Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 7 Coordinates Questions and Answers Notes Pdf to clear their doubts.
SSLC Maths Chapter 7 Coordinates Questions and Answers
Coordinates Class 10 Questions and Answers Kerala State Syllabus
SCERT Class 10 Maths Chapter 7 Coordinates Solutions
Class 10 Maths Chapter 7 Kerala Syllabus – Geometry And Numbers
(Textbook Page No. 135)
Question 1.
Find the following
(i) The y-coordinates of points on the x-axis
(ii) The x-coordinates of points on the y-axis
(iii) The coordinates of the origin
(iv) The y-coordinates of points on the line parallel to the x-axis, through the point (0, 1)
(v) The x-coordinates of points on the line parallel to the y-axis, through the point (1, 0)
Answer:
(i) 0
(ii) 0
(iii) (0,0)
(iv) 1
(v) 1
Question 2.
Find the coordinates of the other three vertices of the rectangle in the picture.
Answer:
(4, 0), (0, 3), (0, 0)
Question 3.
The sides of the rectangle in the picture below are parallel to the axes and the origin is the midpoint (point of intersection of the diagonals) of the rect-angle.
What are the coordinates of the other three vertices of the rectangle?
Answer:
See the diagram
B(-3, 2), C(-3, -2), D(3, -2)
Question 4.
The picture below shows an equilateral triangle
Find the coordinates of the vertices of the triangle.
Answer:
The picture below shows an equilateral triangle
Triangle ODB is a 30° – 60° – 90° triangle
OD = 2,
DB = 2√3
A (4, 0),
B (2, 2√3),
O (0, 0)
Class 10 Maths Kerala Syllabus Chapter 7 Solutions – Rectangle Math
(Textbook Page No. 142)
Question 1.
All these rectangles shown below have their sides parallel to the axes. Find the coordinates of the other two vertices of each:
Answer:
In the first figure 5(2, 3), D(-2, 4)
In the second figure A(-1, -4), C(2, -2)
In the third figure 5(2, 3), D(-1, 6)
Question 2.
Without drawing the axes, mark these points with the left and right, and top and bottom positions correct. Find the coordinates of the other two vertices of the rectangle with each pair as the coordinates of two opposite vertices and sides parallel to the axes.
(i) (3,5), (7,8)
(ii) (6,2), (5, 4)
(iii) (-3, 5), (-7, 1)
(iv) (-1, -2), (-5, -4)
Answer:
See the figure.
SCERT Class 10 Maths Chapter 7 Solutions – Lengths And Distances
(Textbook Page No. 150)
Question 1.
Calculate the lengths of the sides and the diagonals of the quadrilateral in the picture below
Answer:
The side joining (-3 ,-2) and (1,-2) is parallel to x axis The length of the side is |—3 —1| = 4
The side joining (-3,-2) and (-3,1) is parallel toy axis The length of the side is |—2 —1| = 3
Using distance formula we can calculate other two sides as √10 and √5
Length of diagonal √13 and 5
Question 2.
Prove that by joining the points (2, 1), (3, 4), (-3, 6) we get a right triangle
Answer:
A(2, 1), B(3, 4), C(-3, 6)
AB =\(\sqrt{(2-3)^2+(1-4)^2}=\sqrt{10}\)
BC = \(\sqrt{(3-(-3))^2+(4-6)^2}=\sqrt{40}\)
AC = \(\sqrt{(2-(-3))^2+(1-6)^2}=\sqrt{50}\)
\(\sqrt{10}^2+\sqrt{40}^2=\sqrt{50}^2 .\)
10 + 40 = 50
Points are the vertices of a right triangle.
Question 3.
A circle is drawn with centre at the origin and with radius 10.
(i) Check whether each of the points with coordinates (6, 9), (5, 9), (6, 8) is within the circle, outside the circle or on the circle.
(ii) Write the coordinates of 8 points on this circle.
Answer:
(i) (0, 0) is the origin. Distance from (0, 0) to (6, 9) is \(\sqrt{6^2+9^2}=\sqrt{117}>\sqrt{100}\)
That is point is outside the circle (5, 9) is also outside the circle
Distance from (0, 0) to (6,8) is \(\sqrt{6^2+8^2}\) = 10.
The point is on the circle.
(ii) (6,8) (-6,8) (6, -8) (-6, -8) (8,6) (-8,6) (8, -6) (-8, -6) Note that (10, 0) (0, 10) (-10, 0) (0, -10) are also on the circle.
Question 4.
Calculate the coordinates of the points where the circle centred on (1, 1) and with radius √2 intersects the coordinate axes.
Answer:
Centre of the circle is (1, 1) and radius √2
Clearly circle passes through (0, 0).
Other points where the circle cut the axes are (2, 0), (0, 2)
Question 5.
The vertices of a triangle are the points (0,0), (4,0), (1,3). Calculate the centre and radius of its circumcircle.
Answer:
Let (x, y) be the centre
(x – 1)2 + (y – 3)2 = (x – 0)2 + (y – 0)2 = (x – 4)2 + (y – 0)2
(x – 1)2 + (y – 3)2 = (x – 0)2 + (y – 0)2
⇒ 2x + 6y = 10
(x – 0)2 + (y – 0)2 = (x – 4)2 + (y – 0)2
⇒ -8x + 16 = 0, x = 2
x = 2,y = 1. Centre of the circle (2,1).
Radius of the circle √5
Coordinates Class 10 Notes Pdf
Class 10 Maths Chapter 7 Coordinates Notes Kerala Syllabus
Introduction
Coordinate geometry or Cartesian geometry is a vast area of study in higher mathematics. In this unit we discuss how is a point located with reference to two perpendicular lines.
The perpendicular lines are called coordinate axes. The pair of numbers used to locate the point are called coordinates.
The distance between two points is calculated using a formula. This formula is popularly known as distance formula. It is the application of Pythagoras theorem in coordinate geometry.
→ The graph of a polynomial can be drawn by taking a horizontal line and a vertical line as the reference.These lines intersect at a point. All distances are measured from this point. The distance in the horizontal line and vertical line together locate the points to draw the graph.
→ “Geogebra’ is an effective tool for drawing graphs of polynomials as well as the geometric figures.
→ A pair of numbers written in the bracket in the form (a, b) locate the position of the point with reference to the horizontal and vertical lines generally known as coordinate axes. The horizontal line is called x – axis and vertical line is called y – axis. The intersecting point of the axes is the origin of coordinates.
→ The following points will help you to solve the problems related to coordinates
- Opposite sides of a rectangle, square, parallelogram and rhombus are equal
- Diagonals of a rectangle, square, parallelogram and rhombus bisect eachother. For rhombus and square the diagonals bisect perpendicularly.
→ A(1, 1) represents a point at the distance 1 horizontally and at the distance 1 vertically. A will be the right top vertex of the square of side 1 and bottom left vertex as origin.
→ Note that (1, 0), (2, 0), (3, 0) …… are the points on x axis. The points (0, 1), (0, 2), (0, 3) ………. y coordinates of all points on the line parallel to x axis are equal x coordinates of all points on the line parallel to y axis are equal
→ Let A (x1, x2) and B (x2, y2) be two points with reference to coordinate axes.
Draw horizontal line through A and vertical line from S.These lines intersect at P. Coordinates of P are (x2, y1)
Distance AP = |x2 – x1|, Distance BP = |y1 – y1|
By pythagoras theorem AB2 = AP2 + BP2 ⇒ AB2 – (|x2 – x1|)2 + (|y2 – y1|)2
Distance between (x1, y1) and (x2, y2) is \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
→ The Cartesian plane consists of two perpendicular lifies: the x – axis (horizontal line) and the y-axis (vertical line), intersecting at the origin (0, 0).
→ A point’s position is represented in the form of (x, y), where: x is the abscissa (distance from the y – axis).
y is the ordinate (distance from the -axis).
→ If two points have the same x-coordinate, then the line joining them is parallel to the y-axis.
→ If two points have the same y-coordinate, then the line joining them is parallel to the x-axis.
→ In the rectangle with opposite vertices (x1, y1) (x2, y2)and sides parallel to the axes
the length of the top and bottom sides is |x1 – x2|
the length of the left and right sides is |y1 – y2|
→ Distance Formula
(i) Distance between two points:
If the coordinates of two points are (x1, y1) and (x2, y2) such that x1 ≠ x2, and y1 ≠ y2, then the distance between these points is \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
(ii) Distance from the origin:
The distance between the point with coordinates (x, y) and the origin is \(\sqrt{x^2+y^2}\)