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Coordinate Geometry - Distance Formula

Grade 9ICSE

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

🔑Concepts

The Cartesian Plane: A two-dimensional coordinate system formed by the intersection of a horizontal line (xx-axis) and a vertical line (yy-axis). Visually, these axes meet at a point called the Origin (0,0)(0, 0) and divide the plane into four quadrants.

Coordinates of a Point: Every point is represented by an ordered pair (x,y)(x, y), where xx is the abscissa (distance from the yy-axis) and yy is the ordinate (distance from the xx-axis). Visually, to locate (3,4)(3, 4), you move 3 units right and 4 units up from the origin.

Geometric Interpretation of Distance: The distance between two points PP and QQ is the length of the straight line segment joining them. If you draw horizontal and vertical lines through PP and QQ, they form a right-angled triangle where the segment PQPQ is the hypotenuse.

The Distance Formula: Derived from the Pythagorean theorem, the distance dd between points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is calculated as the square root of the sum of the squares of the differences in their xx and yy coordinates.

Distance from the Origin: A specialized case of the distance formula where one point is (0,0)(0, 0). For any point P(x,y)P(x, y), the distance from the origin is simply the square root of the sum of the squares of its coordinates.

Collinearity of Points: Three points AA, BB, and CC are collinear (lying on the same straight line) if the sum of the lengths of any two segments equals the length of the third segment, such as AB+BC=ACAB + BC = AC.

Classification of Triangles: By calculating the distances between vertices, we can identify triangle types. For instance, an equilateral triangle has three equal side lengths, while an isosceles triangle has two.

Properties of Quadrilaterals: The distance formula helps verify shapes. A square is confirmed if all four sides are equal and both diagonals are equal in length. In a rectangle, opposite sides are equal and diagonals are equal.

📐Formulae

d=sqrt(x2x1)2+(y2y1)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).\n2. Apply the distance formula: d=sqrt(x2x1)2+(y2y1)2d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\n3. Substitute the values: d=sqrt(2(3))2+(102)2d = \\sqrt{(2 - (-3))^2 + (-10 - 2)^2}.\n4. Simplify inside the brackets: d=sqrt(5)2+(12)2d = \\sqrt{(5)^2 + (-12)^2}.\n5. Calculate the squares: d=sqrt25+144=sqrt169d = \\sqrt{25 + 144} = \\sqrt{169}.\n6. 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(2k)2+(13)2AB = \\sqrt{(2 - k)^2 + (-1 - 3)^2}.\n2. Given AB=5AB = 5, so 5=sqrt(2k)2+(4)25 = \\sqrt{(2 - k)^2 + (-4)^2}.\n3. Square both sides to remove the square root: 25=(2k)2+1625 = (2 - k)^2 + 16.\n4. Subtract 1616 from both sides: 9=(2k)29 = (2 - k)^2.\n5. Take the square root of both sides: pm3=2k\\pm 3 = 2 - k.\n6. Case 1: 3=2kRightarrowk=13 = 2 - k \\Rightarrow k = -1.\n7. Case 2: 3=2kRightarrowk=5-3 = 2 - k \\Rightarrow k = 5.\n8. 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.