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Coordinate Geometry - Apply distance formula to compute and compare distances between points

Grade 10CBSE

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

🔑Concepts

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The Distance Formula is derived from the Pythagoras Theorem. For any two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) in the Cartesian plane, the horizontal distance is ∣x2−x1∣|x_2 - x_1| and the vertical distance is ∣y2−y1∣|y_2 - y_1|. The distance PQPQ is the hypotenuse of the right-angled triangle formed by these segments.

A right-angled triangle on a coordinate plane showing the distance between two points as the hypotenuse.
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To verify if three points A,B,A, B, and CC form an equilateral triangle, use the distance formula to compute AB,BC,AB, BC, and CACA. If AB=BC=CAAB = BC = CA, the triangle is equilateral.

An equilateral triangle showing three equal sides.
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A point P(x,y)P(x, y) is equidistant from two points AA and BB if PA=PBPA = PB. This condition is often used to find a missing coordinate or the relation between xx and yy for points on the perpendicular bisector of ABAB.

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To check if a quadrilateral with given vertices is a square, show that all four sides are equal (AB=BC=CD=DAAB = BC = CD = DA) AND both diagonals are equal (AC=BDAC = BD).

📐Formulae

Distance between two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2): d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance of a point P(x,y)P(x, y) from the Origin O(0,0)O(0, 0): d=x2+y2d = \sqrt{x^2 + y^2}

Condition for Collinearity of points A,B,CA, B, C: AB+BC=ACAB + BC = AC (or any other combination of segments totaling the third)

💡Examples

Problem 1:

Find the distance between the points A(3,−2)A(3, -2) and B(−1,1)B(-1, 1).

Solution:

  1. Identify the coordinates: (x1,y1)=(3,−2)(x_1, y_1) = (3, -2) and (x2,y2)=(−1,1)(x_2, y_2) = (-1, 1).
  2. Substitute values into the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
  3. Calculate the differences: x2−x1=−1−3=−4x_2 - x_1 = -1 - 3 = -4 and y2−y1=1−(−2)=3y_2 - y_1 = 1 - (-2) = 3.
  4. Square the differences: (−4)2=16(-4)^2 = 16 and (3)2=9(3)^2 = 9.
  5. Add the squares: 16+9=2516 + 9 = 25.
  6. Take the square root: d=25=5d = \sqrt{25} = 5 units.

Explanation:

We apply the distance formula directly by calculating the horizontal and vertical displacements between the two points and then using the Pythagorean approach to find the total distance.

Problem 2:

Determine if the points P(1,5)P(1, 5), Q(2,3)Q(2, 3), and R(−2,−11)R(-2, -11) are collinear.

Solution:

  1. Calculate PQPQ: (2−1)2+(3−5)2=12+(−2)2=1+4=5≈2.23\sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.23.
  2. Calculate QRQR: (−2−2)2+(−11−3)2=(−4)2+(−14)2=16+196=212=253≈14.56\sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53} \approx 14.56.
  3. Calculate PRPR: (−2−1)2+(−11−5)2=(−3)2+(−16)2=9+256=265≈16.28\sqrt{(-2-1)^2 + (-11-5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265} \approx 16.28.
  4. Check if the sum of two distances equals the third: PQ+QR=5+253≠265PQ + QR = \sqrt{5} + 2\sqrt{53} \neq \sqrt{265}.
  5. Since PQ+QR≠PRPQ + QR \neq PR, the points are not collinear.

Explanation:

To check for collinearity, we find the lengths of all possible segments between the three points. If the sum of the two shorter segments equals the longest segment, the points lie on a single line.

Problem 3:

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

Graph showing point P on the x-axis equidistant from points A and B.

Solution:

Let the point on the xx-axis be P(x,0)P(x, 0). Since PP is equidistant from AA and BB, PA=PBPA = PB. (x−2)2+(0−(−5))2=(x−(−2))2+(0−9)2\sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - (-2))^2 + (0 - 9)^2} Squaring both sides: (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 (−7,0)(-7, 0).

Explanation:

Any point on the xx-axis has a yy-coordinate of 0. We equate the distances PAPA and PBPB using the distance formula and solve for xx.

Problem 4:

Check whether the points A(5,−2)A(5, -2), B(6,4)B(6, 4) and C(7,−2)C(7, -2) are the vertices of an isosceles triangle.

An isosceles triangle plotted on a coordinate plane with vertices A, B, and C.

Solution:

We calculate the lengths of the three sides: AB=(6−5)2+(4−(−2))2=12+62=1+36=37AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} BC=(7−6)2+(−2−4)2=12+(−6)2=1+36=37BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} AC=(7−5)2+(−2−(−2))2=22+02=4=2AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 Since AB=BC=37AB = BC = \sqrt{37}, the triangle has two equal sides.

Explanation:

A triangle is isosceles if at least two of its sides are of equal length. By calculating all three side lengths, we find AB=BCAB = BC.