Orienting Yourself The Use of Coordinates Class 9 Extra Questions Maths Chapter 1 - #NCSOLVE 📚

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Class 9 Orienting Yourself The Use of Coordinates Extra Questions

Extra Questions of Orienting Yourself The Use of Coordinates

Class 9 Maths Chapter 1 Extra Questions

Orienting Yourself The Use of Coordinates Class 9 Short Question Answer

Question 1.
What are the coordinates of A, B, C and D in the following figure?

Solution:
The coordinates of A are (-4, 3).
The coordinates of B are (4, 2).
The coordinates of C are (-3, -2).
The coordinates of D are (4, -2).

Question 2.
Look at the following figure and answer the following:

(i) Which point is having its y-coordinate as (-5)?
(ii) Which point is having x-coordinate as 4?
(iii) What are the coordinates of origin?
(iv) Which point is having x-coordinate as (-3)?
Solution:
(i) The point B is having its y-coordinate as (-5).
(ii) The point C is having its x-coordinate as 4.
(iii) The coordinates of O are (0, 0).
(iv) The point A is having x-coordinate as (-3).

Question 3.
Look at the following figure and answer the following:
(i) Which two points have the same x-coordinate?
(ii) Which two points have the same y-coordinate?

Solution:
(i) ∵ The x-coordinate of the point Q is 6.
The x-coordinate of the point R is 6.
The points Q and R have the same x-coordinate.

(ii) ∵ The y-coordinate of P is (-3).
The y-coordinate of R is (-3).
∴ The points P and R have the same y-coordinate.

Question 4.
Find a point on the y-axis equidistant from (- 5, 2) and (9, – 2).
Solution:
Let the required point on the y-axis be P(0, y)
∴ PA = PB
⇒ \(\sqrt{(0+5)^2+(y-2)^2}=\sqrt{(0-9)^2+(y+2)^2}\)
⇒ \(\sqrt{5^2+y^2+4-4 y}=\sqrt{(-9)^2+y^2+4+4 y}\)
⇒ 25 + y² + 4 – 4y = 81 + y² + 4 + 4y
⇒ y² – y² – 4y – 4y = 81 + 4 – 4 – 25
⇒ -8y = 85 – 29
⇒ – 8y = 56

Question 5.
Find the distance of the point (3, – 4) from the origin.
Solution:
The coordinates of origin (0, 0).
∴ Distance of (3, – 4) from the origin
= \(\sqrt{(3-0)^2+(-4-0)^2}\)
= \(\sqrt{(3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}\) = 5 units

Question 6.
Point P (5, -3) is one of the two points of trisection of the line segment joining the points A(7, -2) and B( 1, -5) near to A. Find the coordinates of the other point of trisection.
Solution:

Since P is near to A.
∴ Other point Q is the mid point of PB
⇒ x = \(\frac{1}{2}\) = 3 = -4
Thus, the point Q is (3, – 4).

Question 7.
If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of ΔABC, then find the length of the median through the vertex A.
Solution:
∵ AD is the median on BC.

∴ D is the mid-point of BC.
⇒ Coordinates of D are:
\(\frac{1}{2}\) i.e (2, 1)
Now, the length of the median
AD = \(\sqrt{(2+2)^2+(1-4)^2}\)
= \(\sqrt{(4)^2+(-3)^2}\)
= \(\sqrt{16+9}=\sqrt{25}\)
= 5 units

Question 8.
If the points A(4, 3) and B(x, 5) are on the circle with the centre 0(2, 3), find the value of x.
Solution:
Let 0(2, 3) be the centre of the circle.
∴ OA = OB
⇒ OA2 = OB2
⇒ (4 – 2)² + (3 – 3)² = (x – 2)² + (5 – 3)²
⇒ 2² = (x – 2)² + 22
⇒ (x – 2)² = 0
⇒ x – 2 = 0
⇒ x = 2
Thus, the required value of x is 2.

Orienting Yourself The Use of Coordinates Class 9 Long Question Answer

Question 1.
Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the III quadrant.
Solution:
As the length and breadth of the rectangle are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the III quadrant, so the coordinates of the vertices of a rectangle OABC are 0(0,0), A (- 5, 0), B(- 5,- 3) and C(0, – 3).

Question 2.
From the given figure, answer the following: y-axis

(i) Write the points whose x-coordinate is 0.
(ii) Write the points whose y-coordinate is 0.
(iii) Write the points whose x-coordinate is -5.
Solution:
(i) Clearly, the distance of points A, L and O from y-axis is 0.
So, A(0, 3), L(0, -4) and 0(0, 0) are the points whose x-coordinate is 0.

(ii) Clearly, the distance of points G, I and O from y-axis is 0.
So, G(5, 0), I(—2, 0) and 0(0, 0) are the points whose y-coordinate is 0.

(iii) Clearly, the distance of points H and D from the y-axis is 5 units, and both lie in second and third quadrants respectively.
So, H(— 5, — 3) and D(-5, 1) are the points whose x-coordinate is -5.

Orienting Yourself The Use of Coordinates Class 9 Case Based Questions

Question 1.
KGB schools provide free education to girl students from weaker sections. The local body of a town wants to open a KGB school in the town for which a rectangular plot ABCD, as shown in the following figure, is very suitable. But this plot belongs to Rati Ram. Rati Ram agrees to exchange it with the triangular plot PQR as shown in the same figure. The coordinates of the vertices of both the plots are shown in the figure.

(i) Find the area of rectangular plot ABCD.
(ii) Find the area of triangular plot PQR.
(iii) Compare the areas of both the plots.
OR
If Coordinate A (2, 8) and B (8, 8) shift to A'(2, 7) and B'(8, 7), then find the area of rectangle.
Solution:
(i) For rectangle ABCD:
AB = 8 – 2 = 6 units ← Length
AD = 12 – 8 = 4 units ← Breadth
∴ Area of rectangle ABCD = AB × AD
= 6 units × 4 units
= 24 sq. units. ………(1)

(ii) For triangle PQR:
QR = 12-4 =8 units ← Base
SP = 6 – 0 =6 units ← Height
Area of APQR = \(\frac{1}{2}\) × base × height
= (\(\frac{1}{2}\) × 8 × 6) sq. units
= 24 sq. units. ………..(2)

(iii) From (1) and (2), we get
⇒ area of rect. ABCD = area of APQR
OR
For new rectangle A’B’CD:
A’B’ = 8 – 2 = 6 units
A’D = 12 – 7 = 5 units
∴ Area of rectangle A’B’CD = A’B’ × A’D
= 6 × 5
= 30 sq. units

Orienting Yourself The Use of Coordinates Class 9 Competency Based Questions

Question 1.
A(5, 1), B(l, 4) and C(8, 5) are the coordinates of the vertices of a triangle.
Which of the following types of triangle will AABC be?
(a) Equilateral triangle
(b) Scalene right-angled triangle
(c) Isosceles right-angled triangle
(d) Isosceles acute-angled triangle
Solution:
(c) Isosceles right-angled triangle
AB = 5 units
BC = 5√2 units
CA = 5 units
AB = CA
∆ABC is an isosceles triangle.
AB² + CA² = 5² + 5² = 50 units
BC2 = (5√2)² = 50 units
∴ AB² + CA² = BC²
So, ∆ABC is a right-angled triangle.

Question 2.
Shown below are 2 identical rectangles such that their breadth is half their length.

What arc the coordintes of point A7
(a) (4, -5)
(b) (-4, -6)
(c) (-2, -7)
(d) (-2, -9)
Solution:
(c) (-2, -7)

Question 3.
Graph the points P(2, 0), L(1, 1), A(- 1, 1), N(- 2, 0), E(- 1,-1) and T( 1, -1), and draw the hexagon PLANET.
Solution:
Plot the points P(2, 0) right 2, no vertical movement, L(1, 1) right 1, up 1; A(- 1, 1) left 1, up 1; N(- 2, 0) left 2, no vertical movement; E(- 1, -1) left 1, down 1 and T(1, – 1) right 1, down 1.
Join PL, LA, AN, NE, ET, TP. The graph shows the hexagon PLANET.

Question 4.
A square is inscribed in a circle of radius 2 cm with centre O at the origin. All 4 vertices of the square lie on the coordinate axes.
Use the distance formula to find the length of the side of the square. Show your work.
Solution:
The coordinates of the vertices of the square would be
(2, 0), (0, -2), (-2, 0), (0, 2).

Use the distance formula and any 2 adjacent coordinates of the vertices of the square to find the length of the side of the square
AB = \(\sqrt{(2-0)^2+(0-2)^2}\)
= \(\sqrt{4+4}\) cm.
So, the side of square is 2√2 cm.

Question 5.
The three vertices of a rhombus PQRS are P(2, -3), Q(6,5) and R(-2, 1).

(i) Find the coordinates of the point where both the diagonals PR and QS intersect.
(ii) Find the coordinates of the fourth vertex S.
Show your steps and give valid reasons.
Solution:
(i) The diagonals of a rhombus bisect each other.
The point of intersection of the diagonals by finding the midpoint of P(2,-3) and R(-2, 1) as (0, -1).

(ii) Mid-point of Q(6, 5) and S(x, y) = \(\left(\frac{6+x}{2}, \frac{5+y}{2}\right)\)
where x and y are the coordinates of the fourth vertex S.
Use the above steps and equate the respective coordinates of the mid-points to get the following relationships:
(a) 0 = \(\frac{6+x}{2}\)
(b) -1 = \(\frac{5+y}{2}\)
x = -6 and y = -7
∴ Coordinates of the fourth vertex S are (-6, -7).

Orienting Yourself The Use of Coordinates Extra Questions Class 9

Question 1.
In which quadrant does the point (5, -6) lie?
Solution:
In point (5, -6), x = 5 > 0 an y = -6 < 0
∴ Point (5, -6) lies in the IV quadrant.

Question 2.
If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then find the value of (Abscissa of P) – (Abscissa of Q).
Solution:
We have, points P(-2, 3) and Q(-3, 5).
Here, abscissa of P, i.e., x-coordinate of P, is -2
and abscissa of Q, i.e., x-coordinate of Q, is -3.
∴ (Abscissa of P) – (Abscissa of Q)
= -2 – (-3)
= -2 + 3
= 1

Question 3.
If the coordinates of two points are A(7, -3) and B(7, -5), then find (Ordinate of A) – (Ordinate of B).
Solution:
Given, point A(7, -3) i.e. ordinate of A is -3
and point B (7, -5), i.e., ordinate of B is -5.
∴ Ordinate of A – Ordinate of B = -3 – (-5)
= -3 + 5
= 2

Question 4.
Find the perpendicular distance of the point P(3, 4) from the Y-axis.
Solution:
We know that the abscissa, or the x-coordinate of a point, is the perpendicular distance from the Y-axis.
So, perpendicular distance of the point P(3, 4) from Y-axis = Abscissa = 3 units

Question 5.
In the following figure, O is the intersecting point of OA and OC, and OABC is a square of side 2 units. Find the position of points A, B, and C.

Solution:
Here, position means the coordinates of the points.
As point A lies on the positive X-axis, at a distance of 2 units from the intersection point O, its position will be (2, 0).
Point B lies in the I quadrant, and it is at a distance of 2 units from both the axes, so its position will be (2, 2).
Point C lies on the positive Y-axis, at a distance of 2 units from the intersection point O, so its position will be (0, 2).

Question 6.
The coordinates of points P and Q are (-5, 3) and (-5, m), respectively. If the sum of abscissae and ordinates of both points is equal, then find the possible value of m.
Solution:
Given, coordinates of P = (-5, 3) and Q = (-5, m).
According to the question,
m + 3 = -5 – 5
⇒ m + 3 = -10
⇒ m = -10 – 3
⇒ m = -13

Question 7.
In the given figure, LM is a line parallel to the Y-axis at a distance of 3 units.

(i) What are the coordinates of the points P, R, and Q?
(ii) What is the difference between the abscissae of the points L and M?
Solution:
Given that LM is a line parallel to the Y-axis, its perpendicular distance from the Y-axis is 3 units.
(i) Coordinates of point P = (3, 2)
[∵ Its perpendicular distance from the X-axis is 2 units]
Coordinates of point Q = (3, -1)
[∵ Its perpendicular distance from the X-axis is 1, in the negative direction of the Y-axis]
Coordinates of point R = (3, 0)
[∵ It lies on the X-axis, so its y-coordinate is zero]

(ii) Abscissa of point L = 3 and abscissa of point M = 3
∴ Difference between the abscissae of the points L and M = 3 – 3 = 0.

Question 8.
Find the distance of the point P(5, -12) from the origin.
Solution:
∴ \(\sqrt{x^2+y^2}=\sqrt{5^2+(-12)^2}\) = 13 units

Question 9.
If the distance between the points A(-1, 2) and B(x, 2) is 9 units, find the values of x.
Solution:
∴ \(\sqrt{(x-(-1))^2+(2-2)^2}\) = 9
⇒ (x + 1)² = 81
⇒ x = -10, 8

Question 10.
Find the points on the X-axis, which are at a distance of √73 units from the point (5, -8). How many such points are there?
Solution:
Let the point on the X-axis be (x, 0).
∴ \(\sqrt{(x-5)^2+(0-(-8))^2}=\sqrt{73}\)
⇒ (x – 5)² = 9
⇒ x – 5 = ±3
⇒ x = 8, 2
Therefore, the points on the X-axis are (8, 0) and (2, 0).
∴ Required number of points = 2

Question 11.
The radius of a circle with centre at the origin is \(\frac {1}{2}\) units. Find all the points on the circle that are of the form (-y, y). Show your steps.
Solution:
The distance between the origin and the point (-y, y) is \(\frac {1}{2}\) units.
∵ Using the distance formula, we have
\((-y)^2+y^2=\left(\frac{1}{2}\right)^2\)
⇒ 2y² = \(\frac {1}{4}\)
⇒ y² = \(\frac {1}{8}\)
⇒ y = \(\frac{1}{2 \sqrt{2}}\) and \(-\frac{1}{2 \sqrt{2}}\)
The points are \(\left(\frac{1}{2 \sqrt{2}},-\frac{1}{2 \sqrt{2}}\right)\) and \(\left(-\frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)\).

Question 12.
Find the point on the y-axis, equidistant from two points A(-3, 4) and B(3, 6) on the same plane.
Solution:
Let the point on the y-axis be P(0, y).
Given, P(0, y) is equidistant from the points A(-3, 4) and B(3, 6).
So, AP = BP
⇒ \(\sqrt{[0-(-3)]^2+(y-4)^2}\) = \(\sqrt{(0-3)^2+(y-6)^2}\) [by distance formula]
⇒ 9 + y² + 16 – 8y = 9 + y² + 36 – 12y [squaring both sides]
⇒ 12y – 8y = 20
⇒ 4y = 20
⇒ y = 5
Hence, the point on the y-axis is (0, 5).

Question 13.
Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y). Find the value of y and the radius of the circle.
Solution:
Since O is the centre of the circle and A, B are points on its circumference.
∴ OA = OB
⇒ y = -1, 7
When y = -1, radius = 5 units
and when y = 7, radius = √793 units

Orienting Yourself The Use of Coordinates Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
The point (6, -5) lies in which of the following quadrants
(a) I-quadrant
(b) II-quadrant
(c) III-quadrant
(d) IV-quadrant

Question 2.
The distance between the points P(6, 0) and Q(-2, 0) is
(a) 2 units
(b) 8 units
(c) 6 units
(d) 4 units

Question 3.
Which of the following is the x-coordinate of all points on the y-axis?
(a) 0
(b) 1
(c) 2
(d) any number

Assertion Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): Point A( -2, -6) lies in the III quadrant.
Reason (R): A point both of whose coordinates are negative lies in the III quadrant.

Question 2.
Assertion (A): The value of r is 6, for which the distance between the points P(2, -3) and Q(10, y) is 10.
Reason (R): Distance between two given points A(x₁, y₁) and H(x₂, y₂) is given by AB = \(=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\).

Short Answer Type Questions

Question 1.
Answer the following:
(i) Find the distance of (4, -3) from the origin.
(ii) Find the distance between the points (\(\frac{1}{2}\), 2) and (\(\frac{1}{2}\), 2).

Question 2.
Write whether the following statements are True or False. Justify your answer.
(i) Point (0, – 2) lies on y-axis.
(ii) The perpendicular distance of the point (4, 3) from the x-axis is 4.

Question 3.
Find the coordinates of the midpoint of the line segment joining (4, 3) and (-2, -1).

Question 4.
Plot the following points on a graph and join them in order. Name the figure so obtained and find the area of the figure.
A(0, 2), B(3, 0), C(-3, 0), D(0, -2)

Question 5.
If A(0, 2) is equidistant from B(3, a) and C(a, 5). Find the value of a.

Question 6.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

Long Answer Type Questions

Question 1.
Shown below is a coordinate grid with points A, B, and C plotted on it.

Find the length of all the sides of the triangle formed by A, B, and C. Show your work.

Question 2.
Plot the points (2, 1), (3, -2) and (-1, -2) on graph paper and check whether they are collinear or not. If not, find the area of the figure formed.

The post Orienting Yourself The Use of Coordinates Class 9 Extra Questions Maths Chapter 1 appeared first on Learn CBSE.

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