CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 16

Answer: (a) (2, 5)

Explanation:
1. Let A = (x, y). Since P is the mid-point of AB, \( \frac{x-2}{2}=0 \) and \( \frac{y+3}{2}=4 \).
2. So x = 2 and y = 5, giving A = (2, 5).

Teacher's Note:
1. Apply the mid-point formula to each coordinate separately.
2. Always substitute the known end point and the mid-point, then solve for the unknown end point.

Answer: (d) 1

Explanation:
1. By the section formula, the y-coordinate of P is \( \frac{3y+1\times 5}{3+1} \).
2. Given this equals 2: 3y + 5 = 8, so 3y = 3.
3. Hence y = 1.

Teacher's Note:
1. In the ratio m : n, the point is nearer to the second point when m is larger.
2. Write the section formula for the y-coordinate only, since the x-coordinates already match.

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

Explanation:
1. Distance squared = \( (a\cos\theta+b\sin\theta)^2+(a\sin\theta-b\cos\theta)^2 \).
2. Expanding, the cross terms 2ab sinθ cosθ cancel, leaving \( a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)=a^2+b^2 \).
3. So the distance is \( \sqrt{a^2+b^2} \).

Teacher's Note:
1. The middle terms cancel because one is +2ab sinθ cosθ and the other is −2ab sinθ cosθ.
2. Use sin²θ + cos²θ = 1 to finish.

Answer: (d) −1

Explanation:
1. Using the section formula, \( k=\frac{1\times(-7)+2\times 2}{1+2} \).
2. So k = (−7 + 4)/3 = −1.
3. Check the y-coordinate: (1×4 + 2×(−2))/3 = 0, as required.
4. Hence k = −1.

Teacher's Note:
1. Check the other coordinate to confirm the ratio fits the given point.
2. In the ratio m : n, the first point's coordinate is multiplied by n.

Answer: (a) −2

Explanation:
1. Three points are collinear when the area of the triangle they form is zero.
2. 3(p + 5) + 5(−5 − 1) + 7(1 − p) = 0.
3. 3p + 15 − 30 + 7 − 7p = 0, so −4p − 8 = 0.
4. Hence p = −2.

Teacher's Note:
1. Collinear means zero area; use the determinant (area) formula.
2. Expand carefully with signs: the (−5 − 1) term gives −30.

Answer: (b) \( 5\sqrt{2} \)

Explanation:
1. Distance = \( \sqrt{(0+5)^2+(5-0)^2}=\sqrt{50} \).
2. So the distance is \( 5\sqrt{2} \) units.

Teacher's Note:
1. Distance is never negative, so option (c) cannot be right.
2. Simplify √50 as √(25×2).

Answer: (c) \( \frac{16}{5} \)

Explanation:
1. Using the section formula, \( k=\frac{2\times5+3\times2}{2+3}=\frac{16}{5} \).
2. Check: y = (2×1 + 3×6)/5 = 4, which matches the given point.
3. So \( k=\frac{16}{5} \).

Teacher's Note:
1. The given y-coordinate 4 is a useful check on the ratio.
2. Multiply the first ratio term by the second point's x-coordinate.

Answer: (c) 13 units

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

Teacher's Note:
1. Distance from the origin is √(x² + y²).
2. 5, 12, 13 is a Pythagorean triple worth remembering.

Answer: (c) Origin

Explanation:
1. The x-axis and the y-axis cross at the point (0, 0).
2. This point is called the origin.

Teacher's Note:
1. The origin has both coordinates equal to zero.
2. The axes are lines, while the origin is a single point.

Answer: (b) negative, positive

Explanation:
1. The second quadrant lies to the left of the y-axis and above the x-axis.
2. So x is negative and y is positive: the signs are (−, +).

Teacher's Note:
1. Quadrant signs go anticlockwise: (+, +), (−, +), (−, −), (+, −).
2. Abscissa means the x-coordinate; ordinate means the y-coordinate.

Answer: (d) \( 2\sqrt{17} \)

Explanation:
1. Distance = \( \sqrt{(-5-3)^2+(2-4)^2}=\sqrt{64+4}=\sqrt{68} \).
2. √68 = √(4×17) = \( 2\sqrt{17} \).

Teacher's Note:
1. Reduce the root by pulling out the largest perfect square, here 4.
2. Subtract coordinates in the same order for both x and y.

Answer: (c) 0

Explanation:
1. Collinear points lie on one straight line, so they cannot enclose a triangle.
2. The area is 0.

Teacher's Note:
1. Zero area is the standard test for collinearity.
2. Area can never be negative; the formula uses a modulus.

Answer: (a) 1 : 1

Explanation:
1. Let the x-axis divide AB in the ratio m : n. Points on the x-axis have y = 0.
2. So \( \frac{5m-5n}{m+n}=0 \), giving m = n.
3. The required ratio is 1 : 1.

Teacher's Note:
1. A point on the x-axis has y-coordinate 0, so use the y-formula.
2. Equal and opposite y-values mean the x-axis passes through the mid-point.

Answer: (c) \( \sqrt{34} \)

Explanation:
1. The diagonal is AB, joining A(0, 3) and B(5, 0).
2. AB = \( \sqrt{5^2+3^2}=\sqrt{34} \).
3. So the diagonal is \( \sqrt{34} \) units.

Teacher's Note:
1. Diagonals of a rectangle are equal, so OC works equally well.
2. Here the sides are 5 and 3, so use Pythagoras.

Answer: (a) \( \left(4,\frac{13}{3}\right) \)

Explanation:
1. x = (1×8 + 2×2)/3 = 4.
2. y = (1×3 + 2×5)/3 = 13/3.
3. So Q = \( \left(4,\frac{13}{3}\right) \).

Teacher's Note:
1. The ratio 1 : 2 means Q is nearer to P.
2. Multiply P's coordinates by 2 and R's by 1.

Answer: (a) \( 2\sqrt{10} \) units

Explanation:
1. PR = \( \sqrt{(8-2)^2+(3-5)^2}=\sqrt{40}=2\sqrt{10} \) units.

Teacher's Note:
1. Reduce √40 as √(4×10).
2. The horizontal gap is 6 and the vertical gap is 2.

Answer: (c) \( \left(\frac{13}{9},0\right) \)

Explanation:
1. Let the point be A(x, 0), with Q = (4, 13/3). Then AP² = AQ².
2. (2 − x)² + 25 = (4 − x)² + 169/9.
3. 4 − 4x + 25 = 16 − 8x + 169/9, so 4x = 52/9 and x = 13/9.
4. The point is \( \left(\frac{13}{9},0\right) \).

Teacher's Note:
1. For a point on the x-axis, the y-coordinate is 0.
2. Squaring both distances removes the roots and the x² terms cancel.

Answer: (c) (5, 4)

Explanation:
1. Mid-point = ((2 + 8)/2, (5 + 3)/2) = (5, 4).

Teacher's Note:
1. Average the x-coordinates and the y-coordinates.
2. The mid-point lies between P and R on the line.

Answer: (b) (6, 2)

Explanation:
1. Shifting the origin right by 2 reduces every x-coordinate by 2, and shifting it up by 1 reduces every y-coordinate by 1.
2. R(8, 3) becomes (8 − 2, 3 − 1) = (6, 2).

Teacher's Note:
1. When the origin moves, the point itself does not move; only its coordinates change.
2. Subtract the shift from each coordinate.

Answer: (d) none of these

Explanation:
1. Distance = \( \sqrt{(6-2)^2+(-8-3)^2}=\sqrt{16+121}=\sqrt{137} \) cm.
2. √137 is about 11.7 cm, which matches none of the options 12, 15 and 18, so the answer is none of these.

Teacher's Note:
1. Do not round √137 to 12; it is not exactly 12.
2. Compare exact values when options are close.

Answer: (b) 10 cm

Explanation:
1. Distance = \( \sqrt{(-6)^2+8^2}=\sqrt{100} \) = 10 cm.

Teacher's Note:
1. The house is the origin, so use √(x² + y²).
2. 6, 8, 10 is a Pythagorean triple.

Answer: (c) 1 : 2

Explanation:
1. House to grocery store: \( \sqrt{2^2+3^2}=\sqrt{13} \).
2. House to electrician's shop: \( \sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13} \).
3. The ratio is √13 : 2√13 = 1 : 2.

Teacher's Note:
1. The two points are on opposite sides of the origin along one line, since (−4, −6) = −2 × (2, 3).
2. Find each distance separately and divide.

Answer: (c) 1 : 1

Explanation:
1. House to bus stand = √(36 + 64) = 10 cm.
2. House to food cart = √(36 + 64) = 10 cm.
3. The ratio is 1 : 1. (The house is the mid-point of the bus stand and the food cart.)

Teacher's Note:
1. (−6, 8) and (6, −8) are opposite points about the origin.
2. Equal distances give the ratio 1 : 1.

Answer: (d) none of these

Explanation:
1. For a parallelogram the diagonals must bisect each other, so the mid-points of the diagonals must coincide.
2. The mid-point of the grocery store and electrician's shop is (−1, −3/2), and the mid-point of the bus stand and food cart is (0, 0). These are different.
3. The other pairings also give different mid-points, so the figure is not a parallelogram (hence not a rectangle or square). The answer is none of these.

Teacher's Note:
1. A rectangle or square is a parallelogram, so first test the diagonals' mid-points.
2. Different mid-points rule out all three named shapes.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Using the section formula with ratio 2 : 7: x = (2×6 + 7×(−3))/9 = −1 and y = (2×(−8) + 7×10)/9 = 6.
2. So the Assertion is true. The Reason is also a true statement.
3. The Reason is a test for collinearity and does not explain how the ratio 2 : 7 is found, so it is not the correct explanation.

Teacher's Note:
1. Check the Assertion by actually computing both coordinates.
2. A true Reason is not automatically the explanation of the Assertion.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A point with x-coordinate 0 lies on the y-axis, so (0, 4) is on the y-axis: the Assertion is true.
2. A point with y-coordinate 0 lies on the x-axis: the Reason is true.
3. But the Reason talks about the x-axis and does not explain the Assertion.

Teacher's Note:
1. x = 0 means the y-axis; y = 0 means the x-axis.
2. Do not mix the two statements.

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

Explanation:
1. Using the mid-point formula: y = (4 − 2)/2 = 1 and −1 = (x + 4)/2, so x = −6.
2. The Assertion is true and it follows directly from the Reason, which is also true.

Teacher's Note:
1. Apply the mid-point formula to x and y separately.
2. Here the Reason is exactly the formula used.