| Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
Exercise 10.1
Question 1 A
In which quadrants do the following points lie:
(10, -3)
Sol :(10, -3)
Given coordinate (10,-3) lies in Quadrant IV because as shown in the figure that those coordinate who have (+, -) sign lies in IV quadrants, or can say whose x-axis is “+” and y-axis is “–“ lies in IV Quadrant.
Question 1 B
In which quadrants do the following points lie:
(-4, -6)
Sol :(-4, -6)
Given coordinate (-4,-6) lies in Quadrant III because as shown in the figure that those points which have a sign like this (-, -) or can say whose both x-axis and y-axis is “–“ lies in III quadrants. So (-4,-6) lies in III Quadrant.
Question 1 C
In which quadrants do the following points lie:
(-8, 6)
Sol :(-8, 6)
Given coordinate (-8,6) lies in Quadrant II because as shown in the figure that those points which have a sign like this (-, +) lies in II quadrants.
So (-8,6) lies in III Quadrant. Here also x-axis point is “-” and y-axis point is “+” so (-8,6) lies in Quadrant II.
Given coordinate $\left(\frac{3}{2}, 5\right)$lies, in Quadrant I because as shown in the figure that those coordinate who have (+, +) sign lies in IV quadrants or can say whose x-axis is “+” and y-axis is also “+“ lies in I Quadrant.
Given coordinate is (3,0). This point lies on the x-axis because its y-axis is on origin. Therefore, it lies on the x-axis between the Quadrant I and Quadrant IV.
Given coordinate is (0, -5). This point lies on the y-axis because its x-axis is on origin. Therefore, it lies on the y-axis between the Quadrant III and Quadrant IV.
Here is the graph for coordinate (4,5)
The point having -5 lies on the on the y-axis on the negative side because here x-axis is 0 and when x-axis is 0 then points lies on the y-axis and when y-axis is 0 then point lies on the x-axis.
We can show it on graph with points (0, -5).
Sol :
Here we have three vertices of rectangle say A (-2, 0) B (2, 0) and C (2, 1) so when we start graphing it on the graph as shown in the graph below then after plotting all three vertices you will get something like this.
Therefore, after joining all the vertices with a line segment, we will get our fourth vertex because in rectangle opposites sides are parallel and all the angles are right angle so, by joining all the lines according to properties of the rectangle you will get the fourth vertex. So fourth vertex of a rectangle is (-2, 1).
It is easy to draw a triangle when it’s all vertices are given. We have to just locate all the given points on the graph and join them with a line as shown in the graph below.
Step 1. Locate all the vertices on the graph.
Step 2. Joins all the vertices with a line and it will form a triangle.
Given that the base of the equilateral triangle is on the y-axis and mid-point of the base is at the origin, so its figure will be like this as shown.
So here, O(0,0) is the midpoint of the base.
An equilateral triangle has all sides equal so if O is the mid-point of the base BC, so B and C are the two vertices of the triangle. Now we have two vertices of the triangle, which is the base of equilateral triangle lying on the y-axis. Now if base in on y-axis then x-axis are as bisector of the base and so our third vertices will be on the x-axis either left or right.
So now in right ∆BOA
Pythagoras Theorem: In a right-angled triangle the square of the biggest side(hypotenuse) equals the sum of the squares of the other two sides(Perpendicular and base).
BO2 + OA2 = AB2 { By Pythagoras theorem}
a2 + OA2 = (2a) 2
OA2 = 4a2- a2
OA2 = 3a2
OA = ±a√3
So vertices of triangle are A(±a√3,0) B(0,a)and C(0,-a).
Since AB and AD both have as an endpoint, we can find the coordinate of A by finding the intersection of the two sides.
So the coordinates of A will be where the x-axis and y-axis intersect.
As we know AB lies on the x-axis so the coordinates of B can be found by using the coordinates of A and changing the x-coordinate by the measure of AB.
As we know the opposite side of a rectangle, AD and BC are congruent. Now if we have a measure of BC, we can simply find the y-coordinate of D.
Since AB and AD are the x and y-axes, A is at(0,0), B is at (10,0), C is at (10,8), and D is at (0,8).
Square ABCD. Center = O(0,0) Origin.
AB = BC = 20 units.
Y-coordinates of AB = $\frac{0-20}{2}$ = -10
Y-coordinates of AD = $\frac{0+20}{2}$ = 10
∴the coordinates are :-
A(-10,-10)
B(10,-10)
C(10,10)
D(-10,10)
Question 1 D
In which quadrants do the following points lie:
Sol :Given coordinate $\left(\frac{3}{2}, 5\right)$lies, in Quadrant I because as shown in the figure that those coordinate who have (+, +) sign lies in IV quadrants or can say whose x-axis is “+” and y-axis is also “+“ lies in I Quadrant.
Question 1 E
In which quadrants do the following points lie:
(3, 0)
Sol :(3, 0)
Given coordinate is (3,0). This point lies on the x-axis because its y-axis is on origin. Therefore, it lies on the x-axis between the Quadrant I and Quadrant IV.
Question 1 F
In which quadrants do the following points lie:
(0, -5)
Sol :(0, -5)
Given coordinate is (0, -5). This point lies on the y-axis because its x-axis is on origin. Therefore, it lies on the y-axis between the Quadrant III and Quadrant IV.
Question 2 A
Plot the following points in a rectangular coordinate system:
(4, 5)
Sol :(4, 5)
Here is the graph for coordinate (4,5)
Question 2 B
Plot the following points in a rectangular coordinate system:
(-2,-7)
Sol :(-2,-7)
Question 2 C
Plot the following points in a rectangular coordinate system:
(6,-2)
Sol :(6,-2)
Question 2 D
Plot the following points in a rectangular coordinate system:
(-4, 2)
Sol :(-4, 2)
Question 2 E
Plot the following points in a rectangular coordinate system:
(4, 0)
Sol :(4, 0)
Question 2 F
Plot the following points in a rectangular coordinate system:
(0, 3)
Sol :(0, 3)
Question 3
Where does the point having y-coordinate -5 lie?
Sol :The point having -5 lies on the on the y-axis on the negative side because here x-axis is 0 and when x-axis is 0 then points lies on the y-axis and when y-axis is 0 then point lies on the x-axis.
We can show it on graph with points (0, -5).
Question 4
If three vertices of a rectangle are (-2, 0), (2, 0), (2, 1) find the coordinates of the fourth vertex.
Sol :
Here we have three vertices of rectangle say A (-2, 0) B (2, 0) and C (2, 1) so when we start graphing it on the graph as shown in the graph below then after plotting all three vertices you will get something like this.
Therefore, after joining all the vertices with a line segment, we will get our fourth vertex because in rectangle opposites sides are parallel and all the angles are right angle so, by joining all the lines according to properties of the rectangle you will get the fourth vertex. So fourth vertex of a rectangle is (-2, 1).
Question 5
Draw the triangle whose vertices are (2, 3), (-4, 2) and (3, -1).
Sol :It is easy to draw a triangle when it’s all vertices are given. We have to just locate all the given points on the graph and join them with a line as shown in the graph below.
Step 1. Locate all the vertices on the graph.
Step 2. Joins all the vertices with a line and it will form a triangle.
Question 6
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.
Sol :Given that the base of the equilateral triangle is on the y-axis and mid-point of the base is at the origin, so its figure will be like this as shown.
So here, O(0,0) is the midpoint of the base.
An equilateral triangle has all sides equal so if O is the mid-point of the base BC, so B and C are the two vertices of the triangle. Now we have two vertices of the triangle, which is the base of equilateral triangle lying on the y-axis. Now if base in on y-axis then x-axis are as bisector of the base and so our third vertices will be on the x-axis either left or right.
So now in right ∆BOA
Pythagoras Theorem: In a right-angled triangle the square of the biggest side(hypotenuse) equals the sum of the squares of the other two sides(Perpendicular and base).
BO2 + OA2 = AB2 { By Pythagoras theorem}
a2 + OA2 = (2a) 2
OA2 = 4a2- a2
OA2 = 3a2
OA = ±a√3
So vertices of triangle are A(±a√3,0) B(0,a)and C(0,-a).
Question 7
Let ABCD be a rectangle such that AB = 10 units and BC = 8 units. Taking AB and AD as x and y-axes respectively, find the coordinates of A, B, C and D.
Sol :Since AB and AD both have as an endpoint, we can find the coordinate of A by finding the intersection of the two sides.
So the coordinates of A will be where the x-axis and y-axis intersect.
As we know AB lies on the x-axis so the coordinates of B can be found by using the coordinates of A and changing the x-coordinate by the measure of AB.
As we know the opposite side of a rectangle, AD and BC are congruent. Now if we have a measure of BC, we can simply find the y-coordinate of D.
Since AB and AD are the x and y-axes, A is at(0,0), B is at (10,0), C is at (10,8), and D is at (0,8).
Question 8
ABCD is a square having a length of a side 20 units. Taking the centre of the square as the origin and x and y-axes parallel to AB and AD respectively, find the coordinates of A, B, C and D.
Sol :Square ABCD. Center = O(0,0) Origin.
AB = BC = 20 units.
Y-coordinates of AB = $\frac{0-20}{2}$ = -10
Y-coordinates of AD = $\frac{0+20}{2}$ = 10
∴the coordinates are :-
A(-10,-10)
B(10,-10)
C(10,10)
D(-10,10)
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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