CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 15

Answer: (c) –3

Explanation:
1. PQ² = 169, so (x − 2)² + (−7 − 5)² = 169.
2. (x − 2)² + 144 = 169, so (x − 2)² = 25 and x − 2 = ±5.
3. This gives x = 7 or x = −3. Only −3 is among the options.

Teacher's Note:
1. Always square both sides of the distance formula before solving.
2. A squared equation can give two values, so check which one is in the options.

Answer: (b) ab′ = a′b

Explanation:
1. Take the slopes from (a′, b′) to the other two points and equate them.
2. (b − b′)/(a − a′) = (b − 2b′)/(a − 2a′).
3. Cross-multiplying: ab − 2a′b − ab′ + 2a′b′ = ab − a′b − 2ab′ + 2a′b′.
4. This simplifies to ab′ = a′b.

Teacher's Note:
1. Three points are collinear when the slope between any two pairs is the same.
2. Cancel the common terms carefully.

Answer: (b) –12

Explanation:
1. The midpoint is ((−6 − 2)/2, (5 + 3)/2) = (−4, 4).
2. So a/3 = −4, which gives a = −12.

Teacher's Note:
1. The midpoint formula averages the x-coordinates and the y-coordinates separately.
2. Compare only the x-coordinate to find a.

Answer: (b) –15

Explanation:
1. x = (3 + k)/2 and y = (4 + 7)/2.
2. Put these in 2x + 2y + 1 = 0: (3 + k) + 11 + 1 = 0.
3. So k + 15 = 0 and k = −15.

Teacher's Note:
1. 2x and 2y remove the denominators, so no fractions are left.
2. Substitute the midpoint into the given line equation.

Answer: (b) 3 : 4

Explanation:
1. Let the ratio be k : 1. The x-coordinate of the dividing point is (−4k + 3)/(k + 1).
2. On the y-axis the x-coordinate is 0, so −4k + 3 = 0 and k = 3/4.
3. The ratio is 3/4 : 1, that is 3 : 4.

Teacher's Note:
1. A point on the y-axis has x-coordinate 0.
2. Write the ratio as k : 1 and convert at the end.

Answer: (b) 1 : 6

Explanation:
1. Let the ratio be m : n. Using the x-coordinate, (4m − 3n)/(m + n) = −2.
2. 4m − 3n = −2m − 2n, so 6m = n and m : n = 1 : 6.
3. Check with y: (−9 × 1 + 5 × 6)/7 = 3, which matches.

Teacher's Note:
1. Use either coordinate to find the ratio, then check with the other.
2. The point is closer to the first end, so the ratio is small.

Answer: (c) \( \frac{x}{a} \) + \( \frac{y}{b} \) = 1

Explanation:
1. A, B and P are collinear, so the slopes from A are equal: y/(x − a) = (b − 0)/(0 − a) = −b/a.
2. Then ay = −b(x − a), so bx + ay = ab.
3. Dividing by ab gives x/a + y/b = 1.

Teacher's Note:
1. This is the intercept form of the line with x-intercept a and y-intercept b.
2. Check by putting (a, 0) and (0, b) in the option.

Answer: (b) 12

Explanation:
1. The sides are 4 (from (0, 4) to (0, 0)), 3 (from (0, 0) to (3, 0)) and √(9 + 16) = 5.
2. Perimeter = 4 + 3 + 5 = 12.

Teacher's Note:
1. Two of the sides lie along the axes, so their lengths can be read directly.
2. Use the distance formula only for the slanting side.

Answer: (a) (2, 0)

Explanation:
1. A point equidistant from A and B on the x-axis is the midpoint of AB.
2. Midpoint = ((−1 + 5)/2, 0) = (2, 0).

Teacher's Note:
1. When both points lie on the x-axis, the equidistant point on that axis is simply the midpoint.
2. You can also verify using the distance formula.

Answer: (c) (–3, –5)

Explanation:
1. Reflection in the x-axis keeps the x-coordinate and changes the sign of the y-coordinate.
2. So (−3, 5) becomes (−3, −5).

Teacher's Note:
1. Reflection in the x-axis changes the sign of y; reflection in the y-axis changes the sign of x.
2. Draw a quick sketch if you are unsure.

Answer: (d) 1

Explanation:
1. The y-coordinate of the dividing point is (3 × y + 1 × 5)/(3 + 1) = 2.
2. So 3y + 5 = 8 and y = 1.
3. The x-coordinate check: (3 × 4 + 1 × 6)/4 = 9/2, which agrees with P.

Teacher's Note:
1. With ratio m : n the formula is (m × second point + n × first point)/(m + n).
2. The x-coordinate gives a useful check on the setup.

Answer: (a) (5, 11/2)

Explanation:
1. PR = √(8² + 6²) = 10, so PQ : QR = 2.5 : 7.5 = 1 : 3.
2. Q = ((3 × 3 + 1 × 11)/4, (3 × 4 + 1 × 10)/4) = (5, 11/2).

Teacher's Note:
1. Find the whole length PR first, then the ratio in which Q divides it.
2. Q is one-fourth of the way from P to R.

Answer: (a) – 6 and 1

Explanation:
1. Midpoint: y = (4 − 2)/2 = 1.
2. −1 = (x + 4)/2, so x + 4 = −2 and x = −6.

Teacher's Note:
1. Match the x-coordinates and y-coordinates separately.
2. Each unknown appears in only one equation here.

Answer: (c) 3 : 2, y = 3

Explanation:
1. Let the ratio be k : 1. Then 2 = (6k − 4)/(k + 1), so 2k + 2 = 6k − 4 and k = 3/2.
2. The ratio is 3 : 2.
3. y = (3 × 3 + 2 × 3)/5 = 3.

Teacher's Note:
1. Both end points have y = 3, so every point between them also has y = 3.
2. Use the x-coordinate to get the ratio.

Answer: (c) 4 : 9

Explanation:
1. Let the ratio be k : 1. The point is ((−2k + 1)/(k + 1), (k + 2)/(k + 1)).
2. It lies on the line: 3(1 − 2k) + 4(k + 2) = 7(k + 1).
3. 3 − 6k + 4k + 8 = 7k + 7, so 4 = 9k and k = 4/9. The ratio is 4 : 9.

Teacher's Note:
1. Put the section-formula point into the equation of the line.
2. A positive k means the division is internal.

Answer: (b) (2, 0)

Explanation:
1. Let the point be P(x, 0). Then AP² = BP².
2. (x + 2)² + 9 = (x − 5)² + 16, so 4x + 13 = −10x + 41.
3. 14x = 28, so x = 2. The point is (2, 0).

Teacher's Note:
1. A point on the x-axis has the form (x, 0).
2. Equating the squares of the distances removes the roots.

Answer: (b) \( \left(\frac{1}{3}, \frac{5}{3}\right) \)

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

Teacher's Note:
1. With ratio 1 : 2 the point is closer to A, the first point.
2. Multiply the ratio parts crosswise: m with the second point, n with the first.

Answer: (b) (0, 3)

Explanation:
1. x = (3 − 8 + 5)/3 = 0.
2. y = (−7 + 6 + 10)/3 = 9/3 = 3.
3. The centroid is (0, 3).

Teacher's Note:
1. The centroid is the average of the three x-coordinates and the three y-coordinates.
2. Add the numbers with signs carefully.

Answer: (b) Rectangle

Explanation:
1. AB = √(4 + 9) = √13, BC = √(36 + 16) = √52, CD = √(4 + 9) = √13, DA = √(36 + 16) = √52.
2. Opposite sides are equal. Diagonals: AC = √(64 + 1) = √65 and BD = √(16 + 49) = √65, so the diagonals are equal.
3. A parallelogram with equal diagonals is a rectangle. Adjacent sides are unequal, so it is not a square or a rhombus.

Teacher's Note:
1. Equal diagonals turn a parallelogram into a rectangle.
2. Compare all four sides and both diagonals before deciding.

Answer: (b) bp = aq

Explanation:
1. PA² = PB²: (p − a − b)² + (q − b + a)² = (p − a + b)² + (q − a − b)².
2. Using (u − b)² − (u + b)² = −4ub with u = p − a gives −4b(p − a).
3. Using (v + a)² − (v − a)² = 4av with v = q − b gives 4a(q − b).
4. Sum: −4bp + 4ab + 4aq − 4ab = 0, so bp = aq.

Teacher's Note:
1. Use the difference of squares to expand quickly.
2. Many terms cancel, leaving a simple relation.

Answer: (d) (4, 5)

Explanation:
1. The distance from the x-axis is the absolute value of the y-coordinate, so |y| = 5.
2. The distance from the y-axis is the absolute value of the x-coordinate, so |x| = 4.
3. (4, 5) satisfies both.

Teacher's Note:
1. Distance from the x-axis is the y-value, and distance from the y-axis is the x-value.
2. Do not swap the two.

Answer: (d) \( \frac{1}{a} \) + \( \frac{1}{b} \) = 1

Explanation:
1. Slope from (a, 0) to (1, 1) is 1/(1 − a); slope from (1, 1) to (0, b) is (1 − b)/1.
2. Equal slopes: 1 = (1 − a)(1 − b) = 1 − a − b + ab, so ab = a + b.
3. Dividing by ab: 1 = 1/b + 1/a.

Teacher's Note:
1. Equate the slopes to get the condition for collinearity.
2. Divide by ab at the end to reach the options.

Answer: (b) y = 5x – 7

Explanation:
1. Let P = (h, 5h + 3). Then M = ((h + 3)/2, (5h + 1)/2).
2. Put x = (h + 3)/2, so h = 2x − 3.
3. y = (5h + 1)/2 = (5(2x − 3) + 1)/2 = (10x − 14)/2 = 5x − 7.

Teacher's Note:
1. Write P using the equation of its line, then use the midpoint formula.
2. Eliminate h to get the equation satisfied by M.

Answer: (c) (3, –2)

Explanation:
1. Let the centre be (x, y). Distances to (3, −7) and (3, 3) are equal: (y + 7)² = (y − 3)², so 20y = −40 and y = −2.
2. Distances to (6, −6) and (3, −7) are equal: (x − 6)² + 16 = (x − 3)² + 25, so −12x + 52 = −6x + 34 and x = 3.
3. The centre is (3, −2); the radius is 5 from each point.

Teacher's Note:
1. The centre of a circle is equidistant from all points on it.
2. Use two equations, one for each pair of points.

Answer: (a) (– 6, 8)

Explanation:
1. The distance of (x, y) from the origin is √(x² + y²).
2. For (−6, 8): √(36 + 64) = √100 = 10.
3. The others give √20, √61 and √52.

Teacher's Note:
1. Test each option with the distance formula from the origin.
2. 36 + 64 = 100 is a quick one to spot.

Answer: (b) (0, 4) and (0, 6)

Explanation:
1. (−2, −3) and (−2, −4): distance 1.
2. (0, 4) and (0, 6): distance |6 − 4| = 2.
3. (7, 2) and (6, 2): distance 1. (4, −3) and (2, 3): √(4 + 36) = √40.

Teacher's Note:
1. When two points share a coordinate, the distance is the difference of the other coordinate.
2. Check each option quickly.

Answer: (d) a rhombus, if all the four sides are equal.

Explanation:
1. Options (a), (b) and (c) are the standard tests for three points.
2. A rhombus needs four vertices, so three points can never form a rhombus.

Teacher's Note:
1. A rhombus has four vertices, so it cannot come from three points.
2. The sum of two distances equalling the third gives collinear points.

Answer: (c) a square, if all the four sides and diagonals are equal.

Explanation:
1. A rectangle needs equal opposite sides and equal diagonals, so (a) is wrong.
2. A parallelogram needs equal opposite sides, so (b) is wrong, and four points do not give a triangle, so (d) is wrong.
3. Four equal sides with equal diagonals describe a square, so (c) is correct.

Teacher's Note:
1. Recall the properties: square means equal sides and equal diagonals.
2. Test each statement against the definition of the shape.