CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 10

Answer: PQ = √[(7 − 2)² + (5 − 5)²] = 5 units.

Teacher's Note:
1. Both points have the same y-coordinate, so the distance is just the difference of x-values.
2. The distance formula gives the same result.

Answer: Mid-point x = (−6 − 2)/2 = −4, so \( \dfrac{α}{3} \) = −4 and α = −12. The y-coordinate (5 + 3)/2 = 4 agrees.

Teacher's Note:
1. Equate the given x-coordinate with the mid-point formula.
2. Check the y-coordinate as a quick verification.

Answer: Let P(0, b) and Q(a, 0). Then a/2 = 2 and b/2 = −5, so a = 4 and b = −10. Hence P(0, −10) and Q(4, 0).

Teacher's Note:
1. A point on the y-axis has x = 0 and a point on the x-axis has y = 0.
2. Then apply the mid-point formula.

Answer: (4 − 1)² + (p − 0)² = 25, so p² = 16 and p = ±4.

Teacher's Note:
1. Square both sides of the distance formula to remove the root.
2. Remember that p has two values, +4 and −4.

Answer: Area of triangle = 0: 1(0 − b) + 0(b − 2) + a(2 − 0) = 0, so 2a − b = 0, i.e. b = 2a.

Teacher's Note:
1. Three points are collinear when the area of the triangle formed by them is zero.
2. Expand the determinant carefully with signs.

Answer: The centre is the mid-point of AB. If A = (x, y): (x + 1)/2 = 2 and (y + 4)/2 = −3, so x = 3, y = −10. A is (3, −10).

Teacher's Note:
1. The centre of a circle is the mid-point of every diameter.
2. Solve each coordinate separately.

Answer: Mid-point of BC is D(2, 1). AD = √[(2 + 2)² + (1 − 4)²] = √25 = 5 units.

Teacher's Note:
1. A median joins a vertex to the mid-point of the opposite side.
2. Find the mid-point first and then use the distance formula.

Answer: 64 + (x + 5)² = 100, so (x + 5)² = 36 and x + 5 = ±6. Hence x = 1 or x = −11.

Teacher's Note:
1. Square the distance and isolate the bracket.
2. Take both the positive and negative square roots.

Answer: Radius OA = √(4 + 0) = 2. OB² = (x − 2)² + 4 = 4, so x = 2.

Teacher's Note:
1. Every point of a circle is at the same distance from the centre.
2. Use OA = OB.

Answer: Area = 0: −3(p + 5) + 2(−5 − 9) + 4(9 − p) = 0, so −7p − 7 = 0 and p = −1.

Teacher's Note:
1. Set the area of the triangle formed by the three points equal to zero.
2. Be careful with negative signs while expanding.

Answer: Taking them in order, each side = √18 = 3√2, so all four sides are equal. Both diagonals, (3, 2)–(−3, 2) and (0, 5)–(0, −1), have length 6. Equal sides and equal diagonals make it a square.

Teacher's Note:
1. A quadrilateral with four equal sides and equal diagonals is a square.
2. Compute all four sides and both diagonals.

Answer: AP : PB = 1 : 2, so P = ((1·5 + 2·2)/3, (1·(−8) + 2·1)/3) = (3, −2). Then 6 + 2 + k = 0, so k = −8.

Teacher's Note:
1. Convert AP : AB into the ratio AP : PB before using the section formula.
2. Substitute the point into the line to find k.

Answer: 2(y − 5) + x(5 − 1) + 7(1 − y) = 0, giving 4x − 5y − 3 = 0.

Teacher's Note:
1. Equate the area of the triangle to zero.
2. Collect the terms in x and y.

Answer: Let the ratio be k : 1. The point is ((2k + 1)/(k + 1), (3k + 2)/(k + 1)). Substituting: 2(2k + 1) + 3(3k + 2) = 10(k + 1), so 3k = 2 and k = \( \dfrac{2}{3} \). The ratio is 2 : 3.

Teacher's Note:
1. Express the dividing point using k and put it on the line.
2. Solve the resulting linear equation for k.

Answer: Sides: AB = BC = CD = DA = √5, so it is a rhombus. Diagonals AC = √18 and BD = √2 are unequal, so it is not a square.

Teacher's Note:
1. A rhombus has four equal sides but its diagonals need not be equal.
2. Compare the diagonals to decide if it is a square.

Answer: Mid-points: (1, 0), (1, 2), (0, 1). Area = ½|1(2 − 1) + 1(1 − 0) + 0| = 1 sq. unit. Given triangle: ½|0 + 2(3 + 1) + 0| = 4 sq. units. Ratio = 1 : 4.

Teacher's Note:
1. Find the three mid-points and then use the area formula for each triangle.
2. The mid-point triangle always has one-fourth of the area.

Answer: Ratio k : 1: (−4k + 3)/(k + 1) = −2, so k = \( \dfrac{5}{2} \), i.e. 5 : 2. Then a = (5·2 − 4·2)/7 = \( \dfrac{2}{7} \).

Teacher's Note:
1. Use the y-coordinate first because it is fully known.
2. Then substitute the ratio to get a.

Answer: Let B = (x, y). (3x + 8)/7 = −1 gives x = −5, and (3y + 20)/7 = 2 gives y = −2. B is (−5, −2).

Teacher's Note:
1. Use the internal section formula with B unknown.
2. Solve for x and y one at a time.

Answer: Mid-point of diagonal AC (1, 1)–(4, 8) is (2.5, 4.5) and mid-point of BD (4, 4)–(1, 5) is also (2.5, 4.5). Diagonals bisect each other, so it is a parallelogram.

Teacher's Note:
1. The diagonals of a parallelogram bisect each other.
2. Equal mid-points of the two diagonals prove it.

Answer: −1(p + 1) + 2(−1 − 3) + 5(3 − p) = 0, so 6 − 6p = 0 and p = 1.

Teacher's Note:
1. Use the zero-area condition.
2. Simplify to a simple linear equation.

Answer: Mid-point of the other two vertices = (3, 0). Distance from (−1, 3) = √(16 + 9) = 5 units.

Teacher's Note:
1. Find the mid-point of the opposite side first.
2. Then apply the distance formula.

Answer: Let the centre be (x, y). Equating squared distances to (3, −7) and (3, 3) gives y = −2; equating those to (6, −6) and (3, −7) gives x = 3. Centre = (3, −2); each radius is 5.

Teacher's Note:
1. The centre is equidistant from all three points.
2. Solve the two linear equations obtained.

Answer: 9 + y² = 25, so y² = 16. Since y > 0, y = 4.

Teacher's Note:
1. Use the distance formula and square it.
2. Choose the positive root as stated.

Answer: Area = 0: x(−2 + 5) + (−5)(−5 − y) + 3(y + 2) = 0, i.e. 3x + 25 + 5y + 3y + 6 = 0, so 3x + 8y + 31 = 0.

Teacher's Note:
1. Use the area formula and set it to zero.
2. Collect like terms to reach the required result.

Answer: Ratio k : 1 with x = 0: (−k + 5)/(k + 1) = 0, so k = 5; ratio 5 : 1. y = (5(−4) + (−6))/6 = −\( \dfrac{13}{3} \). The point is (0, −\( \dfrac{13}{3} \)).

Teacher's Note:
1. The y-axis has x = 0, which fixes the ratio.
2. Use this ratio to find the y-coordinate.

Answer: 9 + (4 − k)² = (2 − k)² + 9, so (4 − k)² = (2 − k)², which gives 16 − 8k = 4 − 4k and k = 3.

Teacher's Note:
1. Equidistant means the two squared distances are equal.
2. The constant 9 cancels on both sides.

Answer: 9 + (p − 2)² = p² + 9, so −4p + 4 = 0 and p = 1. AB = √(9 + 1) = √10.

Teacher's Note:
1. Set AB² equal to AC² and simplify.
2. Substitute p = 1 to find AB.

Answer: Ratio k : 1: (−3k + 5)/(k + 1) = 2, so k = \( \dfrac{3}{5} \), i.e. 3 : 5. x = (3·4 + 5·12)/8 = 9.

Teacher's Note:
1. Use the known y-coordinate to find the ratio.
2. Then use the ratio for x.

Answer: Area = 0 gives −2(b + 1) + a(−2) + 4(1 − b) = 0, i.e. a + 3b = 1. With a = b + 1 we get 4b = 0, so b = 0 and a = 1.

Teacher's Note:
1. Form one equation from collinearity and combine it with the given condition.
2. Solve the two equations together.

Answer: Ratio k : 1: (3k − 6)/(k + 1) = −4, so k = \( \dfrac{2}{7} \). The ratio is 2 : 7. The y-coordinate (−16 + 70)/9 = 6 confirms it.

Teacher's Note:
1. Use the x-coordinates to get k.
2. Verify with the y-coordinates.