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Orienting Yourself: The Use of Coordinates - Distance Between Two Points in the 2-D Plane

Grade 9CBSE

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

🔑Concepts

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The distance between two points in a 2-D plane is the length of the straight line segment connecting them. This is calculated using the coordinates (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) of the two points.

A coordinate plane showing a line segment between points (x1, y1) and (x2, y2) forming a right-angled triangle to demonstrate the distance formula.
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The Distance Formula is derived from the Pythagoras Theorem. In a right-angled triangle, the distance dd (hypotenuse) is related to the horizontal difference (x2−x1)(x_2 - x_1) and vertical difference (y2−y1)(y_2 - y_1) as: d2=(x2−x1)2+(y2−y1)2d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.

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When calculating distance, the order of points does not matter because squaring the differences (x2−x1)2(x_2 - x_1)^2 and (y2−y1)2(y_2 - y_1)^2 always results in a non-negative value.

Points A and B in different quadrants showing that distance is a scalar magnitude regardless of direction.
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The distance of any point P(x,y)P(x, y) from the Origin O(0,0)O(0, 0) is a special case of the formula, simplified to x2+y2\sqrt{x^2 + y^2}.

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Distance is always a non-negative quantity. Even if coordinates are negative, the square root of the sum of squares will yield a positive value (or zero if the points coincide).

Distance from origin to point (-4, 3) illustrating that distance is positive even with negative coordinates.

📐Formulae

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

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

d=x2+y2 (Distance of point (x,y) from the origin (0,0))d = \sqrt{x^2 + y^2} \text{ (Distance of point } (x, y) \text{ from the origin } (0, 0)\text{)}

💡Examples

Problem 1:

Find the distance between the points A(3,2)A(3, 2) and B(7,5)B(7, 5).

Solution:

Let (x1,y1)=(3,2)(x_1, y_1) = (3, 2) and (x2,y2)=(7,5)(x_2, y_2) = (7, 5). Using the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d=(7−3)2+(5−2)2d = \sqrt{(7 - 3)^2 + (5 - 2)^2} d=(4)2+(3)2d = \sqrt{(4)^2 + (3)^2} d=16+9d = \sqrt{16 + 9} d=25d = \sqrt{25} d=5 unitsd = 5 \text{ units}

Explanation:

We substitute the coordinates of AA and BB into the distance formula. The horizontal difference is 44 and the vertical difference is 33. Calculating the square root of the sum of their squares gives the distance.

Problem 2:

Calculate the distance of the point P(−6,8)P(-6, 8) from the origin.

Solution:

The coordinates of the origin are (0,0)(0, 0). Using the distance from origin formula: d=x2+y2d = \sqrt{x^2 + y^2} d=(−6)2+(8)2d = \sqrt{(-6)^2 + (8)^2} d=36+64d = \sqrt{36 + 64} d=100d = \sqrt{100} d=10 unitsd = 10 \text{ units}

Explanation:

When calculating distance from the origin, x1x_1 and y1y_1 are both zero, simplifying the formula to the square root of the sum of the squares of the point's coordinates.

Problem 3:

Find the distance between M(−1,−2)M(-1, -2) and N(−4,2)N(-4, 2).

Solution:

Let (x1,y1)=(−1,−2)(x_1, y_1) = (-1, -2) and (x2,y2)=(−4,2)(x_2, y_2) = (-4, 2). d=(−4−(−1))2+(2−(−2))2d = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} d=(−4+1)2+(2+2)2d = \sqrt{(-4 + 1)^2 + (2 + 2)^2} d=(−3)2+(4)2d = \sqrt{(-3)^2 + (4)^2} d=9+16d = \sqrt{9 + 16} d=25d = \sqrt{25} d=5 unitsd = 5 \text{ units}

Explanation:

Care must be taken with negative signs. Subtracting a negative number is the same as adding its absolute value. (−3)2(-3)^2 becomes positive 99 because any real number squared is non-negative.

Problem 4:

Find the distance between the points A(2,−3)A(2, -3) and B(−4,5)B(-4, 5).

Distance between points A and B across different quadrants.

Solution:

  1. Identify the coordinates: (x1,y1)=(2,−3)(x_1, y_1) = (2, -3) and (x2,y2)=(−4,5)(x_2, y_2) = (-4, 5).
  2. Apply the distance formula: d=(−4−2)2+(5−(−3))2d = \sqrt{(-4 - 2)^2 + (5 - (-3))^2}
  3. Simplify the terms: d=(−6)2+(8)2d = \sqrt{(-6)^2 + (8)^2} d=36+64d = \sqrt{36 + 64}
  4. Calculate the final value: d=100=10 unitsd = \sqrt{100} = 10 \text{ units}

Explanation:

We subtract the coordinates to find the horizontal and vertical displacements, then use the square root of the sum of their squares.

Problem 5:

Show that the point P(3,4)P(3, 4) is at a distance of 55 units from the origin.

A line segment of length 5 connecting the origin to point (3,4).

Solution:

  1. The coordinates of the point are (x,y)=(3,4)(x, y) = (3, 4).
  2. Use the distance from origin formula: d=x2+y2d = \sqrt{x^2 + y^2}
  3. Substitute the values: d=32+42d = \sqrt{3^2 + 4^2} d=9+16d = \sqrt{9 + 16}
  4. Final result: d=25=5 unitsd = \sqrt{25} = 5 \text{ units}

Explanation:

The distance of a point from the origin is simply the square root of the sum of the squares of its coordinates.