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Coordinate Geometry - Distance Formula

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Distance Formula calculates the length of the line segment connecting two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) in a Cartesian plane, derived using the Pythagorean Theorem.

A right-angled triangle formed between points P and Q illustrating the horizontal and vertical differences used in the distance formula.
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Distance from the Origin: For any point A(x,y)A(x, y), its distance from the origin O(0,0)O(0, 0) is simplified as the square root of the sum of the squares of its coordinates.

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Collinearity: Three points AA, BB, and CC are collinear if the sum of the distances between two pairs of points equals the distance between the third pair, such as AB+BC=ACAB + BC = AC.

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Properties of Geometrical Figures: The distance formula is used to identify types of triangles (Equilateral, Isosceles, Right-angled) and quadrilaterals (Square, Rectangle, Rhombus) based on side lengths and diagonal equality.

📐Formulae

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

DistancefromOrigin(0,0)to(x,y)=sqrtx2+y2Distance\\ from\\ Origin\\ (0,0)\\ to\\ (x, y) = \\sqrt{x^2 + y^2}

AB+BC=AC(ConditionforcollinearityofA,B,C)AB + BC = AC\\ (Condition\\ for\\ collinearity\\ of\\ A, B, C)

💡Examples

Problem 1:

Find the distance between the points P(−3,2)P(-3, 2) and Q(2,−10)Q(2, -10).

Solution:

  1. Identify the coordinates: (x1,y1)=(−3,2)(x_1, y_1) = (-3, 2) and (x2,y2)=(2,−10)(x_2, y_2) = (2, -10).\
  2. Apply the distance formula: d=sqrt(x2−x1)2+(y2−y1)2d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\
  3. Substitute the values: d=sqrt(2−(−3))2+(−10−2)2d = \\sqrt{(2 - (-3))^2 + (-10 - 2)^2}.\
  4. Simplify inside the brackets: d=sqrt(5)2+(−12)2d = \\sqrt{(5)^2 + (-12)^2}.\
  5. Calculate the squares: d=sqrt25+144=sqrt169d = \\sqrt{25 + 144} = \\sqrt{169}.\
  6. Find the square root: d=13d = 13 units.

Explanation:

This problem uses the standard distance formula to find the length of the segment connecting two points across different quadrants.

Problem 2:

Find the value of kk if the distance between the points A(k,3)A(k, 3) and B(2,−1)B(2, -1) is 55 units.

Solution:

  1. Use the distance formula: AB=sqrt(2−k)2+(−1−3)2AB = \\sqrt{(2 - k)^2 + (-1 - 3)^2}.\
  2. Given AB=5AB = 5, so 5=sqrt(2−k)2+(−4)25 = \\sqrt{(2 - k)^2 + (-4)^2}.\
  3. Square both sides to remove the square root: 25=(2−k)2+1625 = (2 - k)^2 + 16.\
  4. Subtract 1616 from both sides: 9=(2−k)29 = (2 - k)^2.\
  5. Take the square root of both sides: pm3=2−k\\pm 3 = 2 - k.\
  6. Case 1: 3=2−kRightarrowk=−13 = 2 - k \\Rightarrow k = -1.\
  7. Case 2: −3=2−kRightarrowk=5-3 = 2 - k \\Rightarrow k = 5.\
  8. Therefore, k=−1k = -1 or k=5k = 5.

Explanation:

This example demonstrates how to solve for an unknown coordinate when the distance is already known by forming and solving a quadratic equation.

Problem 3:

Show that the points A(1,1)A(1, 1), B(5,1)B(5, 1), and C(3,4)C(3, 4) form an isosceles triangle.

Triangle ABC plotted on a coordinate plane with vertices (1,1), (5,1), and (3,4).

Solution:

  1. Find ABAB: AB=(5−1)2+(1−1)2=42+02=4AB = \sqrt{(5 - 1)^2 + (1 - 1)^2} = \sqrt{4^2 + 0^2} = 4 units.
  2. Find BCBC: BC=(3−5)2+(4−1)2=(−2)2+32=4+9=13BC = \sqrt{(3 - 5)^2 + (4 - 1)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} units.
  3. Find CACA: CA=(3−1)2+(4−1)2=22+32=4+9=13CA = \sqrt{(3 - 1)^2 + (4 - 1)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} units. Since BC=CA=13BC = CA = \sqrt{13}, triangle ABCABC is an isosceles triangle.

Explanation:

To prove a triangle is isosceles, calculate all three side lengths using the distance formula and check if at least two sides are equal.

Problem 4:

Find a point on the x-axis which is equidistant from A(2,−5)A(2, -5) and B(−2,9)B(-2, 9).

Coordinate plot showing point P on the x-axis connected to points A and B by equal length lines.

Solution:

Let the required point on the x-axis be P(x,0)P(x, 0). Given PA=PBPA = PB, then PA2=PB2PA^2 = PB^2. (x−2)2+(0−(−5))2=(x−(−2))2+(0−9)2(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2 (x−2)2+25=(x+2)2+81(x - 2)^2 + 25 = (x + 2)^2 + 81 x2−4x+4+25=x2+4x+4+81x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 −4x+29=4x+85-4x + 29 = 4x + 85 −8x=56-8x = 56 x=−7x = -7 The point is P(−7,0)P(-7, 0).

Explanation:

A point on the x-axis always has a y-coordinate of 00. We use the distance formula to set the distance from PP to AA equal to the distance from PP to BB.