CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 26

Answer: (a) (0, 1), (8, 3) and (6, 7)

Explanation:
1. Areas: (0, 1), (8, 3), (6, 7) give a non-zero value \( (\frac{1}{2}|0 + 8\cdot 6 + 6\cdot (-2)| = 18) \).
2. The other three sets all give zero area, so they are collinear.

Teacher's Note:
1. Area of a triangle = \( \frac{1}{2}|x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| \).

Answer: (c) −1

Explanation:
1. Area = 0: \( -5(p + 2) + 1(-2 - 1) + 4(1 - p) = 0 \), so \( -9p - 9 = 0 \) and \( p = -1 \).

Teacher's Note:
1. Area of a triangle = \( \frac{1}{2}|x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| \).

Answer: (a) 7

Explanation:
1. Diagonals AC and BD bisect each other: \( (6 + 9)/2 = (8 + p)/2 \), so \( p = 7 \).

Teacher's Note:
1. Equate midpoints of the two diagonals.

Answer: (b) 1

Explanation:
1. Area = \( \frac{1}{2}|a(1) + (a + 1)(0) + (a + 2)(-1)| = \frac{1}{2}|a - a - 2| = 1 \).

Teacher's Note:
1. The result does not depend on a, since shifting the triangle does not change its area.

Answer: (c) 29

Explanation:
1. Area = \( \frac{1}{2}|5(-3 + 5) + (-9)(-5 - 2) + (-3)(2 + 3)| = \frac{1}{2}|10 + 63 - 15| = 29 \).

Teacher's Note:
1. Area of a triangle = \( \frac{1}{2}|x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| \).

Answer: (b) 96

Explanation:
1. Shoelace sum: \( (-1)(-9) - (-3)(6) = 27 \); \( (-3)(-8) - 5(-9) = 69 \); \( 5(9) - 3(-8) = 69 \); \( 3(6) - (-1)(9) = 27 \).
2. Total = 192, so the area = \( 192/2 = 96 \).

Teacher's Note:
1. Use the shoelace method taking the vertices in order.

Answer: (b) 1/2 and −1

Explanation:
1. Setting the area determinant to zero gives \( 2k^{2} + k - 1 = 0 \), i.e. \( (2k - 1)(k + 1) = 0 \).
2. So \( k = 1/2 \) or \( k = -1 \).

Teacher's Note:
1. Collinear points give a zero determinant.

Answer: (a) (7/2, 13/2) or (−3/2, 3/2)

Explanation:
1. Let the third vertex be \( (x, x + 3) \). Area = \( \frac{1}{2}|2(-2 - x - 3) + 3(x + 3 - 1) + x(1 + 2)| = \frac{1}{2}|4x - 4| = 2|x - 1| \).
2. \( 2|x - 1| = 5 \) gives \( x = 7/2 \) or \( x = -3/2 \), so the vertex is \( (7/2, 13/2) \) or \( (-3/2, 3/2) \).

Teacher's Note:
1. Area of a triangle = \( \frac{1}{2}|x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| \).

Answer: (a) collinear

Explanation:
1. The first three lie on the line \( y = (b/a)x \), and (\( a^{2} \), ab) satisfies it too, since ab = \( (b/a)a^{2} \).
2. So all four points are collinear.

Teacher's Note:
1. Check whether each point satisfies the same line.

Answer: (a) a²

Explanation:
1. The base is the vertical side of length a between \( (a, c + a) \) and \( (a, c) \).
2. The third vertex is at horizontal distance \( 2a \) from it, so the area = \( \frac{1}{2} \times a \times 2a = a^{2} \).

Teacher's Note:
1. Use base × height.

Answer: (c) 11/8 or −3/8

Explanation:
1. Area ABC = \( \frac{1}{2}|6(7) + (-3)(-5) + 4(-2)| = 49/2 \).
2. Area DBC = \( \frac{1}{2}|x(7) + (-3)(-2 - 3x) + 4(3x - 5)| = \frac{1}{2}|7x + 6 + 9x + 12x - 20| = \frac{1}{2}|28x - 14| \).
3. \( |28x - 14| = 49/2 \), so \( 28x - 14 = \pm 49/2 \), giving \( x = 11/8 \) or −3/8.

Teacher's Note:
1. Area of a triangle = \( \frac{1}{2}|x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| \).

Answer: (d) 6⅔

Explanation:
1. The slope between the first two is \( (1 - 7)/(6 - 2) = -3/2 \).
2. From (6, 1) to \( (x, 0) \): \( (0 - 1)/(x - 6) = -3/2 \), so \( x - 6 = 2/3 \) and \( x \) = 6⅔.

Teacher's Note:
1. Equal slopes for collinear points.

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

Explanation:
1. PA = PB puts \( P \) on the perpendicular bisector of AB: \( (x - 3)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y + 2)^{2} \) simplifies to \( x - 3y = 1 \).
2. Area of PAB = \( \frac{1}{2}|6x + 2y - 26| = 10 \) gives \( |3x + y - 13| = 10 \). With \( x = 3y + 1 \): \( 10y - 10 = \pm 10 \), so \( y = 2 (x = 7) \) or \( y = 0 (x = 1) \). So \( P = (7, 2) \) or (1, 0).

Teacher's Note:
1. Combine the equal-distance condition with the area condition.

Answer: (c) 7

Explanation:
1. \( 4k - 27 + 4 - 3 - 2 = 0 \), so \( 4k = 28 \) and \( k = 7 \).

Teacher's Note:
1. Substitute the point.

Answer: (a) 9x² − 7y² + 63 = 0

Explanation:
1. The locus is a hyperbola with foci (0, ±4) and \( 2a = 6 \), so \( a = 3 \) and \( b^{2} = 16 - 9 = 7 \).
2. \( y^{2}/9 - x^{2}/7 = 1 \), i.e. \( 7y^{2} - 9x^{2} = 63 \), which is \( 9x^{2} - 7y^{2} + 63 = 0 \).

Teacher's Note:
1. The difference of distances from two fixed points is constant for a hyperbola.

Answer: (b) 6x + 4y = 5

Explanation:
1. \( (x - 1)^{2} + (y - 3)^{2} = (x + 2)^{2} + (y - 1)^{2} \), which simplifies to \( -6x - 4y + 5 = 0 \).
2. So \( 6x + 4y = 5 \).

Teacher's Note:
1. The locus is the perpendicular bisector.

Answer: (c) 9x² + 5y² = 45

Explanation:
1. An ellipse with foci (0, ±2) and \( 2a = 6 \): \( a = 3, c = 2, b^{2} = 5 \).
2. \( x^{2}/5 + y^{2}/9 = 1 \), i.e. \( 9x^{2} + 5y^{2} = 45 \).

Teacher's Note:
1. The sum of distances to two fixed points is constant for an ellipse.

Answer: (d) 6x + y + 7 = 0

Explanation:
1. \( y = -6x - 7 \), so \( 6x + y + 7 = 0 \).

Teacher's Note:
1. Use \( y \) = mx \( + c \) with \( c = -7 \).

Answer: (a) √3x + y = 2 + 3√3

Explanation:
1. Slope = \( \tan 120^{\circ} = -\sqrt{3} \).
2. \( y - 2 = -\sqrt{3}(x - 3) \), so \( \sqrt{3x} + y = 2 + 3\sqrt{3} \).

Teacher's Note:
1. Point-slope form.

Answer: (b) 3x − 4y = 12

Explanation:
1. \( x/4 + y/(-3) = 1 \), so \( 3x - 4y = 12 \).

Teacher's Note:
1. Intercept form \( x/a + y/b = 1 \).

Answer: (b) −1/2

Explanation:
1. Slopes: \( -2/(3a) \) and −3/4. Their product must be −1: \( (2/(3a))(3/4) = -1 \), so \( 1/(2a) = -1 \) and \( a = -1/2 \).

Teacher's Note:
1. Perpendicular lines have slopes whose product is −1.

Answer: (c) y − √3x − 14 = 0

Explanation:
1. The line makes \( 90^{\circ} + 150^{\circ} = 240^{\circ} \) with the x-axis (the same as 60°), so its slope is √3.
2. With distance 7 from the origin, \( y = \sqrt{3x} + c \) where \( |c|/2 = 7 \), so \( c = \pm 14 \). Of the options, \( y - \sqrt{3x} - 14 = 0 \) fits.

Teacher's Note:
1. Use the slope and the distance formula \( |c|/\sqrt{1 + m^{2}} \).

Answer: (c) 2x − 3y + 12 = 0

Explanation:
1. The line is \( 2x - 3y = k \). Putting \( x = 0, y = 4 \): \( k = -12 \), so \( 2x - 3y + 12 = 0 \).

Teacher's Note:
1. Parallel lines share the \( x \) and \( y \) coefficients.

Answer: (c) 1

Explanation:
1. The other diagonal passes through the midpoint (3, 2) of the given diagonal: \( 2 - 3 + \lambda = 0 \), so \( \lambda = 1 \).

Teacher's Note:
1. The diagonals of a rectangle bisect each other.

Answer: (d) 2x + y − 4 = 0

Explanation:
1. The intercepts are 2 and 4, so \( x/2 + y/4 = 1 \), i.e. \( 2x + y = 4 \).

Teacher's Note:
1. The midpoint of \( (a, 0) \) and \( (0, b) \) is \( (a/2, b/2) \).

Answer: (c) 3x − 4y + 12 = 0 and 4x + 3y − 6 = 0; (x − x₁)/2 = (y − y₁)/3 and (x − x₂)/3 = (y − y₂)/(−2); (x − x₁)/5 = (y − y₁)/(−4) and (x − x₂)/4 = (y − y₂)/5

Explanation:
1. Pair 1: slopes 3/4 and −4/3, product −1, so perpendicular.
2. Pair 2: direction ratios (2, 3) and (3, −2): \( 2\cdot 3 + 3\cdot (-2) = 0 \), so perpendicular.
3. Pair 3: direction ratios (5, −4) and (4, 5): \( 20 - 20 = 0 \), so perpendicular.
4. The pair \( 4x + 5y - 8 = 0 \) and \( 3x - 2y + 4 = 0 \) has slopes −4/5 and 3/2, product −6/5, so it is not perpendicular. Hence the first three pairs are perpendicular.

Teacher's Note:
1. For direction ratios \( (a, b) \) and \( (c, d) \), perpendicular means ac + bd = 0.

Answer: (b) x + y = 5

Explanation:
1. The midpoint is (3, 2) and the slope of the segment is 1, so the perpendicular has slope −1.
2. \( y - 2 = -(x - 3) \), so \( x + y = 5 \).

Teacher's Note:
1. Use the midpoint and the negative reciprocal slope.