Class 10 Maths Chapter 7 Coordinate Geometry MCQs Quiz

Are you worried about the formulas and concepts in Chapter 7, ‘Coordinate Geometry’? Are problems involving the distance between two points or dividing a line segment causing you trouble? Fear not! We have the perfect solution for you: the Class 10 Maths Chapter 7 Coordinate Geometry MCQs Quiz. This free online test will help you build a strong grasp of this chapter and understand the pattern of questions asked in the exam. Let’s elevate your board exam preparation with this comprehensive Coordinate Geometry quiz!

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Coordinate Geometry MCQs Quiz

Click the Start button to begin the quiz and attempt all questions within the given time. Read each MCQ carefully, select the correct option, and move forward to complete the quiz. Your score will appear at the end to help you check your preparation level..

Q1: What is the distance between points A(2, 3) and B(4, 1)?

A: \( \sqrt{2} \) units
B: \( 2 \) units
C: \( 2\sqrt{2} \) units
D: \( 2\sqrt{2} \) units

Explanation: Distance formula: \( AB = \sqrt{(4-2)^2 + (1-3)^2}
= \sqrt{(2)^2 + (-2)^2}
= \sqrt{4+4}
= \sqrt{8}
= 2\sqrt{2} \) units.

Q2: What are the coordinates of the point that divides the line segment joining the points (7, 1) and (3, 5) in the ratio 2:3 internally?

A: (5, 3)
B: \( \left( \frac{27}{5}, \frac{13}{5} \right) \)
C: (4, 2)
D: \( \left( \frac{17}{5}, \frac{11}{5} \right) \)

Explanation: Section formula: \( x = \frac{m x_2 + n x_1}{m+n} = \frac{2 \times 3 + 3 \times 7}{2+3} = \frac{6+21}{5} = \frac{27}{5} \), \( y = \frac{m y_2 + n y_1}{m+n} = \frac{2 \times 5 + 3 \times 1}{5} = \frac{10+3}{5} = \frac{13}{5} \).

Q3: In what ratio does the point (-1, 6) divide the line segment joining (-3, 10) and (6, -8)?

A: 2:7
B: 7:2
C: 2:5
D: 5:2

Explanation: Let the ratio be k:1. Using section formula for x-coordinate: \( -1 = \frac{k \times 6 + 1 \times (-3)}{k+1} \Rightarrow -1(k+1) = 6k – 3 \Rightarrow -k – 1 = 6k – 3 \Rightarrow -7k = -2 \Rightarrow k = \frac{2}{7} \). Thus the ratio is 2/7 : 1 = 2:7.

Q4: What is the distance of the point (-3, 4) from the origin?

A: 3 units
B: 5 units
C: 4 units
D: 7 units

Explanation: Distance of point (x,y) from origin (0,0) = \( \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \) units.

Q5: What are the coordinates of the midpoint of the line segment joining (6, 8) and (0, 0)?

A: (0, 8)
B: (6, 0)
C: (3, 4)
D: (4, 3)

Explanation: Midpoint formula: \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) = \left( \frac{6+0}{2}, \frac{8+0}{2} \right) = (3, 4) \).

Q6: What is the distance between the points (-4, 7) and (2, -5)?

A: \( 6\sqrt{5} \) units
B: \( 12\sqrt{2} \) units
C: 13 units
D: \( 6\sqrt{3} \) units

Explanation: Distance formula: \( \sqrt{(2 – (-4))^2 + (-5 – 7)^2} = \sqrt{(6)^2 + (-12)^2} = \sqrt{36 + 144} = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \) units.

Q7: What is the relation between x and y so that the point (x, y) is equidistant from the points (7, 1) and (3, 5)?

A: x – y = 2
B: x − y = 2
C: x + y = 4
D: x − y = 4

Explanation: Point (x, y) is equidistant from (7,1) and (3,5). Using distance formula:
\( (x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2 \).
Expanding both sides:
\( x^2 -14x+49 + y^2 -2y+1 = x^2 -6x+9 + y^2 -10y+25 \).
Cancelling common terms and simplifying:
\( -14x -2y +50 = -6x -10y +34 \)
\( \Rightarrow -8x +8y +16 = 0 \)
\( \Rightarrow -x + y + 2 = 0 \)
\( \Rightarrow x – y = 2 \).

Q8: What is the distance between the points (a, b) and (-a, -b)?

A: \( \sqrt{a^2 + b^2} \)
B: \( 2\sqrt{a^2 + b^2} \)
C: \( 2(a^2 + b^2) \)
D: \( a^2 + b^2 \)

Explanation: Distance = \( \sqrt{(-a – a)^2 + (-b – b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2+4b^2} = 2\sqrt{a^2+b^2} \).

Q9: The points (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order. What are the values of x and y?

A: x = 6, y = 3
B: x = 3, y = 6
C: x = 6, y = 3
D: x = 5, y = 4

Explanation: In a parallelogram, the diagonals bisect each other. Let the vertices be A(1,2), B(4,y), C(x,6), D(3,5). The midpoints of diagonals AC and BD will be the same. Midpoint of AC: \( \left( \frac{1+x}{2}, \frac{2+6}{2} \right) = \left( \frac{1+x}{2}, 4 \right) \). Midpoint of BD: \( \left( \frac{4+3}{2}, \frac{y+5}{2} \right) = \left( \frac{7}{2}, \frac{y+5}{2} \right) \). Equating: \( \frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x=7 \Rightarrow x=6 \). And \( 4 = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y=3 \).

Q10: In what ratio does the y-axis divide the line segment joining (2, 3) and (4, 7)?

A: 1:2
B: 1:1
C: 2:1
D: 3:1

Explanation: The point on the y-axis is (0, y). Let it divide the segment in the ratio k:1.
Using section formula for x-coordinate: \( 0 = \frac{k \cdot 4 + 1 \cdot 2}{k+1} \)
\( \Rightarrow 0 = 4k+2 \)
\( \Rightarrow 4k = -2 \)
\( \Rightarrow k = -\frac{1}{2} \).
The negative sign indicates external division, but the magnitude of the ratio is 1:2. The y-axis divides the segment internally in the ratio 1:2 (since the point lies between the two points, we take the absolute ratio).

Q11: What are the coordinates of the midpoint of the line segment joining (-2, 5) and (3, -5)?

A: (1, 0)
B: \( \left( \frac{1}{2}, 0 \right) \)
C: (0.5, 0)
D: (0, 0.5)

Explanation: Midpoint: \( \left( \frac{-2+3}{2}, \frac{5+(-5)}{2} \right) = \left( \frac{1}{2}, \frac{0}{2} \right) = \left( \frac{1}{2}, 0 \right) \).

Q12: What type of triangle is formed by the vertices (5, -2), (6, 4) and (7, -2)?

A: Isosceles triangle
B: Equilateral triangle
C: Right triangle
D: Scalene triangle

Explanation: Let the points be A(5,-2), B(6,4), C(7,-2). Length of AB = \( \sqrt{(6-5)^2 + (4+2)^2} = \sqrt{1 + 36} = \sqrt{37} \).
Length of BC = \( \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1 + 36} = \sqrt{37} \).
Length of AC = \( \sqrt{(7-5)^2 + (-2+2)^2} = \sqrt{4 + 0} = 2 \).
Since AB = BC, it is an isosceles triangle.

Q13: What are the coordinates of the point on the x-axis which is equidistant from the points (7, 6) and (3, 4)?

A: (0, 15/4)
B: (15/4, 0)
C: (15/2, 0)
D: (0, 15/2)

Explanation: The point on the x-axis is (x, 0). Its distances from (7,6) and (3,4) are equal.
\( \sqrt{(x-7)^2 + (0-6)^2} = \sqrt{(x-3)^2 + (0-4)^2} \).
Squaring both sides:
\( (x-7)^2 + 36 = (x-3)^2 + 16 \)
\( \Rightarrow x^2 -14x +49 +36 = x^2 -6x +9 +16 \)
\( \Rightarrow -14x +85 = -6x +25 \)
\( \Rightarrow -8x = -60 \)
\( \Rightarrow x = \frac{60}{8} = \frac{15}{2} \).
Thus the point is \( \left( \frac{15}{2}, 0 \right) \).

Q14: The points A(4, 5), B(7, 6), C(6, 3) and D(3, 2) are the vertices of a parallelogram. What is the midpoint of the line segment joining the midpoints of its diagonals?

A: \( \left( \frac{10}{2}, \frac{8}{2} \right) \)
B: (5, 4)
C: (4, 5)
D: (6, 3)

Explanation: In a parallelogram, the diagonals bisect each other. Hence, the midpoints of both diagonals are the same point, which is the center of the parallelogram. Therefore, the midpoint of the line segment joining the midpoints of the diagonals is the same point. Center = Midpoint of diagonal AC = \( \left( \frac{4+6}{2}, \frac{5+3}{2} \right) = (5, 4) \).

Q15: The distance between points P(5, -3) and Q(3, y) is \( 2\sqrt{5} \) units. What is the possible value of y?

A: -5
B: 1
C: -1 and -5
D: -1 and -7

Explanation: \( PQ = \sqrt{(3-5)^2 + (y+3)^2} = \sqrt{(-2)^2 + (y+3)^2} = \sqrt{4 + (y+3)^2} = 2\sqrt{5} \).
Squaring both sides: \( 4 + (y+3)^2 = 20 \)
\( \Rightarrow (y+3)^2 = 16 \)
\( \Rightarrow y+3 = \pm 4 \).
If y + 3 = 4, then y=1;
If y + 3 = -4, then y = -7. According to the given options, y = -1 and -7 is the correct choice.

Quiz Results

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Class 10 Maths Chapter 7 Coordinate Geometry MCQs Quiz

Chapter 7, ‘Coordinate Geometry,’ connects algebra and geometry. In this chapter, you will learn how to determine the position of points on a Cartesian plane and use this information to solve various geometric problems. You will focus on three main Coordinate Geometry Formulas: the Distance Formula to find the distance between two points, the Section Formula to find the coordinates of a point dividing a line segment in a given ratio, and the formula to find the area of a triangle formed by three points. These topics are very important for the CBSE Class 10 Maths Quiz and board exams.

Conclusion

‘Coordinate Geometry’ is a chapter where you can score full marks by applying the formulas correctly. Our Class 10 Maths Chapter 7 Coordinate Geometry MCQs Quiz is designed to help you master this skill. This quiz gives you repeated practice on important topics like the Distance Formula, the Section Formula, and the Area of a Triangle. By solving these objective questions, you can improve your speed and accuracy in calculations, which is crucial for your Board Exam Preparation. Try this Free Online Maths Quiz for Class 10 now and finalize your preparation!

FAQs on Coordinate Geometry MCQs

1. Question: How many marks from the chapter Coordinate Geometry are asked in the board exam?
Answer: This chapter typically carries 6-7 marks in the CBSE board exam, including MCQs, short answer, and long answer questions.

2. Question: What are the most important formulas in this chapter?
Answer: The most important formulas are the distance formula between two points, the section formula for a line segment, and the formula for the area of a triangle.

3. Question: Is this Coordinate Geometry quiz based on the NCERT syllabus?
Answer: Yes, our Coordinate Geometry Class 10 quiz is fully based on the NCERT curriculum and the CBSE syllabus.

4. Question: What is the distance formula between two points (x₁, y₁) and (x₂, y₂)?
Answer: The distance formula is: √[(x₂ – x₁)² + (y₂ – y₁)²].

5. Question: How do you find the coordinates of a point that divides a line segment in a ratio m:n?
Answer: If a point (x, y) divides the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n, then its coordinates are: x = (mx₂ + nx₁)/(m+n) and y = (my₂ + ny₁)/(m+n).

6. Question: Is this online quiz free?
Answer: Yes, this Free Online Maths Quiz for Class 10 is absolutely free. You can take it anytime and anywhere for practice.

7. Question: How should I prepare this chapter for the board exam?
Answer: First, memorize all the formulas well. Then, solve the examples and exercise questions from the NCERT book. Finally, test your skills with our Coordinate Geometry Online Test.

8. Question: How do you find the area of a triangle formed by three points?
Answer: For points (x₁, y₁), (x₂, y₂), and (x₃, y₃), the area is ½ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|.

9. Question: How can you check if three points are collinear?
Answer: Three points are collinear if the area of the triangle formed by them is zero.

10. Question: What is the benefit of taking this online quiz?
Answer: This quiz helps you improve time management, check the accuracy of your formulas, and understand the exam pattern, which boosts your confidence and leads to better performance.

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