Coordinate Geometry Class 9 MCQ Maths Chapter 4

Students often review Class 9 Advanced Maths Book Solutions and Class 9 Advanced Maths Chapter 4 Coordinate Geometry MCQ Questions for quick revisions before tests.

Class 9 Coordinate Geometry MCQ

Coordinate Geometry MCQ Class 9

Question 1.
A point is located at (-3, 4). In which quadrant does it lie?
A. Quadrant III
B. Quadrant I
C. Quadrant IV
D. Quadrant II
Answer:
(D) Since, the x-coordinate is negative and the y-coordinate is positive, the point lies in the II qudrant.

Question 2.
Which of the following points lies on the X-axis?
A. (5, 5)
B. (5, 0)
C. (-5, -5)
D. (0, 5)
Answer:
(B) A point lies on the X-axis if its ordinate (y-coordinate) is 0.
Therefore, the point (5, 0) lies on X-axis.

Question 3.
The reflection of the point A(-4, 3) in the Y-axis is
A. (4, 3)
B. (4, -3)
C. (-4, -3)
D. (0, 3)
Answer:
(A) The reflection of the point A(-4, 3) in the Y-axis is (4, 3).

Question 4.
The coordinates of the point P (-3, 5) on reflecting on the X-axis are
A. (3, 5)
B. (-3, -5)
C. (3, -5)
D. (-3, 5)
Answer:
(B) The reflection of the point (-3, 5) in the X-axis is the point (-3, -5).

Question 5.
If the image of the point P under reflection in the X-axis is (-6, 2) then the coordinates of the point P are
A. (6, 2)
B. (-6, -2)
C. (6, -2)
D. (-6, 0)
Answer:
(B) If the image of point P under reflection in the X-axis is (-6, 2) then the coordinate of the point P are (-6, -2).

Question 6.
The reflection of the point (-6, 0) in the origin is the point
A. (0, -6)
B. (0, 6)
C. (6, 0)
D. None of these
Answer:
(C) We know that the reflection of point P(x, y) in the origin is P'(-x, -y).
∴ The reflection of the point (-6, 0) in the origin is the point (6, 0).

Question 7.
When reflecting a point in the Y-axis, which rule is correct?
A. Swap the x and y values.
B. Keep the sign of the abscissa unchanged; change the sign of the ordinate.
C. Keep the sign of the ordinate unchanged; change the sign of the abscissa.
D. Change the signs of both coordinates.
Answer:
(C) The correct rule is to keep the sign of the ordinate unchanged and change the sign of the abscissa.

Question 8.
Find the slope of the line passing through the points A(2, 3) and B(4, 11).
A. 2
B. 3
C. 4
D. 8
Answer:
(C) The slope of a line passing through (x₁, y₁,) and
(x₂, y₂) is m = \(\frac{y_2-y_1}{x_2-x_1}\).
On substituting the values, we get
m = \(\frac{11 – 3}{4 – 2}\) = \(\frac{8}{2}\) = 4

Question 9.
The line through the point (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
A. 2
B. 4
C. 6
D. 8
Answer:
(B) Slope of the line through the points (-2, 6) and (4, 8) is
m₁ = \(\frac{8 – 6}{4 – (-2)}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Slope of the line through the points (8, 12) and (x, 24) is
m₂ = \(\frac{24 – 12}{x – 8}\) = \(\frac{12}{x – 8}\)
Since, two lines are perpendicular.
∴ m₁m₂ = -1
⇒ \(\frac{1}{3}\) x \(\frac{12}{x – 8}\) = -1 ⇒ 4 = – (x – 8) ⇒ x = 4

Question 10.
If two lines with slopes m₁ and m₂ are perpendicular to each other, which of the following relations is true?
A. m₁ = m₂
B. m₁ + m₂ = 0
C. m₁ . m₂ = 1
D. m₁ . m₂ = -1
Answer:
(D) For two lines to be perpendicular, the product of their slopes must be -1.
Therefore, the correct relation is m₁.m₂ = -1.

Question 11.
What is the slope of a line parallel to the Y-axis?
A. 0
B. 1
C. Undefined
D. -1
Answer:
(C) A line parallel to the Y-axis is vertical and makes an angle of 90° with the X-axis.
Since, the horizontal change (Δx) is 0, the slope is undefined.

Question 12.
A line has the equation y = 3x – 5. What are its slope and y-intercept?
A. Slope = 3, y-intercept = 5
B. Slope = -5, y-intercept = 3
C. Slope = 3, y-intercept = -5
D. Slope = -3, y-intercept = 5
Answer:
(C) Comparing the equation with the slope-intercept form y = mx + c, we get m= 3 and c = -5
Thus, the slope = 3 and y-intercept = -5.

Question 13.
For the general equation of a line Ax + By + C = 0, what is the formula for its slope (m)?
A. m = [ltaex]\frac{A}{B}[/latex]
B. m = -[ltaex]\frac{A}{B}[/latex]
C. m = [ltaex]\frac{B}{A}[/latex]
D. m = -[ltaex]\frac{C}{A}[/latex]
Answer:
(B) On rewriting Ax + By + C = 0 in the form y = mx + c, we get
By = – Ax – C
⇒ y = (-\(\frac{A}{B}\))x – \(\frac{C}{B}\)
The formula for the slope is m = –\(\frac{A}{B}\).

Question 14.
If the slope of a line is \(\frac{2}{3}\), what is the slope of a line perpendicular to it
A. \(\frac{2}{3}\)
B. –\(\frac{2}{3}\)
C. \(\frac{3}{2}\)
D. –\(\frac{3}{2}\)
Answer:
(D) The slope of a perpendicular line is the negative reciprocal of the original slope.
Perpendicular slope = –\(\) = –\(\frac{3}{2}\)

Question 15.
What is the equation of a line passing through the origin with a slope of-2?
A. y = -2x
B. y = 2x
C. x + 2y = 0
D. y = x – 2
Answer:
(A) A line passing through the origin has the form
y = mx ……. (i) [∵ c = 0]
On substituting m = -2 in the Eq. (i), we get
y = -2x

Question 16.
Which of the following represents a line in intercept form that cuts the X -axis at 4 and the Y-axis at – 3?
A. \(\frac{x}{4}\) + \(\frac{y}{3}\) = 1
B. \(\frac{x}{4}\) – \(\frac{y}{3}\) = 1
C. \(\frac{x}{-3}\) + \(\frac{y}{4}\)
D. 4x – 3y = 1
Answer:
(B) In the intercept form, a is the x-intercept and b is the y-intercept.
On substituting a = 4 and b = -3 in \(\frac{x}{a}\) + \(\frac{x}{b}\) = 1,
we get
\(\frac{x}{4}\) – \(\frac{y}{3}\) = 1,

Question 17.
In the equation \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, the term ‘intercept’ refers to
A. the slope of the line
B. the angle the line makes with the axis
C. the specific location, where the graph cuts through an axis
D. the distance between two points on the line
Answer:
(C) The term intercept refers to the specific location, where a graph makes contact with or cuts through an axis.

Coordinate Geometry Class 9 Assertion and Reason Questions

Direction (Q. Nos. 1-20) In the questions given below, there are two statements marked as Assertion (A) and Reason (R). Read the statements and choose the correct option.
Options
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.

Question 1.
Assertion (A) : If a point P(4, -7) is reflected in the X-axis, then its image is (4, 7).
Reason (R) : Reflection in the X-axis changes the sign of the y-coordinate while the x-coordinate remains unchanged.
Answer:
(A) Given point is P(4, -7).
We know that on reflection in the X-axis the x-coordinate remains the same and the sign of the y-coordinate changes.
Therefore, the reflection of (4, -7) in the X-axis is (4, 7).

Hence, Assertion is true.
Also, the Reason correctly states the rule of reflection in the X-axis.
Therefore, both Assertion and Reason are true and Reason correctly explains the Assertion.

Question 2.
Assertion (A) : The values of a for which the straight line 2x + 3y + 4 + a (6x – y + 12) = 0 is perpendicular to the line 7x + 5y – 4 = 0 is \(\frac{-29}{37}\)
Reason (R) : Two line having slopes m₁ and m₂ are perpendicular, if m₁ = m₂.
Answer:
(C) We have, 2x + 3y + 4 + a (6x – y + 12) = 0
⇒ (2 + 6a) x + (3 – a) y + (4 + 12a) = 0 …….. (i)
and 7x + 5y – 4 = 0 ……… (ii)
Slope of line (i), m₁ = –\(\frac{2+6 a}{3-a}\)
and slope of line (ii), m₂ = –\(\frac{7}{5}\)
Since, lines (i) and (ii) are perpendicular.
∴ m₁ . m₂ = -1
⇒ –\(\frac{2+6 a}{3-a}\) . (-\(\frac{7}{5}\)) = -1
⇒ \(\frac{14+42 a}{15-5 a}\) = -1 ⇒ 14 + 42a = – 15 + 5a
⇒ 37a = -29 ⇒ a = – \(\frac{29}{37}\)
Hence, Assertion is true but Reason is false.

Question 3.
Assertion (A) : Equation of line for which slope = \(\frac{1}{5}\) and x-intercept is –\(\frac{4}{5}\) is x – 5y = 4.
Reason (R) : Equation i fline in slope intercept form is y = mx + c, where c is the y-intercept
Answer:
(A) Hint ∴ Equation of required line,
y = \(\frac{1}{5}\) x – \(\frac{4}{5}\)
⇒ 5y = x – 4
⇒ x – 5y = 4
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.

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