JEE Mathematics Coordinates and Straight Lines MCQs Set 08

Answer: (b) second quadrant

Explanation:
1. The x-coordinate is negative and the y-coordinate is positive.
2. Negative x and positive y is quadrant II.

Teacher's Note:
1. Remember the order I (+, +), II (−, +), III (−, −), IV (+, −).
2. Quadrants are numbered anticlockwise.

Answer: (a) 13

Explanation:
1. Distance = \( \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).

Teacher's Note:
1. Squares remove the negative sign.
2. Do not stop at 169; take the square root.

Answer: (d) 10

Explanation:
1. AB = \( \sqrt{(-4-2)^2 + (-5-3)^2} = \sqrt{36+64} = 10 \).

Teacher's Note:
1. Subtract coordinates in the same order for x and y.
2. The result is the hypotenuse of a 6-8-10 right triangle.

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

Explanation:
1. x = r cos θ = 5 cos 60° = \( \frac{5}{2} \).
2. y = r sin θ = 5 sin 60° = \( \frac{5\sqrt{3}}{2} \).

Teacher's Note:
1. Cosine goes with x and sine with y.
2. cos 60° = 1/2 and sin 60° = √3/2.

Answer: (b) (2, 60°)

Explanation:
1. r = \( \sqrt{1 + 3} \) = 2.
2. tan θ = √3 and the point is in quadrant I, so θ = 60°.

Teacher's Note:
1. Always check the quadrant when using tan θ.
2. Verify with x = r cos θ = 2 × 1/2 = 1.

Answer: (a) (\( \sqrt{2} \), 135°)

Explanation:
1. r = \( \sqrt{1 + 1} = \sqrt{2} \).
2. tan θ = −1 and the point lies in quadrant II, so θ = 180° − 45° = 135°.

Teacher's Note:
1. tan θ = −1 also holds for 315°, which lies in quadrant IV.
2. Use the signs of x and y to choose the correct quadrant.

Answer: (d) \( (2\sqrt{3}, 2) \)

Explanation:
1. x = 4 cos 30° = \( 2\sqrt{3} \).
2. y = 4 sin 30° = 2.

Teacher's Note:
1. Check: x² + y² = 12 + 4 = 16 = r².
2. cos 30° = √3/2 and sin 30° = 1/2.

Answer: (c) (−7, 0)

Explanation:
1. Let the point be (x, 0): (x − 2)² + 25 = (x + 2)² + 81.
2. −4x + 29 = 4x + 85, so x = −7.

Teacher's Note:
1. Point on the x-axis has y = 0.
2. Check: distances² are 81 + 25 = 106 and 25 + 81 = 106.

Answer: (b) (0, 9)

Explanation:
1. Let the point be (0, y): 36 + (y − 5)² = 16 + (y − 3)².
2. 61 − 10y = 25 − 6y, so y = 9.

Teacher's Note:
1. Point on the y-axis has x = 0.
2. Check: 36 + 16 = 52 and 16 + 36 = 52.

Answer: (a) an equilateral triangle

Explanation:
1. Sides: 4, √(4 + 12) = 4 and √(4 + 12) = 4.
2. All three sides are equal.

Teacher's Note:
1. Compare the three squared distances.
2. An equilateral triangle needs all three sides equal.

Answer: (d) collinear

Explanation:
1. AB = \( \sqrt{8} = 2\sqrt{2} \), BC = \( 2\sqrt{2} \), AC = \( \sqrt{32} = 4\sqrt{2} \).
2. AB + BC = AC, so the points lie on one line.

Teacher's Note:
1. For collinearity the longest distance equals the sum of the other two.
2. Here B is the midpoint of AC.

Answer: (c) a right-angled isosceles triangle

Explanation:
1. Sides: 4, 4 and √32.
2. The two equal sides satisfy 4² + 4² = 32 = (√32)², so it is right-angled and isosceles.

Teacher's Note:
1. Check the Pythagoras relation after sorting the sides.
2. The right angle is at the origin.

Answer: (b) 5

Explanation:
1. (k − 1)² + 9 = 25, so (k − 1)² = 16.
2. k − 1 = ±4, and k > 1 gives k = 5.

Teacher's Note:
1. The other root k = −3 is rejected by the condition k > 1.
2. Square both sides to remove the root.

Answer: (a) 12

Explanation:
1. Sides: √(9 + 16) = 5, 4 and 3.
2. Perimeter = 5 + 4 + 3 = 12.

Teacher's Note:
1. This is a right triangle with sides 3, 4, 5.
2. Compute each side separately.

Answer: (d) (3, 5)

Explanation:
1. The diagonals AC and BD have the same midpoint.
2. Midpoint of AC is (7/2, 4), so D = (7 − 4, 8 − 3) = (3, 5).

Teacher's Note:
1. Use D = A + C − B.
2. Check: AB and DC are both (3, 1) as displacements.

Answer: (c) (1, 2)

Explanation:
1. Midpoint = ((−2 + 4)/2, (6 − 2)/2) = (1, 2).

Teacher's Note:
1. The midpoint is the section formula with ratio 1 : 1.
2. Average the coordinates separately.

Answer: (b) (5, 6)

Explanation:
1. x = (2 × 7 + 1 × 1)/3 = 5.
2. y = (2 × 8 + 1 × 2)/3 = 6.

Teacher's Note:
1. The point is closer to the second point because it gets the larger multiplier.
2. Pair each ratio term with the opposite end point.

Answer: (a) (10, 11)

Explanation:
1. x = (3 × 4 − 2 × 1)/(3 − 2) = 10.
2. y = (3 × 5 − 2 × 2)/(3 − 2) = 11.

Teacher's Note:
1. External division uses a minus sign in numerator and denominator.
2. The point lies outside the segment beyond (4, 5).

Answer: (d) (4, 1)

Explanation:
1. Let B = (h, k): (3h − 12)/5 = 0 gives h = 4.
2. (3k − 18)/5 = −3 gives k = 1.

Teacher's Note:
1. Write each coordinate equation separately.
2. Check: (3 × 4 + 2 × (−6))/5 = 0.

Answer: (c) 1 : 2

Explanation:
1. Let the ratio be k : 1; the y-coordinate is (6k − 3)/(k + 1) = 0.
2. So k = 1/2, a ratio of 1 : 2.

Teacher's Note:
1. Put the coordinate that must vanish equal to zero.
2. A positive k means internal division.