CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 22

Answer: (d) a right-angled isosceles triangle

Explanation:
1. Sides: √50, √50 and 10. Since 50 + 50 = 100, it is right-angled, and two equal sides make it isosceles.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (d) parallelogram

Explanation:
1. A(3, -2), B(4, 0), C(6, -3), D(5, -5). AB = (1, 2) and DC = (1, 2), BC = (2, -3) and AD = (2, -3). Opposite sides are equal and parallel, and AB = √5 differs from BC = √13, so it is a parallelogram that is not a rhombus.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (c) an equilateral triangle

Explanation:
1. The first two points are 2a apart. The distance from (2a, 4a) to the third is √(3a² + a²) = 2a, and from (2a, 6a) to the third is √(3a² + a²) = 2a. All sides equal 2a.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (b) \( \left(\frac{1+\sqrt3}{2}, \frac{7-5\sqrt3}{2}\right) \)

Explanation:
1. Let the third vertex be (x, y). The midpoint of the given vertices is (1/2, 7/2) and the half side vector is (5/2, 1/2). The third vertex is the midpoint plus √3 times the perpendicular vector (1/2, -5/2) or minus it: (1/2 ± √3/2, 7/2 ∓ 5√3/2). The option (1 + √3)/2, (7 - 5√3)/2 is one of them; the other is ((1 - √3)/2, (7 + 5√3)/2).

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (a) \( \left(2-\frac{\sqrt{11}}{2}, \frac52\right) \)

Explanation:
1. The base joins (2, 0) and (2, 5) and has length 5, so the apex lies on the perpendicular bisector y = 5/2. With equal sides 3, the horizontal offset is √(9 - 25/4) = √11/2. The apex is (2 ± √11/2, 5/2).

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (a) \( 1 \)

Explanation:
1. 9 + (k - 2)² = k² + 9, so (k - 2)² = k², giving -4k + 4 = 0 and k = 1.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (d) \( (\frac{8}{7}, - \frac{9}{7}) \)

Explanation:
1. x = (3 x 4 + 4 x (-1))/7 = 8/7 and y = (3 x (-7) + 4 x 3)/7 = -9/7.

Teacher's Note:
1. The section formula for division in ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)).

Answer: (c) \( \left(\frac{13}{3},\frac{13}{3}\right), \left(\frac{23}{3},\frac{20}{3}\right) \)

Explanation:
1. The first point divides in the ratio 1 : 2: ((11 + 2)/3, (9 + 4)/3) = (13/3, 13/3). The second divides in 2 : 1: ((22 + 1)/3, (18 + 2)/3) = (23/3, 20/3).

Teacher's Note:
1. The section formula for division in ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)).

Answer: (b) \( \left(1,-\frac83\right), \left(-1,-\frac{10}{3}\right) \)

Explanation:
1. With A(3, -2) and B(-3, -4), the point dividing in 1 : 2 is ((-3 + 6)/3, (-4 - 4)/3) = (1, -8/3), and in 2 : 1 is ((-6 + 3)/3, (-8 - 2)/3) = (-1, -10/3).

Teacher's Note:
1. The section formula for division in ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)).

Answer: (b) \( (1, 2) \)

Explanation:
1. With A(-2, -1), B(1, 0), C(4, 3), D = A + C - B = (-2 + 4 - 1, -1 + 3 - 0) = (1, 2).

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (a) \( 5 \)

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

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (d) \( -15 \)

Explanation:
1. x = (3 + k)/2 and y = 11/2. Then 2x + 2y + 1 = (3 + k) + 11 + 1 = 0, so k = -15.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (c) 2 : 3 internally

Explanation:
1. Let the ratio be k : 1. The point is ((8k + 3)/(k + 1), (9k - 1)/(k + 1)). Substituting: (8k + 3) - (9k - 1) - 2(k + 1) = 0, i.e. -3k + 2 = 0, so k = 2/3, the ratio 2 : 3 internally.

Teacher's Note:
1. The section formula for division in ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)).

Answer: (c) \( -\frac25 \)

Explanation:
1. From the y-coordinate: (8k + 3)/(k + 1) = 6, giving k = 3/2. Then m = (2k - 4)/(k + 1) = (3 - 4)/(5/2) = -2/5.

Teacher's Note:
1. The section formula for division in ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)).

Answer: (b) \( \frac{29}{5} \)

Explanation:
1. From the x-coordinate: (-8k - 3)/(k + 1) = -6, so k = 3/2 (ratio 3 : 2). Then a = (9k + 1)/(k + 1) = (27/2 + 1)/(5/2) = 29/5.

Teacher's Note:
1. The section formula for division in ratio m : n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)).

Answer: (a) \( 80 \) sq. units

Explanation:
1. Split along the diagonal from (-3, 2) to (7, -6). Triangle (-3, 2), (5, 4), (7, -6): area = (1/2) x 84 = 42, since -3(4 + 6) + 5(-6 - 2) + 7(2 - 4) = -30 - 40 - 14 = -84. Triangle (-3, 2), (7, -6), (-5, -4): area = (1/2) x 76 = 38, since -3(-6 + 4) + 7(-4 - 2) + (-5)(2 + 6) = 6 - 42 - 40 = -76. Total area = 42 + 38 = 80 square units.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Answer: (d) \( \frac{11}{2} \) sq. units

Explanation:
1. Shoelace sum: (1 x 2 - 6 x 2) + (6 x 3 - 5 x 2) + (5 x 4 - 3 x 3) + (3 x 2 - 1 x 4) = -10 + 8 + 11 + 2 = 11. Area = 11/2 square units.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Answer: (d) \( 3 \)

Explanation:
1. Shoelace sum: (1 x 6 - (-5) x 2) + ((-5)(-4) - 7 x 6) + (7(-2) - k(-4)) + (k x 2 - 1(-2)) = 16 - 22 + (-14 + 4k) + (2k + 2) = -18 + 6k. This is zero for k = 3.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Answer: (c) \( (2, -4) \)

Explanation:
1. The centre lies on the perpendicular bisector of (2, -9) and (2, 1), which is y = -4. Equating distances to (5, -8) and (2, 1): (x - 5)² + 16 = (x - 2)² + 25, i.e. -10x + 41 = -4x + 29, so x = 2. The centre is (2, -4).

Teacher's Note:
1. The centre of a circle through three points is equidistant from all of them.

Answer: (b) \( (3, -2) \)

Explanation:
1. The perpendicular bisector of (3, -7) and (3, 3) is y = -2. Equating distances from (6, -6) and (3, -7): (x - 6)² + 16 = (x - 3)² + 25 gives -12x + 52 = -6x + 34, so x = 3. The centre is (3, -2).

Teacher's Note:
1. The centre of a circle through three points is equidistant from all of them.

Answer: (a) \( 5 \)

Explanation:
1. PQ² = 25 + 16 = 41. QR² = (x - 1)² + 25. So (x - 1)² = 16 and x = 5 or -3. Among the options only 5 is possible.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (a) \( (1, 0) \) and \( (1, 4) \)

Explanation:
1. The diagonal has length 4 and its midpoint is (1, 2). The other diagonal is perpendicular, of length 4, through (1, 2): its end points are (1, 0) and (1, 4).

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Answer: (d) \( \left(\frac72,\frac{13}{2}\right) \)

Explanation:
1. Let the third vertex be (x, x + 3). Area = (1/2)[2(-2 - x - 3) + 3(x + 3 - 1) + x(1 + 2)] = (1/2)(-10 - 2x + 3x + 6 + 3x) = (1/2)(4x - 4) = 2x - 2 in absolute value. Setting |2x - 2| = 5 gives x = 7/2 or -3/2. For x = 7/2, y = 13/2.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Answer: (c) \( \frac{11}{8} \)

Explanation:
1. Area of ABC = 49/2. Area of DBC = (1/2) x absolute value of [x(5 + 2) + (-3)(-2 - 3x) + 4(3x - 5)] = (1/2) x absolute value of (28x - 14) = absolute value of (14x - 7). Setting this equal to 49/4 gives 14x - 7 = 49/4 or -49/4, so x = 11/8 or x = -3/8. Of the options, 11/8 is possible.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Answer: (b) \( 5 \)

Explanation:
1. Slope of (1, -1) to (11, 4) = 5/10 = 1/2. Slope of (1, -1) to (a, 1) = 2/(a - 1). Setting 2/(a - 1) = 1/2 gives a = 5.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Answer: (b) \( \sqrt{85} \)

Explanation:
1. The midpoint of AB is (-1, -1). The distance from C(5, -8) is √(36 + 49) = √85.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.