CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 19

Answer: (a) \( 9\sqrt{2} \)

Explanation:
1. d² = (6 + 3)² + (7 + 2)² = 81 + 81 = 162.
2. d = \( \sqrt{162} = 9\sqrt{2} \).

Teacher's Note:
1. Use d = √[(x₂ − x₁)² + (y₂ − y₁)²].
2. Simplify the root by taking out perfect squares.

Answer: (b) \( \left(\frac{10}{7}, \frac{33}{7}\right) \)

Explanation:
1. By the section formula x = (3 × 2 + 4 × 1)/7 = 10/7.
2. y = (3 × 7 + 4 × 3)/7 = 33/7.

Teacher's Note:
1. Section formula: x = (m₁x₂ + m₂x₁)/(m₁ + m₂).
2. The first number of the ratio multiplies the second point.

Answer: (c) 29 square units

Explanation:
1. Area = ½ |5(−3 + 5) + (−9)(−5 − 2) + (−3)(2 + 3)|.
2. = ½ |10 + 63 − 15| = ½ × 58 = 29 square units.

Teacher's Note:
1. Area = ½
2. x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)
3. .
4. Always take the modulus because area is positive.

Answer: (d) 6x − 5y + 30 = 0

Explanation:
1. Intercept form: x/(−5) + y/6 = 1.
2. Multiplying by 30: −6x + 5y = 30, i.e. 6x − 5y + 30 = 0.

Teacher's Note:
1. Intercept form: x/a + y/b = 1.
2. Clear the denominators carefully including the sign.

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

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

Teacher's Note:
1. Parallel lines have the same x and y coefficients.
2. Only the constant term changes.

Answer: (b) x² + y² + 6x − 8y − 24 = 0

Explanation:
1. (x + 3)² + (y − 4)² = 49.
2. x² + 6x + 9 + y² − 8y + 16 = 49, so x² + y² + 6x − 8y − 24 = 0.

Teacher's Note:
1. Circle: (x − h)² + (y − k)² = r².
2. Expand and collect the constants: 9 + 16 − 49 = −24.

Answer: (c) \( \left(-\frac{5}{3}, 2\right) \)

Explanation:
1. The point nearer to (−3, 4) divides the segment from (1, −2) in the ratio 2 : 1.
2. x = (2 × (−3) + 1 × 1)/3 = −5/3 and y = (2 × 4 + 1 × (−2))/3 = 2.

Teacher's Note:
1. Two trisection points divide the segment in the ratios 1 : 2 and 2 : 1.
2. The nearer point to B divides AB in 2 : 1.

Answer: (d) x + y = 7

Explanation:
1. (x − 2)² + (y − 3)² = (x − 4)² + (y − 5)².
2. −4x − 6y + 13 = −8x − 10y + 41, so 4x + 4y = 28, i.e. x + y = 7.

Teacher's Note:
1. The locus is the perpendicular bisector of the join.
2. Its slope is −1 because the join has slope 1.

Answer: (a) 96 square units

Explanation:
1. Shoelace sum = (−1)(−9) − (−3)(6) + (−3)(−8) − 5(−9) + 5(9) − 3(−8) + 3(6) − (−1)(9) = 192.
2. Area = ½ × 192 = 96 square units.

Teacher's Note:
1. Use the shoelace formula for polygons, listing vertices in order.
2. Check by splitting into two triangles.

Answer: (b) right-angled

Explanation:
1. Squares of the sides: AB² = 16 + 49 = 65, BC² = 36 + 9 = 45, CA² = 4 + 16 = 20.
2. 65 = 45 + 20, so the triangle is right-angled at C.

Teacher's Note:
1. Test the converse of the Pythagoras theorem.
2. The longest side must equal the sum of squares of the other two.

Answer: (c) 29 square units

Explanation:
1. Area = ½ |10(2 + 5) + 3(−5 − 5) + 6(5 − 2)| = ½ |70 − 30 + 18|.
2. = ½ × 58 = 29 square units.

Teacher's Note:
1. Substitute carefully and watch the signs.
2. The answer need not be an integer.

Answer: (d) 1 : 4

Explanation:
1. The mid-point triangle has sides half those of the triangle, so it is similar with ratio 1 : 2.
2. Area ratio = 1² : 2² = 1 : 4 (check: the areas are 5 and 20 square units).

Teacher's Note:
1. The medial triangle divides the triangle into four equal triangles.
2. Area ratio is the square of the side ratio.

Answer: (a) two possible values

Explanation:
1. Area = ½ |−2(p − 9) + 1(9 − 1) + 4(1 − p)| = ½ |30 − 6p| = 3.
2. So |30 − 6p| = 6, giving p = 4 or p = 6: two values.

Teacher's Note:
1. A modulus equation gives two cases, ±.
2. Check that both values give non-zero area.

Answer: (b) −1

Explanation:
1. Slope between (5, 7) and (8, 11) = 4/3.
2. Slope between (x, −1) and (5, 7) = 8/(5 − x) = 4/3, so 5 − x = 6 and x = −1.

Teacher's Note:
1. Collinear points have equal slopes.
2. You may also set the area to zero.

Answer: (c) \( \sqrt{(a-b)^2 + (b-c)^2} \)

Explanation:
1. Difference of x: (b + c) − (c + a) = b − a. Difference of y: (c + a) − (a + b) = c − b.
2. d = \( \sqrt{(a-b)^2 + (b-c)^2} \).

Teacher's Note:
1. Cancel common terms before squaring.
2. The distance formula always has a plus sign inside the root.

Answer: (d) (3, 1)

Explanation:
1. x = (1 × 5 + 2 × 2)/3 = 3.
2. y = (1 × (−3) + 2 × 3)/3 = 1.

Teacher's Note:
1. Point is nearer the first point when the ratio is 1 : 2.
2. Check that x lies between 2 and 5.

Answer: (a) (3, 0)

Explanation:
1. Let the point be (x, 0): (x − 7)² + 36 = (x + 3)² + 16.
2. −14x + 85 = 6x + 25, so x = 3.

Teacher's Note:
1. A point on the x-axis has y = 0.
2. Equate the squared distances.

Answer: (b) \( \left(4\sqrt{2}, 45^\circ\right) \)

Explanation:
1. r = √(4² + 4²) = 4√2.
2. tan θ = 4/4 = 1, so θ = 45°.

Teacher's Note:
1. r = √(x² + y²) and tan θ = y/x.
2. Check the quadrant for the angle.

Answer: (c) \( \left(3, 3\sqrt{3}\right) \)

Explanation:
1. x = r cos θ = 6 × ½ = 3.
2. y = r sin θ = 6 × (√3/2) = 3√3.

Teacher's Note:
1. x = r cos θ and y = r sin θ.
2. Learn the values of sin and cos for 30°, 45° and 60°.

Answer: (d) \( 3 \pm 2\sqrt{15} \)

Explanation:
1. (x₁ − 3)² + (2 − 4)² = 64, so (x₁ − 3)² = 60.
2. x₁ − 3 = ±√60 = ±2√15, so x₁ = 3 ± 2√15.

Teacher's Note:
1. Square both sides to remove the root.
2. Remember the ± sign when taking a square root.