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

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)
Answer: b) (0, b)

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
Answer: a) 30 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
Answer: b) 13 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)
Answer: c) (-2, 4)

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
Answer: a) 0

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
Answer: b) p=10, q=6

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

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
Answer: 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
Answer: a) (a+c)/2

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
Answer: c) -1/3

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).
    Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.1
  • 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:
    Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.2
  • 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:
Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.3
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
Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.4
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)

Area of a quadrilateral, ABCD = ar(∆ABC) + ar(∆ADC)
Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.5

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