CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 13

Practice MCQs for Class 10 Mathematics Chapter 07 Coordinate Geometry

Explore reliable objective questions for Chapter 07 Coordinate Geometry tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Navigate directly to the 50 objective questions for Chapter 07 Coordinate Geometry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question: If the distance between the points \((3, -5)\) and \((x, -5)\) is 15 units, then the values of \(x\) are
(a) \(12, -18\)
(b) \(-12, 18\)
(c) \(18, 5\)
(d) \(-9, -12\)
Show Answer & Explanation

Answer: (b) \(-12, 18\)

Explanation:
1. \(15^2 = (x - 3)^2\), so \(x - 3 = \pm 15\).
2. So \(x = 18\) or \(x = -12\).

Question: \(AD\) is a median of \(\triangle ABC\) with vertices \(A(5, -6)\), \(B(6, 4)\) and \(C(0, 0)\). The length of \(AD\) is
(a) \(\sqrt{68}\) units
(b) \(2\sqrt{15}\) units
(c) \(\sqrt{101}\) units
(d) 10 units
Show Answer & Explanation

Answer: (a) \(\sqrt{68}\) units

Explanation:
1. \(D\) is the midpoint of \(BC\), so \(D = (3, 2)\).
2. \(AD = \sqrt{(3 - 5)^2 + (2 + 6)^2} = \sqrt{4 + 64} = \sqrt{68}\).

Question: The centre of a circle is at \((2, 0)\). If one end of a diameter is at \((6, 0)\), then the other end is at
(a) \((0, 0)\)
(b) \((4, 0)\)
(c) \((-2, 0)\)
(d) \((-6, 0)\)
Show Answer & Explanation

Answer: (c) \((-2, 0)\)

Explanation:
1. The centre is the midpoint: \(\frac{6 + x}{2} = 2\), so \(x = -2\), and \(y = 0\).

Question: The distance of the point \((-6, 8)\) from the \(x\)-axis is
(a) 6 units
(b) -6 units
(c) 8 units
(d) 10 units
Show Answer & Explanation

Answer: (c) 8 units

Explanation:
1. Distance from the \(x\)-axis is \(|y| = 8\).

Teacher's Note:
1. The source's working says 'from the y-axis', but the question asks about the x-axis. The answer 8 is correct for the x-axis.

Question: The point of intersection of the line \(3x - y = 3\) and the \(y\)-axis is
(a) \((0, -3)\)
(b) \((0, 3)\)
(c) \((2, 0)\)
(d) \((-2, 0)\)
Show Answer & Explanation

Answer: (a) \((0, -3)\)

Explanation:
1. On the \(y\)-axis, \(x = 0\), so \(-y = 3\) and \(y = -3\).

Question: If \(P(k, 0)\) divides the line segment joining \(A(2, -2)\) and \(B(-7, 4)\) in the ratio \(1 : 2\), then the value of \(k\) is
(a) 1
(b) 2
(c) -2
(d) -1
Show Answer & Explanation

Answer: (d) -1

Explanation:
1. \(k = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-3}{3} = -1\).

Question: The coordinates of a point \(A\) on the \(y\)-axis, at a distance of 4 units from the \(x\)-axis and below it, are
(a) \((4, 0)\)
(b) \((0, 4)\)
(c) \((-4, 0)\)
(d) \((0, -4)\)
Show Answer & Explanation

Answer: (d) \((0, -4)\)

Explanation:
1. The point is on the \(y\)-axis, so \(x = 0\), and it is 4 units below the \(x\)-axis, so \(y = -4\).

Question: The distance of the point \((-12, 5)\) from the origin is
(a) 12
(b) 5
(c) 13
(d) 169
Show Answer & Explanation

Answer: (c) 13

Explanation:
1. \(d = \sqrt{(-12)^2 + 5^2} = \sqrt{169} = 13\) units.

Question: The distance of point \(P(3, 4)\) from the \(x\)-axis is
(a) 3 units
(b) 4 units
(c) 5 units
(d) 1 unit
Show Answer & Explanation

Answer: (b) 4 units

Explanation:
1. The distance from the \(x\)-axis is \(|y| = 4\).

Question: The distance of the point \(P(-3, -4)\) from the \(x\)-axis (in units) is
(a) 3
(b) -3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. The distance from the \(x\)-axis is \(|y| = 4\).

Question: If \(A\left(\frac{m}{3}, 5\right)\) is the midpoint of the line segment joining \(Q(-6, 7)\) and \(R(-2, 3)\), then the value of \(m\) is
(a) -12
(b) -4
(c) 12
(d) -6
Show Answer & Explanation

Answer: (a) -12

Explanation:
1. Midpoint of \(QR\) is \((-4, 5)\). So \(\frac{m}{3} = -4\), giving \(m = -12\).

Question: The point on the \(x\)-axis which is equidistant from the points \(A(-2, 3)\) and \(B(5, 4)\) is
(a) \((0, 2)\)
(b) \((2, 0)\)
(c) \((3, 0)\)
(d) \((-2, 0)\)
Show Answer & Explanation

Answer: (b) \((2, 0)\)

Explanation:
1. Let \(P(x, 0)\). \(AP^2 = BP^2\): \((x + 2)^2 + 9 = (x - 5)^2 + 16\).
2. This gives \(14x = 28\), so \(x = 2\).

Question: \(C\) is the midpoint of \(PQ\). If \(P\) is \((4, x)\), \(C\) is \((y, -1)\) and \(Q\) is \((-2, 4)\), then \(x\) and \(y\) respectively are
(a) -6 and 1
(b) -6 and 2
(c) 6 and -1
(d) 6 and -2
Show Answer & Explanation

Answer: (a) -6 and 1

Explanation:
1. \(\frac{4 - 2}{2} = y\), so \(y = 1\).
2. \(\frac{x + 4}{2} = -1\), so \(x = -6\).

Question: If \(x - 2y + k = 0\) is a median of the triangle with vertices \(A(-1, 3)\), \(B(0, 4)\) and \(C(-5, 2)\), then the value of \(k\) is
(a) 2
(b) 4
(c) 6
(d) 8
Show Answer & Explanation

Answer: (d) 8

Explanation:
1. Centroid \(G = \left(\frac{-1 + 0 - 5}{3}, \frac{3 + 4 + 2}{3}\right) = (-2, 3)\).
2. Since \(G\) lies on the median, \(-2 - 6 + k = 0\), so \(k = 8\).

Question: In the given figure, the area of \(\triangle ABC\) (in square units) is, where \(A(1, 3)\), \(B(-1, 0)\) and \(C(4, 0)\)
(a) 15
(b) 10
(c) 7.5
(d) 2.5
Show Answer & Explanation

Answer: (c) 7.5

Explanation:
1. Base \(BC = 5\) and height \(= 3\). Area \(= \frac{1}{2} \times 5 \times 3 = 7.5\) square units.

Question: The distance of the point \(P(2, 3)\) from the \(x\)-axis is
(a) 2
(b) 3
(c) 1
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. The distance from the \(x\)-axis is \(y = 3\).

Question: The point which lies on the perpendicular bisector of the line segment joining \(A(-2, -5)\) and \(B(2, 5)\) is
(a) \((0, 0)\)
(b) \((0, 2)\)
(c) \((2, 0)\)
(d) \((-2, 0)\)
Show Answer & Explanation

Answer: (a) \((0, 0)\)

Explanation:
1. The perpendicular bisector passes through the midpoint of \(AB\), which is \((0, 0)\).

Question: If \(P\left(\frac{a}{3}, 4\right)\) is the midpoint of the line segment joining \(Q(-6, 5)\) and \(R(-2, 3)\), then the value of \(a\) is
(a) -4
(b) -12
(c) 12
(d) -6
Show Answer & Explanation

Answer: (b) -12

Explanation:
1. Midpoint of \(QR\) is \((-4, 4)\). So \(\frac{a}{3} = -4\), giving \(a = -12\).

Question: The perpendicular bisector of the line segment joining \(A(1, 5)\) and \(B(4, 6)\) cuts the \(y\)-axis at
(a) \((0, 13)\)
(b) \((0, -13)\)
(c) \((0, 12)\)
(d) \((13, 0)\)
Show Answer & Explanation

Answer: (a) \((0, 13)\)

Explanation:
1. Let \(P(0, b)\). \(PA = PB\): \(1 + (b - 5)^2 = 16 + (b - 6)^2\).
2. This gives \(2b = 26\), so \(b = 13\).

Question:
Assertion (A): The value of \(y\) is 6, for which the distance between the points \(P(2, -3)\) and \(Q(10, y)\) is 10.
Reason (R): The distance between \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
(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.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. \(8^2 + (y + 3)^2 = 100\), so \((y + 3)^2 = 36\), giving \(y = 3\) or \(y = -9\).
2. So the assertion (\(y = 6\)) is false, and the distance formula is true.

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Download Chapter MCQs: Class 10 Mathematics Chapter 07 Coordinate Geometry

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Concept Clarification for Chapter 07 Coordinate Geometry

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