CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 23

Answer: (a) collinear

Explanation:
1. AB = sqrt(16 + 16) = 4 sqrt2, BC = sqrt(64 + 64) = 8 sqrt2 and AC = sqrt(144 + 144) = 12 sqrt2. Since AB + BC = AC the points lie on a line.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (d) (2, 0)

Explanation:
1. Let P = (x, 0). (x - 5)^2 + 16 = (x + 2)^2 + 9 gives -10x + 41 = 4x + 13, so x = 2.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (c) scalene

Explanation:
1. AB = 5, BC = sqrt(1 + 36) = sqrt37 and AC = sqrt(9 + 9) = sqrt18. All three sides are different, so the triangle is scalene.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (b) 3 or -9

Explanation:
1. (10 - 2)^2 + (y + 3)^2 = 100 gives (y + 3)^2 = 36, so y + 3 = 6 or -6, that is y = 3 or y = -9.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (b) right-angled isosceles, with the right angle at A

Explanation:
1. AB^2 = 25 + 81 = 106, AC^2 = 81 + 25 = 106, BC^2 = 16 + 196 = 212 = AB^2 + AC^2. So the right angle is at A and AB = AC.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (a) 3x = 2y

Explanation:
1. (x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2 gives -10x - 2y + 26 = 2x - 10y + 26, so 12x = 8y, i.e. 3x = 2y.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (d) \( \left(0,\ \frac{21}{5}\right) \)

Explanation:
1. x = (3(-4) + 2(6))/5 = 0 and y = (3(5) + 2(3))/5 = 21/5.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (c) (-24, 9)

Explanation:
1. x = (3(-4) - 2(6))/(3 - 2) = -24 and y = (3(5) - 2(3))/(3 - 2) = 9.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (c) 5 : 8 externally

Explanation:
1. Let the ratio be k : 1. -1 = (7k + 4)/(k + 1) gives k = -5/8. A negative ratio means external division, in the ratio 5 : 8.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (b) 1 : 2

Explanation:
1. Let the ratio be k : 1. The y-coordinate (6k - 3)/(k + 1) = 0 gives k = 1/2, so the ratio is 1 : 2.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (a) (4, -11)

Explanation:
1. D divides AB externally in the ratio AD : BD = 3 : 2. x = (3(2) - 2(1))/1 = 4 and y = (3(-3) - 2(1))/1 = -11.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (d) 3 : 4 internally

Explanation:
1. The point dividing in ratio k : 1 is ((2k + 1)/(k + 1), (7k + 3)/(k + 1)). Substituting: 3(2k + 1) + 7k + 3 - 9(k + 1) = 0 gives 4k - 3 = 0, so k = 3/4. The ratio is 3 : 4 internally.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (d) 20 square units

Explanation:
1. Area = (1/2)|-2(4 + 3) + 5(-3 - 1) + 2(1 - 4)| = (1/2)|-14 - 20 - 6| = 20.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (c) \( \left(\frac72,\ \frac{13}{2}\right)\ \text{or}\ \left(-\frac32,\ \frac32\right) \)

Explanation:
1. Area = (1/2)|2(-2 - y) + 3(y - 1) + x(1 + 2)| = (1/2)|3x + y - 7| = 5, so 3x + y = 17 or -3. With y = x + 3: x = 7/2, y = 13/2 or x = -3/2, y = 3/2.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (b) 80 square units

Explanation:
1. Split by the diagonal from (-3, 2) to (7, -6). Triangle with (5, 4) has area 42 and triangle with (-5, -4) has area 38, so the total is 80.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (a) collinear

Explanation:
1. All four points satisfy y = (b/a)x, so they lie on one straight line.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (a) 5

Explanation:
1. Slope between (5, 1) and (4, 2) is -1. Then (p - 1)/(1 - 5) = -1 gives p = 5.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (d) vertices of a rectangle which is not a square

Explanation:
1. The diagonals AC and BD have the same mid-point (3, 3) and the same length 10, so it is a rectangle. The sides are sqrt20 and sqrt80, which are unequal, so it is not a square.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (c) (-1, 0)

Explanation:
1. The diagonals bisect each other. The mid-point of the given diagonal is (-3/2, 1/2). The fourth vertex is the reflection of (-2, 1) in it: (2(-3/2) + 2, 2(1/2) - 1) = (-1, 0).

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (b) a^2

Explanation:
1. Area = (1/2)|a(c - (c - a)) + a((c - a) - (c + a)) + (-a)((c + a) - c)| = (1/2)|a^2 - 2a^2 - a^2| = a^2.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (b) an equilateral triangle

Explanation:
1. The first two points are 2 sqrt2 a apart. The distance from each to the third is sqrt((1 + sqrt3)^2 + (sqrt3 - 1)^2) a = sqrt8 a = 2 sqrt2 a. All sides are equal.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (a) collinear

Explanation:
1. Area = (1/2)|1(7 + 3) + (-2)(-3 - 1) + 3(1 - 7)| = (1/2)|10 + 8 - 18| = 0, so they are collinear.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (d) \( \frac{25}{2}\ \text{square units and }5\sqrt2 \)

Explanation:
1. The sides are 5 (to (-2, 1)), 5 (to (5, 2)) and sqrt50 between (-2, 1) and (5, 2). The legs are 5 and 5 with the right angle at (2, -2), area (1/2)(5)(5) = 25/2 and hypotenuse sqrt50 = 5 sqrt2.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (c) 5 units

Explanation:
1. The mid-point of BC is (3, 0). The distance from A(-1, 3) is sqrt(16 + 9) = 5.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (c) a square

Explanation:
1. AB = BC = CD = DA = sqrt20 and the diagonals AC = BD = sqrt40, so all sides are equal and the diagonals are equal: a square.

Teacher's Note:
1. Distance formula: PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2). Three points are collinear if the sum of two of the distances equals the third. A triangle is isosceles (two equal sides), equilateral (three equal sides) or right-angled (sum of squares of two sides equals the square of the third).

Answer: (b) (1, -3)

Explanation:
1. x3 = 3(1) - 3 + 1 = 1 and y3 = 3(3) - 7 - 5 = -3.

Teacher's Note:
1. Centroid of a triangle = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3); it divides each median in the ratio 2 : 1.

Answer: (a) -15

Explanation:
1. a = (K - 7)/2 and b = 9. So K - 7 + 27 = 5, giving K = -15.

Teacher's Note:
1. Section formula: the point dividing (x1, y1) and (x2, y2) internally in the ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). For external division use m - n in the denominators. The mid-point has ratio 1 : 1.

Answer: (d) \( \frac1a+\frac1b=1 \)

Explanation:
1. The area is zero: a(b - 1) + 0 + 1(0 - b) = ab - a - b = 0. Dividing by ab gives 1/a + 1/b = 1.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (d) (7, 2) or (1, 0)

Explanation:
1. PA = PB gives a = 3b + 1 for P = (a, b). The area is (1/2)|20b - 20| = 10, so b = 2 or 0. Hence P = (7, 2) or (1, 0).

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.

Answer: (c) \( \frac{11}{8}\ \text{or}\ -\frac38 \)

Explanation:
1. Area ABC = 49/2. Area DBC = (1/2)|x(5 + 2) - 3(-2 - 3x) + 4(3x - 5)| = (1/2)|28x - 14|. Setting this equal to half of 49/2, that is 49/4, gives |28x - 14| = 49/2, so 28x - 14 = 49/2 or -49/2, i.e. x = 11/8 or x = -3/8.

Teacher's Note:
1. Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)
2. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
3. . Three points are collinear exactly when this area is zero.