Orienting Yourself The Use of Coordinates Class 9 MCQ Maths Chapter 1

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MCQ on Orienting Yourself The Use of Coordinates Class 9

Orienting Yourself The Use of Coordinates MCQ Class 9

Class 9 Maths Orienting Yourself The Use of Coordinates MCQ

Question 1.
If the point A(a, b) lies in II quadrant then which of the following is correct?
(a) a>b
(b) a <b
(c) a = b
(d) None of these
Answer:
(b) a <b

Explanation:
The point A (a, b) lies in II quadrant.
Thus, x-coordinate i.e. a is negative and y-coordinate i.e. b is positive.
a < b

Question 2.
The perpendicular distance of a point from 7-axis is known as
(a) intersection point
(b) origin
(c) abscissa
(d) ordinate
Answer:
(c) abscissa

Explanation:
Perpendicular distance of a point from Y-axis is called abscissa or x-coordinate.

Question 3.
Abscissa of all the points on the X-axis is
(a) 0
(b) 1
(c) 2
(d) any number
Answer:
(d) any number

Explanation:
Abscissa of all the points on the A-axis is any real number because X-axis is a number line, which contains infinite real numbers on it.

Question 4.
The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is
(a) (0, 5)
(b) (5, 0)
(c) (0, -5)
(d) (-5, 0)
Answer:
(c) (0, -5)

Explanation:
Given, the point lies on Y-axis this shows that its x-coordinate is zero.
Also, it is at a distance of 5 units in negative direction of y-axis, so its y-coordinate is negative. Hence, the required point is (0, -5).

Question 5.
If the coordinates of two points are A (4, 5) and B (- 3, 6) then the value of (Abscissa of A) – (Ordinate of B) is
(a) 3
(b) 5
(c) -2
(d) 9
Answer:
(c) -2

Explanation:
Given, point A(4, 5) i.e. abscissa of A is 4 and point B (- 3, 6) i.e. ordinate of B is 6.
∴ Abscissa of A – Ordinate of B
= 4 – 6
= -2

Question 6.
The distance of point C (8,9) from Y-axis is
(a) 7 units
(b) 8 units
(c) 9 units
(d) 1 unit
Answer:
(b) 8 units

Explanation:
Given, point is C (8,9).
Thus, distance of C from Y-axis = x – corrdinate of the point = 8 units

Question 7.
From the figure given below, the coordinates of point M are

(a) (-4, -3)
(b) (4, -3)
(c) (-4, 3)
(d) (4, 3)
Answer:
(b) (4, -3)

Explanation:
In the given figure, point M lies at a distance of 4 units from Y-axis and at a distance of 3 units from X-axis along the negative direction of Y-axis, x-coordinate is 4 and y-coordinate is -3. Thus, the coordinates of point M are (4, -3).

Question 8.
If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis then the point P has
(a) x-coordinate = – 5
(b) y-coordinate = 5 only
(c) y-coordinate = – 5 only
(d) y-coordinate = 5 or – 5
Answer:
(d) y-coordinate = 5 or – 5

Explanation:
We know that the perpendicular distance of a point from the X-axis gives y-coordinate at that point. Here, the foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure from II quadrant or III quadrant.
Hence, the point P has y-coordinate = 5 or – 5.

Question 9.
The point D lies at a distance of 3 units from X-axis along the positive direction of7-axis and 6 units from 7-axis along the negative direction of X-axis. The coordinates of the point D are
(a) (-6, 3)
(b) (6, 3)
(c) (3, -6)
(d) (-3, 6)
Answer:
(a) (-6, 3)

Explanation:
The point D lies at a distance of 3 units from X-axis i.e. y-coordinate is 3 and at a distance of 6 units from 7-axis along the negative direction of X-axis i.e. x coordinate is -6.
Thus, the coordinates of point D are (-6, 3).

Question 10.
The distance between the points (-7,8) and (1,6) is
(a) 2\(\sqrt{17}\) units
(b) 4\(\sqrt{17}\) units
(c) 2√5 units
(d) 2\(\sqrt{15}\) units
Answer:
(a) 2\(\sqrt{17}\) units

Explanation:
Let x₁ = -7, y₁ = 8, x₂ = 1 and y₂ = 6
Using distance formula,
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(1-(-7))^2+(6-8)^2}\)
= \(\sqrt{8^2+(-2)^2}=\sqrt{64+4}\)
= \(\sqrt{68}\)
= 2\(\sqrt{17}\) units

Question 11.
The distance of point (2a, – b) form X-axis is
(a) a
(b) -a
(c) b
(d) -b
Answer:
(c) b

Explanation:
Let the given point be A(2a, -b) and the perpendicular from A on the X-axis intersects at B.
Points lies on X-axis, so the coordinate of points is (2a, 0).
Distance of given point from X-axis = AB
Using distance formula, we have
AB = \(\sqrt{(2 a-2 a)^2+(0-(-b))^2}\)
= \(\sqrt{0+b^2}\)
= b units

Question 12.
ABCD is a rectangle with its vertices at (2, -2), (8, 4), (4, 8) and (-2,2) taken in order. Length of its diagonal is
(a) 4√2 units
(b) 6√2 units
(c) 4√26 units
(d) 2√26 units
Answer:
(d) 2√26 units

Explanation:
ABCD is a rectangle with its vertices at (2, – 2), (8, 4), (4, 8) and (-2, 2).

Now, length of diagonal AC by using distance formula,
\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(4-2)^2+(8+2)^2}\)
[∵ x₁ = 2, x₂ = 4, y₁ = -2 and y₂ = 8]
= \(\sqrt{100}\)
= \(\sqrt{104}\)
= 2\(\sqrt{26}\) units

Question 13.
If the distances of the point P(x, y) from (2, 0) and (0, 2) are equal then which of the following is true?
(a) x + y = 0
(b) x = y + 2
(c) y = x + 2
(d) x = y
Answer:
(d) x = y

Explanation:
Let the points be A(2, 0) andS (0, 2).
Given that the distance of the point P(x, y) from (2, 0) and (0, 2) are equal.
∴ PA = PB
⇒ PA² = PB²
⇒ (x – 2)² + (y – 0)² = (x – 0)² + (y – 2)²
⇒ x² + 4 – 4x + y² = x² + y² + 4 – 4y
⇒ -4x = -4y
⇒ x = y

Question 14.
The distance of which of the following points from origin is less than 5 units?
(a) (3, 4)
(b) (2, 6)
(c) (-3, -4)
(d) (1, 4)
Answer:
(d) (1, 4)

Explanation:
We know that distance from origin,
d = \(\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}\)
Now, we calculate the distance from each point.
(a) (3,4)
∴ d = \(\sqrt{3^2+4^2}=\sqrt{25}\) = 5 units

(b) (2, 6)
∴ d = \(\sqrt{2^2+6^2}=\sqrt{40}\) >5 units

(c) (-3,-4)
∴ d = \(\sqrt{(-3)^2+(-4)^2}=\sqrt{25}\) = 5 units

(d) (1, 4)
∴ d = \(\sqrt{1^2+4^2}=\sqrt{17}\) <5 units
Hence, the only point (1, 4), whose distance from the origin is less than 5 units.

Question 15.
M is a point on 7-axis at a distance of 4 units from X-axis and it lies below the X-axis. The distance of point M from point Q(5, 1) is
(a) √2 units
(b) \(\sqrt{34}\) units
(c) \(\sqrt{50}\) units
(d) \(\sqrt{90}\) units
Answer:
(c) \(\sqrt{50}\) units

Explanation:
Coordinates of point M are (0, – 4).
MQ = \(\sqrt{(5-0)^2+(1+(-4))^2}\)
[by distance formula]
= \(\sqrt{50}\) units

Question 16.
If (3, 5) is the mid-point of the line segment joining the points (2,1) and (a, 9) then the value of a is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(b) 4

Explanation:
Given, mid-point is (3, 5) and the points are (2, 1) and (a, 9).
Using the mid-point formula,
\(\frac{2+a}{2}\) = 3
⇒ 2 + a = 6
⇒ a = 4

Question 17.
The vertices of a AOAB are O (0, 0), A (4, 0) and B(0,6). The median AD is drawn on OB. The length of AD is

(a) \(\sqrt{52}\) units
(b) 5 units
(c) 25 units
(d) 10 units
Answer:
(b) 5 units

Explanation:
AD is median on OB in AAOB.
So, D is the mid-point of OB.
Coordinates of D = \(\left(\frac{0+0}{2}, \frac{0+6}{2}\right)\) = (0, 3)
Length of AD = \(\sqrt{(4-0)^2+(0-3)^2}\)
= \(\sqrt{16+9}\)
= 5 units

Question 18.
A(-5, 6) and C(9, 1) are the two opposite vertices of a parallelogram ABCD. Its diagonals intersect each other at P(a, b). The relation between a and
(a) b = a – 1.5
(b) b = a + 1.5
(c) b = a – 4.5
(d) b = a + 4.5
Answer:
(b) b = a + 1.5

Explanation:
Hint Mid-point of AC is P(a, b).
Then, a = 2 and b = 3.5
(a) b = a – 1.5, not satisfied
(b) b = a + 1.5, satisfied
(c) b = a – 4.5, not satisfied
(d) b = a + 4.5, not satisfied

Orienting Yourself The Use of Coordinates Class 9 Assertion and Reason Questions

Direction (Q. Nos. 1-5) In the questions given below, there are two statements marked as Assertion (A) and Reason (R). Read the statements and choose the correct option.

Question 1.
Assertion (A): The point (-6, 0) lies on T-axis and the point (0, -7) lies on X-axis.
Reason (R): Every point on the X-axis has zero distance from X-axis and every point on the Y-axis has zero distance from Y-axis.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
Assertion The point (-6, 0) lies on X-axis and point (0, -7) lies on Y-axis. So, Assertion is false.
Reason It is true that every point on the X-axis and Y-axis has zero distance from X-axis and Y-axis, respectively.

Question 2.
Assertion (A): In the following graph, the coordinates of points P and Q are (3,2) and (-1,-2), respectively.

Reason (R): Coordinates of any point can be determined by drawing a perpendicular line to the coordinate axes.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
Assertion The coordinates of points P and Q are (3, 2) and (-1, -2), respectively, which is true.
Reason It is true that the coordinates of any point can be determined by drawing a perpendicular to the coordinate axes. It is also the correct explanation for Assertion.

Question 3.
Assertion (A) The distance of the point (- 4, 5) from the X-axis is 4 units.
Reason (R) Abscissa of a point gives the distance from the Y -axis.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:

Clearly, we see that the distance of the given point (- 4, 5) is 5 units from the X-axis.
Hence, Assertion is false but Reason is true.

Question 4.
Assertion (A) The points (2, 4), (4, 8) and (6,12) are collinear.
Reasoning (R) Three points are said to be collinear, when they all points lie on the same straight line.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
Let A = (2, 4), B = (4, 8) and C = (6, 12).
Then, AB = \(\sqrt{(4-2)^2+(8-4)^2}\)
[by distance formula]
= \(\sqrt{(2)^2+(4)^2}\)
= \(\sqrt{4+16}=\sqrt{20}\)
= 2√5 units

BC = \(\sqrt{(6-4)^2+(12-8)^2}\)
= \(\sqrt{(2)^2+(4)^2}=\sqrt{4+16}\)
= √20
= 2√5 units

and AC = \(\sqrt{(6-2)^2+(12-4)^2}\)
= \(\sqrt{(4)^2+(8)^2}=\sqrt{16+64}=\sqrt{80}\)
= 4√5 units,
∵ AC = AB + BC
Therefore, A, B and C are collinear points.
So, Assertion is true.
Clearly, Reason is true.
Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 5.
Assertion (A) If the points A (4, 5) and B(x, y) lie on a circle with centre O (2, 3) then the values of x and y are 0 and 1, respectively.
Reason (R) Centre of a circle is the mid-point of each chord of the circle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(c) A is true but R is false.

Explanation:
Assertion Given, AB is diameter and O is the centre of the circle.
Since, O is the mid -point of AB.

Coordinate of O = \(\left(\frac{x+4}{2}, \frac{y+5}{2}\right)\)
[by mid point formula]
\(\left(\frac{x+4}{2}, \frac{y+5}{2}\right)=\) = (2, 3)
⇒ \(\frac{x+}{2}\) = 2 and \(\frac{y+5}{2}\) = 3
⇒ x + 4 = 4 and y + 5 = 6
⇒ x = 0 and y = 1
So, Assertion is true.
Clearly, Reason is false.