Chapter 11 of Class 10 Mathematics, "Coordinate Geometry," is an exploration of a fundamental branch of mathematics that unites algebra and geometry. This chapter introduces students to the concept of representing points and shapes on a coordinate plane, providing a bridge between abstract algebraic equations and visual geometric shapes. Here's a concise coordinate geometry class 10 ICSE important questions.
"In Chapter 11, 'Coordinate Geometry,' of Class 10 Mathematics, students embark on a captivating journey that connects the world of algebra with the realm of geometry. This chapter unveils the power of the coordinate plane, where points are represented by ordered pairs (x, y), and equations become geometric shapes.
Coordinate geometry class 10 ICSE is an exciting journey that bridges the gap between algebra and geometry. This chapter introduces students to the fascinating world of plotting points on a coordinate plane and discovering the geometric shapes that emerge from algebraic equations.
Coordinate Geometry allows students to represent points in two-dimensional space using ordered pairs (x, y) and explore how these points can create lines, shapes, and patterns. By understanding concepts like slope, distance, and equations of lines, students gain the ability to analyze and describe geometric figures mathematically.
Coordinate Geometry, often referred to as Cartesian Geometry, as we study in Coordinate geometry class 10 ICSE is a branch of mathematics that combines algebra and geometry. It focuses on the study of geometric shapes and figures using a coordinate system, which assigns numerical coordinates to points in space. This system allows for the precise representation of points, lines, curves, and shapes on a graph.
Here are the key components of Coordinate Geometry:
Coordinate Plane: The fundamental idea in coordinate geometry is the use of a two-dimensional plane known as the Cartesian plane. It consists of two perpendicular lines called the x-axis and the y-axis, which intersect at a point called the origin. The axes divide the plane into four quadrants.
Ordered Pairs: To locate a point on the plane, a pair of numbers (x, y) is used, where "x" represents the horizontal position along the x-axis, and "y" represents the vertical position along the y-axis.
Points: Any location in the plane can be represented by a unique ordered pair (x, y). The point (0, 0) is the origin, where both x and y are zero.
Ans. (b)
Explanation:
The co-ordinates of the centroid of a triangle are:
=\(\begin{pmatrix} \frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \end{pmatrix}\)
=\(\begin{pmatrix} \frac{4+5+3}{3},\frac{-3+5-3}{3} \end{pmatrix}\)i.e.(4,0).
Ans. (b) (– 3, 2)
Explanation:
If A’ (x,y) be the reflection of point A(5, - 3) in the point P(3,-2), then P will be mid - point of AA’.
x=\(\frac{x_1+x_2}{2}\) and y=\(\frac{y_1+y_2}{2}\)
3= \(\frac{x+5}{2}\) and - 2 = \(\frac{y-3}{2}\)
⇒ x=1 and y =-1
∴ Reflection point will be (1,-1).
Explanation:
Let m1 be the slope of the line joining at the points (a, 2a) and (– 2, 3), then
m1 = \(\frac{2a-3}{a+2}\)
Also, slope of the line 4x + 3y + 5 = 0
m2 = -\(\frac{4}{3}\)
Since, both the lines are perpendicular.
So, m1 m2 = -1
∴\(\frac{2a-3}{a+2}\)×\(\frac{(-4)}{3}\) = -1
⇒ 8a – 12 = 3a + 6
⇒ 8a – 3a = 18
⇒ 5a = 18
⇒ a =\(\frac{18}{5}\)
⇒ a = 3\(\frac{3}{5}\)
Explanation:
Let m be the slope of required line
Slope of the given line = \(\frac{2+3}{4-2}\)=\(\frac{5}{2}\)
But the required line is perpendicular to the given line.
Hence,
m × Slope of the given line = – 1
⇒ m× \(\frac{5}{2}\)=-1
⇒ m =\(\frac{-2}{5}\)
∴ Y-intercept, c = 3
Hence, equation of the required line is given by
y = mx + c
i.e., y = \(\frac{-2}{5}\)x+3
⇒ 5y = – 2x + 15
⇒ 2x + 5y – 15 = 0.
Explanation:
Given, A (3, 8), B (– 1, 2) and C (6, – 6).
(a) Slope of BC(m1)= \(\frac{y2-y1}{x2-x1}\)
=\(\frac{-6-2}{6-(-1)}\)
=\(\frac{-8}{7}\)
(b) Slope of a line perpendicular to BC (m)
=\(\frac{1}{m_1}\)
=-\(\frac{1}{-8/7}\)=\(\frac{7}{8}\)
Let, the equation of the line perpendicular to BC and through A be
y – y1 = m (x – x1)
⇒ y-8 =\(\frac{7}{8}\)(x-3)
⇒ 8(y – 8) = 7 (x – 3)
⇒ 8y – 64 = 7x – 21
⇒ 7x – 8y – 21 + 64 = 0
⇒ 7x – 8y + 43 = 0
Which is the required equation.
In conclusion, the study of Coordinate Geometry in ICSE Class 10 Mathematics offers students a powerful bridge between algebra and geometry, allowing them to precisely represent and analyze geometric shapes and patterns in a two-dimensional plane.
By mastering the use of coordinates and equations, students gain essential problem-solving skills that extend far beyond mathematics.
Throughout this chapter, students have explored the principles of the Cartesian plane, learned to plot points, find distances between them, calculate slopes of lines, and work with equations that represent geometric figures. These skills are not only valuable for academic success but also for understanding the real-world applications of coordinate geometry in fields such as engineering, physics, and computer science.
If you're eager to sharpen your skills further and develop a deeper grasp of the concepts covered in this chapter, you'll find coordinate geometry class 10 ICSE important questions available on oswal.io. These resources are designed to assist you in your journey toward a more profound understanding of the subject matter.
Ans: Coordinate Geometry is a branch of mathematics that uses a coordinate system to represent points, lines, and shapes on a two-dimensional plane.
Ans: The Cartesian Plane is the grid used in coordinate geometry, consisting of two perpendicular axes, the x-axis and the y-axis, which intersect at the origin (0,0)
Ans: To plot a point, use an ordered pair (x, y), where "x" represents the horizontal position (along the x-axis) and "y" represents the vertical position (along the y-axis).
Ans: The origin is the point where both x and y coordinates are zero, serving as a reference point for all other coordinates on the plane.
Ans: You can use the distance formula: √((x2 - x1)² + (y2 - y1)²), where (x1, y1) and (x2, y2) are the coordinates of the two points.