Position of a point P in the Cartesian plane with respect to co-ordinate axes is represented by the ordered pair (x, y).
• The line X’OX is called the X-axis and YOY’ is called the Y-axis.
• The part of intersection of the X-axis and Y-axis is called the origin O and the co-ordinates of O are (0, 0).
• The perpendicular distance of a point P from the Y-axis is the ‘x’ co-ordinate and is called the abscissa.
• The perpendicular distance of a point P from the X-axis is the ‘y’ co-ordinate and is called the ordinate.
• Signs of abscissa and ordinate in different quadrants are as given in the diagram:
• Any point on the X-axis is of the form (x, 0).
• Any point on the Y-axis is of the form (0, y).
• The distance between two points P(x1, y1) and Q (x2, y2) is given by
PQ = (x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
Note. If O is the origin, the distance of a point P(x, y) from the origin O(0, 0) is given by
OP = x2+y2−−−−−−√

Section formula. The coordinates of the point which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m : n are:

The above formula is section formula. The ratio m: n can also be written as mn : 1 or k : 1, The
co-ordinates of P can also be written as P(x,y) = kx2+x1k+1,ky2+y1k+1

The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is

Here m : n = 1 :1.

Area of a Triangle. The area of a triangle formed by points A(x1 y1), B(x2, y2) and C(x3, y3) is given by | ∆ |,
where ∆ = 12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
where ∆ represents the absolute value.

• Three points are collinear if |A| = 0.
• If P is centroid of a triangle then the median divides it in the ratio 2 :1. Co-ordinates of P are given by
P=(x1+x2+x33,y1+y2+y33)