CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 18

Answer: (a) True, it is the midpoint

Explanation:
1. The midpoint of the points is \( (\frac{2 - 2}{2}, \frac{4 - 4}{2}) = (0, 0) \).
2. The origin is therefore the midpoint, so it lies on the segment.

Teacher's Note:
1. The midpoint formula averages the coordinates.
2. Points that are negatives of each other are symmetric about the origin.

Answer: (b) the y-axis

Explanation:
1. \( (x - b)^2 + y^2 = (x + b)^2 + y^2 \Rightarrow -4bx = 0 \Rightarrow x = 0 \).
2. The locus is x = 0, the y-axis.

Teacher's Note:
1. It is the perpendicular bisector of the segment joining the two points.
2. The segment lies on the x-axis, so its bisector is the y-axis.

Answer: (b) \( \tfrac{11}{8} \) sq. units

Explanation:
1. Area = \( \tfrac12 \left| -2(0 - \tfrac12) + 5(\tfrac12 - \tfrac34) + 4(\tfrac34 - 0) \right| \).
2. = \( \tfrac12 |1 - \tfrac54 + 3| = \tfrac12 \cdot \tfrac{11}{4} = \tfrac{11}{8} \).

Teacher's Note:
1. Use the determinant-style area formula and keep the fractions exact.
2. The area is not 1, so the claim of 1 sq. unit would be false.

Answer: (a) They are collinear

Explanation:
1. Slope of AB = \( \tfrac{2 + 2}{5 + 3} = \tfrac12 \) and slope of BC = \( \tfrac{4 - 2}{9 - 5} = \tfrac12 \).
2. Equal slopes through the common point B mean the three points lie on one line.

Teacher's Note:
1. Equal slopes for AB and BC show collinearity.
2. The area of triangle ABC is zero as well.

Answer: (b) equal to the sum of the diagonals of the first

Explanation:
1. Each side of the midpoint quadrilateral is half of a diagonal of the original and parallel to it.
2. Its four sides are two halves of AC and two halves of BD, so the perimeter = AC + BD.

Teacher's Note:
1. This is the midpoint (Varignon) theorem for quadrilaterals.
2. Check: AC = \( \sqrt{100 + 16} \) and BD = \( \sqrt{100 + 64} \); the perimeter equals their sum.

Answer: (a) \( 3\sqrt{10} \)

Explanation:
1. \( d = \sqrt{(-6 + 3)^2 + (7 + 2)^2} = \sqrt{9 + 81} = \sqrt{90} \).
2. \( \sqrt{90} = 3\sqrt{10} \).

Teacher's Note:
1. Use the distance formula with the differences -3 and 9.
2. Simplify the root by taking out the perfect square 9.

Answer: (a) \( (\tfrac{10}{7}, \tfrac{33}{7}) \)

Explanation:
1. \( x = \dfrac{3 \cdot 2 + 4 \cdot 1}{7} = \dfrac{10}{7} \).
2. \( y = \dfrac{3 \cdot 7 + 4 \cdot 3}{7} = \dfrac{33}{7} \).

Teacher's Note:
1. In the section formula the first ratio number multiplies the second point.
2. Do not swap m and n.

Answer: (a) 29 sq. units

Explanation:
1. Area = \( \tfrac12 |5(-3 + 5) + (-9)(-5 - 2) + (-3)(2 + 3)| \).
2. = \( \tfrac12 |10 + 63 - 15| = \tfrac12 \cdot 58 = 29 \).

Teacher's Note:
1. Take the modulus of the sum before halving.
2. Forgetting the factor 1/2 would give 58.

Answer: (a) 6x + 5y = 30

Explanation:
1. Intercept form: \( \dfrac{x}{5} + \dfrac{y}{6} = 1 \).
2. Multiply by 30: 6x + 5y = 30.

Teacher's Note:
1. The x-intercept goes under x and the y-intercept under y.
2. Check: the line meets the axes at (5, 0) and (0, 6).

Answer: (a) 3x + 4y + 8 = 0

Explanation:
1. A parallel line has the form 3x + 4y + k = 0.
2. Substituting (4, -5): 12 - 20 + k = 0, so k = 8.

Teacher's Note:
1. Parallel lines differ only in the constant term.
2. Check the point in the final equation.

Answer: (a) x² + y² + 6x - 8y = 24

Explanation:
1. \( (x + 3)^2 + (y - 4)^2 = 49 \).
2. Expanding: \( x^2 + y^2 + 6x - 8y + 25 = 49 \), so \( x^2 + y^2 + 6x - 8y = 24 \).

Teacher's Note:
1. Remember (x - a)² + (y - b)² = r² with the centre (a, b).
2. Move the constant 25 to the right side.

Answer: (a) \( (-\tfrac53, 2) \)

Explanation:
1. The point nearer to (-3, 4) divides the segment from (1, -2) in the ratio 2 : 1.
2. \( x = \dfrac{2(-3) + 1}{3} = -\dfrac53 \), \( y = \dfrac{2(4) - 2}{3} = 2 \).

Teacher's Note:
1. Decide which end the point is closer to before choosing the ratio.
2. The other trisection point is \( (-\tfrac13, 0) \).

Answer: (a) 4x - 9y + 9 = 0

Explanation:
1. \( y - 5 = \tfrac49 (x - 9) \).
2. \( 9y - 45 = 4x - 36 \), so \( 4x - 9y + 9 = 0 \).

Teacher's Note:
1. Use the point-slope form.
2. Clear the fraction by multiplying by 9.