krit.club logo

Coordinate Geometry - Distance Formula

Grade 10CBSE

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

🔑Concepts

•

The Cartesian Plane is defined by two perpendicular lines: the horizontal xx-axis and the vertical yy-axis. Their intersection point is the Origin O(0,0)O(0, 0). Any point P(x,y)P(x, y) is located xx units from the yy-axis (abscissa) and yy units from the xx-axis (ordinate).

•

The Distance Formula measures the straight-line length between any two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2). This can be visualized as the hypotenuse of a right-angled triangle where the horizontal base is ∣x2−x1∣|x_2 - x_1| and the vertical height is ∣y2−y1∣|y_2 - y_1|.

•

The distance of a point P(x,y)P(x, y) from the Origin (0,0)(0, 0) is a simplified case of the distance formula, represented visually as the length of the line segment connecting the center of the coordinate system to the point PP.

•

Three points AA, BB, and CC are said to be collinear if they lie on the same straight line. This is verified using the distance formula by checking if the sum of the lengths of any two segments equals the length of the third segment (e.g., AB+BC=ACAB + BC = AC).

•

Geometric shapes can be identified using distances: For a triangle, we check if all sides are equal (Equilateral), two sides are equal (Isosceles), or if the squares of the sides satisfy the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2 (Right-angled).

•

To distinguish between specific quadrilaterals, we calculate the lengths of the four sides and the two diagonals. For example, a Square has four equal sides and two equal diagonals, whereas a Rhombus has four equal sides but unequal diagonals.

•

The distance between two points is always non-negative. Even if the coordinates are negative, the squared differences (x2−x1)2(x_2 - x_1)^2 and (y2−y1)2(y_2 - y_1)^2 will always result in positive values under the square root.

📐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.