Download CBSE MCQs for Class 10 Mathematics: Chapter 07 Coordinate Geometry
Access targeted multiple-choice questions for Chapter 07 Coordinate Geometry designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
Chapter-wise Objective Questions: Chapter 07 Coordinate Geometry
View or download the dedicated Chapter 07 Coordinate Geometry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.
Multiple Choice Questions
(a) 38
(b) 16
(c) 42
(d) none of these
Show Answer & Explanation
Answer: (d) none of these
Explanation:
1. Side 1: distance between (-1, 4) and (-4, -2) = √(9 + 36) = √45 = 3√5.
2. Side 2: between (-4, -2) and (3, -4) = √(49 + 4) = √53.
3. Side 3: between (3, -4) and (-1, 4) = √(16 + 64) = √80 = 4√5.
4. Perimeter = 7√5 + √53 ≈ 15.65 + 7.28 = 22.9, which matches none of 38, 16 or 42.
Teacher's Note:
1. Use the distance formula for all three sides.
2. Estimate the sum of the surds.
(a) \( \left(1, \frac{\sqrt{3}}{2}\right) \)
(b) \( \left(\frac{2}{3}, \frac{1}{\sqrt{3}}\right) \)
(c) \( \left(\frac{2}{3}, \frac{\sqrt{3}}{2}\right) \)
(d) \( \left(1, \frac{1}{\sqrt{3}}\right) \)
Show Answer & Explanation
Answer: (d) \( \left(1, \frac{1}{\sqrt{3}}\right) \)
Explanation:
1. The sides are 2, 2 and 2 (distance between (0,0) and (1,√3) is 2, between (1,√3) and (2,0) is 2), so the triangle is equilateral.
2. In an equilateral triangle the incentre equals the centroid.
3. Centroid = ((1 + 0 + 2)/3, (√3 + 0 + 0)/3) = (1, 1/√3).
Teacher's Note:
1. Check whether the triangle is equilateral.
2. Then the incentre is the centroid.
(a) a = 2, b = 4
(b) a = 3, b = 4
(c) a = 2, b = 3
(d) a = 3, b = 5
Show Answer & Explanation
Answer: (c) a = 2, b = 3
Explanation:
1. The diagonals PR and QS bisect each other.
2. Midpoint of PR = (3, 4.5) and midpoint of QS = ((4 + a)/2, (6 + b)/2).
3. So 4 + a = 6 and 6 + b = 9, giving a = 2 and b = 3.
Teacher's Note:
1. Equate the midpoints of the two diagonals.
2. PQRS lists the vertices in order.
(a) \( \frac{1}{a^2} + \frac{1}{b^2} = 1 \)
(b) \( \frac{1}{a^2} - \frac{1}{b^2} = 1 \)
(c) \( \frac{1}{a} + \frac{1}{b} = 1 \)
(d) \( \frac{1}{a} - \frac{1}{b} = 1 \)
Show Answer & Explanation
Answer: (c) \( \frac{1}{a} + \frac{1}{b} = 1 \)
Explanation:
1. The line through (a, 0) and (0, b) is x/a + y/b = 1.
2. (1, 1) lies on it, so 1/a + 1/b = 1.
Teacher's Note:
1. Use the intercept form of the line.
2. Substitute the third point.
(a) (-1, 0)
(b) (0, -1)
(c) (-1, 1)
(d) none of these
Show Answer & Explanation
Answer: (a) (-1, 0)
Explanation:
1. The diagonals bisect each other. Midpoint of the given diagonal = (-1.5, 0.5).
2. The other diagonal joins (-2, 1) and (x, y): (x - 2)/2 = -1.5 and (y + 1)/2 = 0.5.
3. x = -1 and y = 0.
Teacher's Note:
1. The midpoint of one diagonal is the midpoint of the other.
2. Use the midpoint formula.
(a) 0
(b) a + b + c
(c) ab + bc + ca
(d) none of these
Show Answer & Explanation
Answer: (a) 0
Explanation:
1. Area = ½|a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)|.
2. = ½|a(c - b) + b(a - c) + c(b - a)| = ½|ac - ab + ab - bc + bc - ac| = 0.
Teacher's Note:
1. Each y-coordinate is (a + b + c) minus the x-coordinate.
2. So the three points lie on x + y = a + b + c.
(a) \( 2\sqrt{p^2 + q^2} \) units
(b) \( \sqrt{2(p^2 + q^2)} \) units
(c) \( 2\sqrt{p^2 - q^2} \) units
(d) none of these
Show Answer & Explanation
Answer: (b) \( \sqrt{2(p^2 + q^2)} \) units
Explanation:
1. A = (p + q, 0) and B = (0, p - q).
2. AB² = (p + q)² + (p - q)² = 2p² + 2q².
3. AB = √(2(p² + q²)).
Teacher's Note:
1. Write the coordinates of A and B first.
2. Apply the distance formula.
(a) always coincident
(b) always collinear
(c) always the vertices of a triangle
(d) coincident in an equilateral triangle and otherwise collinear
Show Answer & Explanation
Answer: (d) coincident in an equilateral triangle and otherwise collinear
Explanation:
1. The centroid, circumcentre and orthocentre always lie on the Euler line, so they are collinear.
2. In an equilateral triangle all three coincide.
3. Hence the full statement is the last option; 'always coincident' and 'always form a triangle' are false.
Teacher's Note:
1. Think of the Euler line.
2. The correct option must cover both cases.
(a) (-4, 2)
(b) (4, 2)
(c) (3, -3)
(d) (0, -4)
Show Answer & Explanation
Answer: (b) (4, 2)
Explanation:
1. AB² = 4 + 16 = 20, so AC² = BC² - AB² = 25 - 20 = 5.
2. AC must be perpendicular to AB (direction (-2, -4)), so C = A ± t(2, -1) with |t|√5 = √5.
3. C = (4, 2) or (0, 4).
Teacher's Note:
1. Use Pythagoras to find AC.
2. AC is perpendicular to AB.
(a) 4
(b) 2
(c) 8
(d) none of these
Show Answer & Explanation
Answer: (b) 2
Explanation:
1. The lines enclose a square with vertices (1, 0), (0, 1), (-1, 0) and (0, -1).
2. Its diagonals are both 2 units long, so area = ½ × 2 × 2 = 2.
Teacher's Note:
1. Find the four vertices.
2. Area of a square = ½ × (diagonal)².
(a) 10.5
(b) 24
(c) 32
(d) 40
Show Answer & Explanation
Answer: (a) 10.5
Explanation:
1. Area = ½|2(0 + 4) + (-1)(-4 - 3) + 2(3 - 0)|.
2. = ½|8 + 7 + 6| = ½ × 21 = 10.5.
Teacher's Note:
1. Use the determinant formula for the area.
2. Keep track of signs.
(a) 10.5
(b) 24
(c) 32
(d) 40
Show Answer & Explanation
Answer: (c) 32
Explanation:
1. Area = ½|(-5)(-5 - 2) + 3(2 + 1) + 5(-1 + 5)|.
2. = ½|35 + 9 + 20| = 32.
Teacher's Note:
1. Use the area formula with the vertices in order.
2. Take the absolute value.
(a) 10.5
(b) 24
(c) 32
(d) 40
Show Answer & Explanation
Answer: (b) 24
Explanation:
1. Area = ½|1(6 + 5) + (-4)(-5 + 1) + (-3)(-1 - 6)|.
2. = ½|11 + 16 + 21| = 24.
Teacher's Note:
1. Apply the area formula carefully.
2. Check the signs of each bracket.
(a) 10.5
(b) 24
(c) 32
(d) 40
Show Answer & Explanation
Answer: (d) 40
Explanation:
1. It is a right triangle with legs 8 and 10.
2. Area = ½ × 8 × 10 = 40.
Teacher's Note:
1. The legs lie on the axes.
2. Area = ½ × base × height.
(a) 12 square units
(b) 9 square units
(c) 24 square units
(d) 6 square units
Show Answer & Explanation
Answer: (d) 6 square units
Explanation:
1. Put y = 0: x = 4, so A = (4, 0). Put x = 0: y = 3, so B = (0, 3).
2. Area = ½ × 4 × 3 = 6.
Teacher's Note:
1. Find the intercepts.
2. The triangle is right-angled at the origin.
(a) 183 square units
(b) 91.5 square units
(c) 124 square units
(d) none of these
Show Answer & Explanation
Answer: (b) 91.5 square units
Explanation:
1. Third vertex = (3 × 1 - 4 + 9, 3 × 4 + 3 - 7) = (8, 8).
2. Area = ½|4(7 - 8) + (-9)(8 + 3) + 8(-3 - 7)| = ½|-4 - 99 - 80| = ½ × 183 = 91.5.
Teacher's Note:
1. Find the third vertex from the centroid formula.
2. Then use the area formula.
(a) \( (4\sqrt{2}, 30^\circ) \)
(b) \( (4\sqrt{2}, 45^\circ) \)
(c) \( (2\sqrt{2}, 45^\circ) \)
(d) \( (2\sqrt{2}, 30^\circ) \)
Show Answer & Explanation
Answer: (b) \( (4\sqrt{2}, 45^\circ) \)
Explanation:
1. r = √(16 + 16) = 4√2.
2. tan θ = 4/4 = 1 and the point is in the first quadrant, so θ = 45°.
Teacher's Note:
1. r = √(x² + y²).
2. θ = tan⁻¹(y/x).
(a) \( b = \pm\sqrt{2}a \)
(b) \( a = \pm\sqrt{2}b \)
(c) b = -2a
(d) a = -2b
Show Answer & Explanation
Answer: (b) \( a = \pm\sqrt{2}b \)
Explanation:
1. D = (a/2, 0) and E = (a/2, b/2).
2. Slope of AD = (0 - b)/(a/2 - 0) = -2b/a. Slope of BE = (b/2)/(a/2) = b/a.
3. Perpendicular: (-2b/a)(b/a) = -1, so a² = 2b² and a = ±√2 b.
Teacher's Note:
1. Find the midpoints D and E.
2. Product of slopes of perpendicular lines is -1.
(a) 5
(b) 3
(c) 5/2
(d) 1
Show Answer & Explanation
Answer: (a) 5
Explanation:
1. The triangle has its right angle at the origin and equal legs, so the altitude from the origin to the hypotenuse is half the hypotenuse.
2. Distance from the origin to 2x + y = 5 is 5/√5 = √5, so the hypotenuse is 2√5.
3. Area = ½ × 2√5 × √5 = 5.
Teacher's Note:
1. Use the perpendicular distance from the origin to the line.
2. In a right isosceles triangle, the altitude on the hypotenuse is half the hypotenuse.
(a) abc
(b) 0
(c) a + b + c
(d) 3abc
Show Answer & Explanation
Answer: (d) 3abc
Explanation:
1. The centroid at the origin gives a + b + c = 0.
2. Using a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) = 0, we get a³ + b³ + c³ = 3abc.
Teacher's Note:
1. Find a + b + c first.
2. Use the identity for a³ + b³ + c³.
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation
Answer: (c) 3
Explanation:
1. The fourth vertex is the sum of two vertices minus the third.
2. (-3, 4) + (0, -4) - (5, 2) = (-8, -2); (-3, 4) + (5, 2) - (0, -4) = (2, 10); (0, -4) + (5, 2) - (-3, 4) = (8, -6).
3. There are three different positions.
Teacher's Note:
1. Any of the three points can be opposite to the fourth vertex.
2. Compute the three possibilities.
(a) \( (4 - \sqrt{3}, 3 + \sqrt{3}) \)
(b) \( (4 + \sqrt{3}, 3 - \sqrt{3}) \)
(c) \( (3 - \sqrt{3}, 4 + \sqrt{3}) \)
(d) \( (3 + \sqrt{3}, 4 - \sqrt{3}) \)
Show Answer & Explanation
Answer: (a) \( (4 - \sqrt{3}, 3 + \sqrt{3}) \)
Explanation:
1. AB = √(4 + 4) = 2√2 and the midpoint of AB is M(4, 3).
2. The distance from M to P is (√3/2) × 2√2 = √6, along the perpendicular direction (-1, 1)/√2.
3. P = (4 - √6/√2, 3 + √6/√2) = (4 - √3, 3 + √3).
Teacher's Note:
1. P lies on the perpendicular bisector of AB.
2. The altitude of an equilateral triangle is (√3/2) times its side.
(a) \( (1, \sqrt{3}) \)
(b) \( (-1, \sqrt{3}) \)
(c) \( (-1, -\sqrt{3}) \)
(d) \( (1, -\sqrt{3}) \)
Show Answer & Explanation
Answer: (a) \( (1, \sqrt{3}) \)
Explanation:
1. x = r cos θ = 2 cos 60° = 1.
2. y = r sin θ = 2 sin 60° = √3.
Teacher's Note:
1. Use x = r cos θ and y = r sin θ.
2. The point is in the first quadrant.
(a) 3 : 4
(b) 3 : 2
(c) 4 : 3
(d) none of these
Show Answer & Explanation
Answer: (a) 3 : 4
Explanation:
1. Let the ratio be k : 1. The point (2k + 1)/(k + 1), (7k + 3)/(k + 1) lies on the line.
2. 3(2k + 1) + (7k + 3) - 9(k + 1) = 0 gives 6k + 3 + 7k + 3 - 9k - 9 = 0, so 4k = 3.
3. k = 3/4, so AP : PB = 3 : 4.
Teacher's Note:
1. Use the section formula with ratio k : 1.
2. Substitute the point in the line.
(a) \( \left(\frac{10}{7}, \frac{-33}{7}\right) \)
(b) \( \left(\frac{10}{7}, \frac{33}{7}\right) \)
(c) \( \left(\frac{33}{7}, \frac{10}{7}\right) \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \left(\frac{10}{7}, \frac{33}{7}\right) \)
Explanation:
1. With ratio 3 : 4: x = (3 × 2 + 4 × 1)/7 = 10/7.
2. y = (3 × 7 + 4 × 3)/7 = 33/7.
3. Check: 3(10/7) + 33/7 - 9 = 63/7 - 9 = 0.
Teacher's Note:
1. Use the section formula with ratio 3 : 4.
2. Verify in the equation of the line.
(a) 4
(b) 17
(c) \( \sqrt{17} \)
(d) \( \sqrt{109} \)
Show Answer & Explanation
Answer: (c) \( \sqrt{17} \)
Explanation:
1. AB = √((2 - 1)² + (7 - 3)²) = √(1 + 16) = √17.
Teacher's Note:
1. Use the distance formula.
2. Do not forget the square root.
Free study material for Mathematics
Chapter 07 Coordinate Geometry Objective Questions & Solutions for Class 10 Mathematics
Class 10 Mathematics Chapter 07 Coordinate Geometry Objective Test Questions
Review structured objective questions for Class 10 Mathematics Chapter 07 Coordinate Geometry. Built according to official CBSE guidelines, these MCQ sets support daily revision and core concept reinforcement.
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FAQs
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