Coordinate Geometry

Study Action Plan (Sorted by Difficulty)
Collinearity of Points
Accuracy: 61%
Total Attempted44
Total Correct27
Total Incorrect17
Difficulty Level:
Medium
Section Formula
Accuracy: 63%
Total Attempted209
Total Correct132
Total Incorrect77
Difficulty Level:
Medium
Parallelogram Diagonal Bisection
Accuracy: 67%
Total Attempted6
Total Correct4
Total Incorrect2
Difficulty Level:
Medium
Trisection of a Line Segment
Accuracy: 70%
Total Attempted91
Total Correct64
Total Incorrect27
Difficulty Level:
Medium
Equidistant Points
Accuracy: 81%
Total Attempted42
Total Correct34
Total Incorrect8
Difficulty Level:
Low
Midpoint Formula
Accuracy: 86%
Total Attempted99
Total Correct85
Total Incorrect14
Difficulty Level:
Low
Distance Formula
Accuracy: 89%
Total Attempted105
Total Correct93
Total Incorrect12
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Calculate the distance between two points using the distance formula: (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
  • Apply the section formula to find the coordinates of a point dividing a line segment internally in a given ratio: x=m1x2+m2x1m1+m2,  y=m1y2+m2y1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \; y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}.
  • Determine the midpoint of a line segment using the midpoint formula: (x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).
  • Assess the collinearity of three points by verifying if the sum of the distances between two pairs equals the distance between the third pair.
  • Find the points of trisection of a line segment using the section formula.
  • Establish a relation between xx and yy for a point (x,y)(x, y) to be equidistant from two given points, resulting in a linear equation.
  • Utilize the property that diagonals of a parallelogram bisect each other to find unknown coordinates.

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Revision Notes & Summary

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Chapter Notes

Distance Formula

  • Formula: The distance between two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) in a Cartesian plane is given by: (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • Example: For points A(4,0)A(4, 0) and B(6,0)B(6, 0), the distance AB=6−4=2AB = 6 - 4 = 2 units.

Section Formula

  • Formula: To find the coordinates of a point P(x,y)P(x, y) that divides a line segment joining two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) internally in the ratio m1:m2m_1:m_2: x=m1x2+m2x1m1+m2,y=m1y2+m2y1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}
  • Example: For a point dividing the segment joining (4,−3)(4, -3) and (8,5)(8, 5) in the ratio 3:13:1, the coordinates are (7,3)(7, 3).

Midpoint Formula

  • Formula: The midpoint of a line segment joining two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) is: (x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Collinearity of Points

  • Condition: Three points are collinear if the sum of the distances between two pairs of points equals the distance between the third pair.
  • Example: Points A(3,1)A(3, 1), B(6,4)B(6, 4), and C(8,6)C(8, 6) are collinear if AB+BC=ACAB + BC = AC.

Trisection of a Line Segment

  • Concept: To find points that divide a line segment into three equal parts, use the section formula for ratios 1:21:2 and 2:12:1.
  • Example: For segment joining A(2,−2)A(2, -2) and B(−7,4)B(-7, 4), trisection points are (−1,0)(-1, 0) and (−4,2)(-4, 2).

Equidistant Points

  • Condition: A point (x,y)(x, y) is equidistant from two given points if it satisfies the equation derived from setting the distances equal.
  • Example: For points (7,1)(7, 1) and (3,5)(3, 5), the relation is x−y=2x - y = 2.

Parallelogram Diagonal Bisection

  • Property: Diagonals of a parallelogram bisect each other.
  • Example: For vertices A(6,1)A(6, 1), B(8,2)B(8, 2), C(9,4)C(9, 4), and D(p,3)D(p, 3), the midpoint condition gives p=7p = 7.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

x−y=2x - y = 2

B.

x+y=10x + y = 10

C.

x−y=−2x - y = -2

D.

x+y=8x + y = 8
Correct Answer: A

Solution:

Since QQ is equidistant from AA and BB, AP=BPAP = BP. Solving (x−7)2+(y−1)2=(x−3)2+(y−5)2\sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2} gives x−y=2x - y = 2.

Chapter Concept:

Trisection of a Line Segment

A.

222\sqrt{2} units

B.

44 units

C.

55 units

D.

323\sqrt{2} units
Correct Answer: A

Solution:

Using the distance formula: PQ=(6−4)2+(8−6)2=4+4=22PQ = \sqrt{(6-4)^2 + (8-6)^2} = \sqrt{4 + 4} = 2\sqrt{2} units.

Chapter Concept:

Collinearity of Points

A.

x=m1x2+m2x1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} and y=m1y2+m2y1m1+m2y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}

B.

x=m1x1+m2x2m1+m2x = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} and y=m1y1+m2y2m1+m2y = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}

C.

x=x1+x22x = \frac{x_1 + x_2}{2} and y=y1+y22y = \frac{y_1 + y_2}{2}

D.

x=m2x2+m1x1m1+m2x = \frac{m_2x_2 + m_1x_1}{m_1 + m_2} and y=m2y2+m1y1m1+m2y = \frac{m_2y_2 + m_1y_1}{m_1 + m_2}
Correct Answer: A

Solution:

The section formula for internal division is given by x=m1x2+m2x1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} and y=m1y2+m2y1m1+m2y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}.

Chapter Concept:

Section Formula

A.

(0,5)(0, 5)

B.

(0,6)(0, 6)

C.

(0,9)(0, 9)

D.

(0,8)(0, 8)
Correct Answer: C

Solution:

The point on the y-axis equidistant from (6,5)(6, 5) and (−4,3)(-4, 3) is (0,9)(0, 9).

Chapter Concept:

Equidistant Points

A.

The points are collinear because PQ+QR=PRPQ + QR = PR.

B.

The points form a right triangle at QQ.

C.

The points form an equilateral triangle.

D.

The points are not collinear because PQ+QR≠PRPQ + QR \neq PR.
Correct Answer: A

Solution:

Calculate the distances: PQ=(−3−0)2+(4−0)2=5PQ = \sqrt{(-3-0)^2 + (4-0)^2} = 5, QR=(0−3)2+(0+4)2=5QR = \sqrt{(0-3)^2 + (0+4)^2} = 5, and PR=(−3−3)2+(4+4)2=10PR = \sqrt{(-3-3)^2 + (4+4)^2} = 10. Since PQ+QR=PRPQ + QR = PR, the points are collinear.

Chapter Concept:

Collinearity of Points

A.

(2,2)(2, 2)

B.

(4,0)(4, 0)

C.

(3,1)(3, 1)

D.

(2,−2)(2, -2)
Correct Answer: A

Solution:

The midpoint formula is (x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right). Thus, (6+(−2)2,−4+82)=(2,2)\left(\frac{6 + (-2)}{2}, \frac{-4 + 8}{2}\right) = (2, 2).

Chapter Concept:

Section Formula

A.

(x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

B.

(x2+x1)2+(y2+y1)2\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2}

C.

x2−x1+y2−y1\sqrt{x_2 - x_1 + y_2 - y_1}

D.

x2+x1+y2+y1\sqrt{x_2 + x_1 + y_2 + y_1}
Correct Answer: A

Solution:

The distance formula is derived from the Pythagorean theorem and is given by (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Chapter Concept:

Distance Formula

A.

0

B.

1

C.

2

D.

-1
Correct Answer: A

Solution:

Using the section formula x=2(−3)+3(3)2+3=0x = \frac{2(-3) + 3(3)}{2 + 3} = 0.

Chapter Concept:

Trisection of a Line Segment

A.

(-1, 5)

B.

(1, 5)

C.

(3, 4)

D.

(0, 4)
Correct Answer: A

Solution:

Using the midpoint formula, (5+x2,−3+y2)=(2,1)\left( \frac{5+x}{2}, \frac{-3+y}{2} \right) = (2, 1). Solving these equations, x=−1x = -1 and y=5y = 5. Thus, FF is (−1,5)(-1, 5).

Chapter Concept:

Midpoint Formula

A.

The midpoint divides the line segment into two equal parts.

B.

The midpoint is always closer to one endpoint than the other.

C.

The midpoint is the average of the coordinates of the endpoints.

D.

The midpoint is the geometric center of the line segment.
Correct Answer: A

Solution:

The midpoint divides the line segment into two equal parts, making option a the correct choice.

Chapter Concept:

Midpoint Formula

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Coordinate Geometry

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True or False

Correct Answer: True

Solution:

When two points lie on the x-axis, their y-coordinates are the same, so the distance is the absolute difference of their x-coordinates.

Chapter Concept :

Distance Formula

Correct Answer: True

Solution:

Collinearity of points can be verified by checking if the sum of the distances between two pairs of points equals the distance between the third pair.

Chapter Concept :

Collinearity of Points

Correct Answer: True

Solution:

Using the section formula, the coordinates of the points dividing the line segment in the ratio 1:2 and 2:1 are (−1,0)(-1, 0) and (−4,2)(-4, 2) respectively.

Chapter Concept :

Trisection of a Line Segment

Correct Answer: True

Solution:

For points A, B, and C to be collinear, the sum of distances AB and BC must equal the distance AC, indicating they lie on the same line.

Chapter Concept :

Collinearity of Points

Correct Answer: True

Solution:

The concept of collinearity involves checking if the sum of the distances between two pairs of points equals the distance between the third pair, confirming they lie on the same straight line.

Chapter Concept :

Collinearity of Points

Correct Answer: False

Solution:

A point that is equidistant from two given points lies on the perpendicular bisector of the segment joining the two points, not necessarily on a line perpendicular to the segment.

Chapter Concept :

Equidistant Points

Correct Answer: True

Solution:

The section formula is applicable for dividing a line segment into any ratio, including equal parts such as trisection.

Chapter Concept :

Trisection of a Line Segment

Correct Answer: True

Solution:

For points on the x-axis, the y-coordinates are zero, so the distance is simply the difference in their x-coordinates.

Chapter Concept :

Distance Formula

Correct Answer: False

Solution:

The midpoint formula is used for dividing a line segment into two equal parts. For trisection, the section formula is used with appropriate ratios.

Chapter Concept :

Trisection of a Line Segment

Correct Answer: True

Solution:

The distance formula for two points in a Cartesian plane is derived from the Pythagorean theorem and is given by (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Chapter Concept :

Distance Formula

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