CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 14

Answer: (d) \( \left(-6, \frac{5}{2}\right) \)

Explanation:
1. Radius = distance of \( \left(\frac{13}{2}, 0\right) \) from the origin = \( \frac{13}{2} \), so \( r^2 = \frac{169}{4} = 42.25 \).
2. A point is in the interior when its (distance from O)² is less than 42.25.
3. (a) 0.5625 + 1 = 1.5625, (b) 4 + 5.44 = 9.44, (c) 25 + 0.25 = 25.25. All are less than 42.25, so they lie inside.
4. (d) 36 + 6.25 = 42.25 = r², so \( \left(-6, \frac{5}{2}\right) \) lies on the circle and not in the interior.

Teacher's Note:
1. Compare (distance)² with r² instead of taking square roots.
2. A point on the circle is not in the interior.

Answer: (b) \( \left(-\frac{5}{3}, \frac{11}{3}\right) \)

Explanation:
1. Area ratio 1 : 9 gives radius ratio 1 : 3, so OA : OB = 1 : 3 and A divides OB in the ratio 1 : 2.
2. A = \( \left(\frac{1 \times 3 + 2 \times (-4)}{3}, \frac{1 \times 5 + 2 \times 3}{3}\right) = \left(-\frac{5}{3}, \frac{11}{3}\right) \).

Teacher's Note:
1. Areas are in the ratio of the squares of radii, so take the square root first.
2. A lies on segment OB, so use the section formula with ratio 1 : 2.

Answer: (d) a = 20, b = 2

Explanation:
1. x-coordinate: \( \frac{9 + b + 7}{3} = 6 \Rightarrow b = 2 \).
2. y-coordinate: \( \frac{a - 4 + 8}{3} = 8 \Rightarrow a = 20 \).

Teacher's Note:
1. Centroid = average of the three x-coordinates and of the three y-coordinates.
2. Solve the two equations separately.

Answer: (b) (1, −6)

Explanation:
1. Let the third vertex be C(x, y). \( \frac{-5 - 2 + x}{3} = -2 \Rightarrow x = 1 \).
2. \( \frac{2 + 7 + y}{3} = 1 \Rightarrow y = -6 \). So C = (1, −6).

Teacher's Note:
1. Third vertex = 3 × centroid − sum of the other two vertices.
2. Do x and y separately to avoid sign slips.

Answer: (a) (0, 0)

Explanation:
1. Midpoint of AB = \( \left(\frac{-2 + 2}{2}, \frac{-5 + 5}{2}\right) = (0, 0) \).
2. Slope of AB = 5/2, so the perpendicular bisector is 2x + 5y = 0.
3. Only (0, 0) satisfies 2x + 5y = 0.

Teacher's Note:
1. The perpendicular bisector always passes through the midpoint.
2. Check each option in the equation of the bisector if needed.

Answer: (c) 12 units

Explanation:
1. AB = 4, BC = 3, and \( AC = \sqrt{3^2 + 4^2} = 5 \).
2. Perimeter = 4 + 3 + 5 = 12 units.

Teacher's Note:
1. B is the origin, so AB and BC lie along the axes.
2. The triangle is a 3-4-5 right triangle.

Answer: (d) AP = \( \frac{1}{2} \)AB

Explanation:
1. \( AP = \sqrt{(4-2)^2 + (2-1)^2} = \sqrt{5} \).
2. \( AB = \sqrt{(8-4)^2 + (4-2)^2} = \sqrt{20} = 2\sqrt{5} \).
3. So AP = \( \frac{1}{2} \)AB. Also PB = \( \sqrt{36 + 9} = 3\sqrt{5} \), which is neither equal to AP nor one third of AB.

Teacher's Note:
1. Find the three lengths with the distance formula and compare.
2. P lies on the line AB, on the side of A away from B.

Answer: (a) (x, y)

Explanation:
1. The angle at O is 90°, so the point equidistant from O, A and B is the circumcentre, the midpoint of the hypotenuse AB.
2. Midpoint of AB = \( \left(\frac{0 + 2x}{2}, \frac{2y + 0}{2}\right) = (x, y) \).

Teacher's Note:
1. In a right triangle the circumcentre is the midpoint of the hypotenuse.
2. Here the hypotenuse is AB.

Answer: (c) (2, −1)

Explanation:
1. Diagonals PR and QS bisect each other, so \( \frac{3 - 3}{2} = \frac{-2 + x}{2} \Rightarrow x = 2 \).
2. \( \frac{4 - 2}{2} = \frac{3 + y}{2} \Rightarrow y = -1 \). So S = (2, −1).

Teacher's Note:
1. In a parallelogram the diagonals have the same midpoint.
2. S = P + R − Q.

Answer: (a) 1 : 2

Explanation:
1. Let the ratio be k : 1. The y-coordinate of the dividing point is 0.
2. \( \frac{6k - 3}{k + 1} = 0 \Rightarrow k = \frac{1}{2} \), so the ratio is 1 : 2.

Teacher's Note:
1. A point on the x-axis has y = 0.
2. Use the y-coordinates in the section formula.

Answer: (d) (3, 5)

Explanation:
1. Diagonals AC and BD bisect each other.
2. \( \frac{1 + 6}{2} = \frac{4 + x}{2} \Rightarrow x = 3 \) and \( \frac{2 + 6}{2} = \frac{3 + y}{2} \Rightarrow y = 5 \). So D = (3, 5).

Teacher's Note:
1. D = A + C − B.
2. Check by finding the midpoint of AC and of BD.

Answer: (b) multiples of 3

Explanation:
1. \( AB^2 = 149 \), so \( (x + 4)^2 + (y - 3)^2 = 149 \).
2. With integer values, 149 = 100 + 49 is the only way to write 149 as a sum of two squares, so (x + 4) is 10 or 7.
3. x = 6 gives y = 3 + 7 = 10 (the other value y = −4 is outside the first quadrant); x = 3 gives y = 3 + 10 = 13.
4. So B is (6, 10) or (3, 13), and the possible x-coordinates 6 and 3 are multiples of 3.

Teacher's Note:
1. Write 149 as a sum of two perfect squares.
2. Reject points that fall outside the first quadrant.

Answer: (b) isosceles triangle

Explanation:
1. Distance between (−4, 0) and (4, 0) = 8.
2. Distance between (−4, 0) and (0, 3) = \( \sqrt{16 + 9} = 5 \); distance between (4, 0) and (0, 3) = 5.
3. Two sides are equal (5, 5) and 5² + 5² ≠ 8², so the triangle is isosceles but not right-angled.

Teacher's Note:
1. Find all three sides before deciding the type.
2. Check the Pythagoras relation before calling it a right triangle.

Answer: (d) infinitely many

Explanation:
1. The points at distance 5 from the origin lie on the circle x² + y² = 25.
2. A quarter of the circle lies in the third quadrant and it contains infinitely many points, for example (−3, −4), (−4, −3) and (−5/√2, −5/√2).

Teacher's Note:
1. Points at a fixed distance from a point form a circle.
2. Do not count only the integer points.

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. \( P = \left(\frac{9k - 1}{k + 1}, \frac{8k + 3}{k + 1}\right) \).
2. Substituting in x − y + 2 = 0: (9k − 1) − (8k + 3) + 2(k + 1) = 0 ⇒ 3k − 2 = 0, so \( k = \frac{2}{3} \).

Teacher's Note:
1. Write P in terms of k and substitute it in the equation of the line.
2. Multiply by (k + 1) to clear the denominator.

Answer: (b) 12

Explanation:
1. \( \sqrt{5^2 + k^2} = 13 \Rightarrow 25 + k^2 = 169 \Rightarrow k^2 = 144 \).
2. So k = 12 or k = −12. Among the options only 12 is possible.

Teacher's Note:
1. Square both sides to remove the root.
2. A square equation gives two values, so check which appears among the options.

Answer: (b) 1 : 7

Explanation:
1. Let the ratio be k : 1. For a point on the y-axis, x = 0: \( \frac{7k - 1}{k + 1} = 0 \Rightarrow k = \frac{1}{7} \).
2. So the ratio is 1 : 7.

Teacher's Note:
1. A point on the y-axis has x = 0.
2. Use the x-coordinates in the section formula.

Answer: (c) −2

Explanation:
1. \( \frac{3c - 4}{2} = -\frac{1}{2} \Rightarrow 3c = 3 \Rightarrow c = 1 \).
2. \( \frac{2 - d}{2} = 2 \Rightarrow d = -2 \).
3. So c × d = 1 × (−2) = −2.

Teacher's Note:
1. Equate the midpoint formula to the given midpoint, coordinate by coordinate.
2. Be careful with the sign of d.

Answer: (a) \( -\frac{1}{5} \)

Explanation:
1. \( x = \frac{3 \times 5 + 2 \times (-3)}{5} = \frac{9}{5} \) and \( y = \frac{3 \times (-6) + 2 \times 4}{5} = -2 \).
2. Sum = \( \frac{9}{5} - 2 = -\frac{1}{5} \).

Teacher's Note:
1. Use m : n = 3 : 2 with the first point (−3, 4) as (x₁, y₁).
2. Add the two coordinates only after finding both.

Answer: (d) 500 m

Explanation:
1. Distance in units = \( \sqrt{(2 + 4)^2 + (0 - 8)^2} = \sqrt{36 + 64} = 10 \) units.
2. Actual distance = 10 × 50 m = 500 m.

Teacher's Note:
1. The shortest distance is the straight line distance.
2. Convert units to metres using the scale at the end.

Answer: (a) ratio 2 : 5, k = 6

Explanation:
1. Let the ratio be m : n. From the y-coordinate: \( \frac{2m + 9n}{m + n} = 7 \Rightarrow 5m = 2n \), so m : n = 2 : 5.
2. Then \( k = \frac{1 \times 2 + 8 \times 5}{7} = 6 \).

Teacher's Note:
1. Use the known coordinate (y = 7) first to find the ratio.
2. Then use the ratio in the x-coordinate.

Answer: (a) −8

Explanation:
1. AP : PB = 1 : 2, so \( P = \left(\frac{1 \times 5 + 2 \times 2}{3}, \frac{1 \times (-8) + 2 \times 1}{3}\right) = (3, -2) \).
2. P lies on 2x − y + k = 0: 6 + 2 + k = 0, so k = −8.

Teacher's Note:
1. AP/AB = 1/3 means AP : PB = 1 : 2.
2. Substitute the point into the line equation.

Answer: (a) (3, −1)

Explanation:
1. Let the centre be (x, y). Equal distances from (3, 3) and (3, −5): \( (y - 3)^2 = (y + 5)^2 \Rightarrow y = -1 \).
2. Equal distances from (−1, −1) and (3, 3): \( (x + 1)^2 + 0 = (x - 3)^2 + 16 \Rightarrow 8x = 24 \Rightarrow x = 3 \).
3. Centre = (3, −1), and the radius is 4.

Teacher's Note:
1. The centre is equidistant from all three points on the circle.
2. Use the pair that gives the simplest equation first.

Answer: (a) (5, 0) and (9, 0)

Explanation:
1. Let the point be (x, 0). \( (x - 7)^2 + (0 + 4)^2 = 20 \).
2. \( (x - 7)^2 = 4 \Rightarrow x = 5 \) or x = 9. So there are two such points.

Teacher's Note:
1. A point on the x-axis has y = 0.
2. A squared equation gives two solutions.

Answer: (b) the vertices of a triangle of area 11 square units

Explanation:
1. Area = \( \frac{1}{2} |1(3 + 11) + 2(-11 - 5) + (-2)(5 - 3)| = \frac{1}{2} |14 - 32 - 4| = 11 \).
2. The area is not zero, so the points are not collinear and form a triangle of area 11 square units.

Teacher's Note:
1. Three points are collinear only if the area of the triangle is zero.
2. Compute the area with the determinant formula.

Answer: (a) 3x + y = 5

Explanation:
1. \( (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \).
2. Expanding and cancelling x² and y²: −6x − 12y + 45 = 6x − 8y + 25, i.e. 12x + 4y = 20.
3. So 3x + y = 5.

Teacher's Note:
1. Square both distances so no roots remain.
2. The x² and y² terms cancel.

Answer: (a) \( \left(2, -\frac{5}{3}\right) \) and \( \left(0, -\frac{7}{3}\right) \)

Explanation:
1. The first point divides the segment in the ratio 1 : 2: \( \left(\frac{-2 + 8}{3}, \frac{-3 - 2}{3}\right) = \left(2, -\frac{5}{3}\right) \).
2. The second point divides it in the ratio 2 : 1: \( \left(\frac{-4 + 4}{3}, \frac{-6 - 1}{3}\right) = \left(0, -\frac{7}{3}\right) \).

Teacher's Note:
1. Trisection means two points dividing the segment in the ratios 1 : 2 and 2 : 1.
2. Apply the section formula twice.

Answer: (a) (7, 3)

Explanation:
1. \( x = \frac{3 \times 8 + 1 \times 4}{4} = 7 \), \( y = \frac{3 \times 5 + 1 \times (-3)}{4} = 3 \).
2. So the point is (7, 3).

Teacher's Note:
1. m : n = 3 : 1, so the point is nearer to (8, 5).
2. Multiply each ratio part with the opposite end point.

Answer: (a) (1, −12) and (5, −10)

Explanation:
1. The diagonals bisect each other, so the intersection point (2, −5) is the midpoint of each diagonal.
2. Opposite of (3, 2): (2 × 2 − 3, 2 × (−5) − 2) = (1, −12). Opposite of (−1, 0): (2 × 2 + 1, 2 × (−5) − 0) = (5, −10).

Teacher's Note:
1. Vertex + opposite vertex = 2 × intersection point.
2. Do each vertex separately.

Answer: (a) (0, −2)

Explanation:
1. Let the point be (0, y). \( 25 + (y + 2)^2 = 9 + (y - 2)^2 \).
2. 25 + y² + 4y + 4 = 9 + y² − 4y + 4, so 8y = −16 and y = −2.

Teacher's Note:
1. A point on the y-axis has x = 0.
2. The y² terms cancel.

Answer: (a) \( \left(-1, \frac{9}{2}\right) \)

Explanation:
1. AR = 3/4 AB, so AR : RB = 3 : 1.
2. \( R = \left(\frac{3 \times 0 + 1 \times (-4)}{4}, \frac{3 \times 6 + 1 \times 0}{4}\right) = \left(-1, \frac{9}{2}\right) \).

Teacher's Note:
1. Convert AR = 3/4 AB into the ratio AR : RB.
2. R is closer to B.

Answer: (a) 5 : 1

Explanation:
1. Let the ratio be k : 1. \( x = \frac{-k + 5}{k + 1} = 0 \Rightarrow k = 5 \).
2. So the y-axis divides the segment in the ratio 5 : 1.

Teacher's Note:
1. On the y-axis x = 0.
2. k : 1 is the ratio in which the first point is joined to the dividing point.

Answer: (a) 29

Explanation:
1. \( -1 = \frac{3x + 4 \times 2}{7} \Rightarrow x = -5 \).
2. \( 2 = \frac{3y + 4 \times 5}{7} \Rightarrow y = -2 \).
3. x² + y² = 25 + 4 = 29.

Teacher's Note:
1. Here C is the dividing point and B is the unknown end.
2. Use the section formula for x and y separately.

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

Explanation:
1. Points equidistant from (2, 3) and (−2, −3) form the perpendicular bisector 2x + 3y = 0, which has infinitely many points, so A is false.
2. The midpoint of the two points is (0, 0), so R is true.

Teacher's Note:
1. The set of points equidistant from two points is a line, not a single point.
2. Midpoint alone does not give uniqueness.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Using R: \( x = \frac{7 \times 4 + 2 \times (-5)}{9} = 2 \) and \( y = \frac{7 \times (-7) + 2 \times 11}{9} = -3 \), so A is true.
2. R is the section formula, which is true, and it is exactly what gives A.

Teacher's Note:
1. Check A by actually applying the formula in R.
2. When the formula in R explains the result, choose (a).

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

Explanation:
1. \( \sqrt{9 + p^2} = 5 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4 \), not 5, so A is false.
2. The point is \( \left(\frac{3 + 14}{3}, \frac{4 - 12}{3}\right) = \left(\frac{17}{3}, -\frac{8}{3}\right) \) with x > 0 and y < 0, so it is in the fourth quadrant and R is true.

Teacher's Note:
1. Solve for p instead of trusting the stated value.
2. Fourth quadrant means x positive and y negative.

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

Explanation:
1. By the formula, \( x = \frac{1 \times (-1) + 2 \times 1}{3} = \frac{1}{3} \) and \( y = \frac{1 \times 1 + 2 \times 2}{3} = \frac{5}{3} \).
2. The point is \( \left(\frac{1}{3}, \frac{5}{3}\right) \), so A (with −1/3) is false, and R is true.

Teacher's Note:
1. Recompute the point; do not trust the value in the assertion.
2. A sign error can make an assertion false.

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

Explanation:
1. The midpoint of PQ is (0, 2), which is equidistant, but all points on the perpendicular bisector 2x − 3y + 6 = 0 are equidistant too, so A is false.
2. R states this correctly, so R is true.

Teacher's Note:
1. The condition PX = QX gives a line.
2. The word only makes the assertion false.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Draw the right triangle with the two points and the horizontal and vertical lines through them; its legs are |x₂ − x₁| and |y₂ − y₁|.
2. The Pythagorean theorem on this triangle gives the distance formula, so both are true and R explains A.

Teacher's Note:
1. The distance formula comes from the Pythagorean theorem.
2. Both statements are true and linked.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Put m = n = 1 in the section formula: \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \), the midpoint formula.
2. Both are true and R explains A.

Teacher's Note:
1. The midpoint is the special case m : n = 1 : 1.
2. R directly justifies A.