CBSE Class 10 Mathematics Coordinate Geometry Notes Set 07

Introduction and Distance Formula

Rectangular Coordinates

Take two perpendicular lines X'OX and Y'OY intersecting at the point O. X'OX and Y'OY are called the co-ordinate axes. X'OX is called the X-axis, Y'OY is called the Y-axis and O is called the origin. Lines X'OX and Y'OY are sometimes also called rectangular axes.

Co-ordinates of a Point

Let P be any point as shown in the figure. Draw PL and PM perpendicular to the Y-axis and X-axis, respectively. The length LP (or OM) is called the x-coordinate or the abscissa of point P and MP is called the y-coordinate or the ordinate of point P. A point whose abscissa is x and ordinate is y is named as the point (x, y) or P(x, y).

The two lines X'OX and Y'OY divide the plane into four parts called quadrants. XOY, YOX', X'OY' and Y'OX are, respectively, called the first, second, third and fourth quadrants. The following table shows the signs of the coordinates of points situated in different quadrants.

Note.
1. Abscissa is the perpendicular distance of a point from the y-axis (positive to the right of the y-axis and negative to the left of the y-axis).
2. Ordinate is positive above the x-axis and negative below the x-axis.
3. Abscissa of any point on the y-axis is zero.
4. Ordinate of any point on the x-axis is zero.
5. Co-ordinates of the origin are (0, 0).

Distance Formula

Let two points be P(x₁, y₁) and Q(x₂, y₂). Take two mutually perpendicular lines as the coordinate axes with O as origin. Mark the points P(x₁, y₁) and Q(x₂, y₂). Draw lines PA and QB perpendicular to the X-axis, which meet the X-axis in points A and B, respectively. Draw lines PC and QD perpendicular to the Y-axis, which meet the Y-axis in C and D, respectively. Produce CP to meet BQ in R.

Now OA = abscissa of P = x₁. Similarly, OB = x₂, OC = y₁ and OD = y₂. Therefore, PR = AB = OB − OA = x₂ − x₁. Similarly, QR = QB − RB = QB − PA = y₂ − y₁. Now, using the Pythagoras theorem in the right-angled triangle PRQ, we have

\[ PQ^2 = PR^2 + RQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]

Since the distance or length of the line segment PQ is always non-negative, on taking the positive square root, we get the distance as

\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This result is known as the distance formula.

Corollary. The distance of a point P(x₁, y₁) from the origin (0, 0) is given by \( OP = \sqrt{x_1^2 + y_1^2} \).

Some Useful Points

1. In questions relating to geometrical figures, take the given vertices in the given order and proceed as indicated.
(i) For an isosceles triangle, we have to prove that at least two sides are equal.
(ii) For an equilateral triangle, we have to prove that three sides are equal.
(iii) For a right-angled triangle, we have to prove that the sum of the squares of two sides is equal to the square of the third side.
(iv) For a square, we have to prove that the four sides are equal and the two diagonals are equal.
(v) For a rhombus, we have to prove that the four sides are equal (and there is no need to establish that the two diagonals are unequal as the square is also a rhombus).
(vi) For a rectangle, we have to prove that the opposite sides are equal and the two diagonals are equal.
(vii) For a parallelogram, we have to prove that the opposite sides are equal (and there is no need to establish that the two diagonals are unequal as the rectangle is also a parallelogram).
2. For three points to be collinear, we have to prove that the sum of the distances between two pairs of points is equal to the third pair of points.

Solved Examples

Example 1. Find the distance between the points (8, −2) and (3, −6).
Answer:
Let P(8, −2) and Q(3, −6). By the distance formula,
\( PQ = \sqrt{(3 - 8)^2 + (-6 + 2)^2} = \sqrt{25 + 16} = \sqrt{41} \) unit.

Example 2. Prove that the points \( (1, -1) \), \( \left(-\dfrac{1}{2}, \dfrac{1}{2}\right) \) and \( (1, 2) \) are the vertices of an isosceles triangle.
Answer:
Let P(1, −1), \( Q\left(-\tfrac{1}{2}, \tfrac{1}{2}\right) \) and R(1, 2).
\( PQ = \sqrt{\left(-\tfrac{1}{2} - 1\right)^2 + \left(\tfrac{1}{2} + 1\right)^2} = \sqrt{\tfrac{9}{4} + \tfrac{9}{4}} = \sqrt{\tfrac{18}{4}} = \tfrac{3}{2}\sqrt{2} \)
\( QR = \sqrt{\left(1 + \tfrac{1}{2}\right)^2 + \left(2 - \tfrac{1}{2}\right)^2} = \sqrt{\tfrac{9}{4} + \tfrac{9}{4}} = \tfrac{3}{2}\sqrt{2} \)
\( PR = \sqrt{(1 - 1)^2 + (2 + 1)^2} = 3 \)
Since PQ = QR, the triangle is isosceles.

Example 3. Using the distance formula, show that the points (−3, 2), (1, −2) and (9, −10) are collinear.
Answer:
Let A(−3, 2), B(1, −2) and C(9, −10). The points are collinear if the sum of the lengths of two line segments is equal to the third.
\( AB = \sqrt{(1 + 3)^2 + (-2 - 2)^2} = \sqrt{32} = 4\sqrt{2} \)
\( BC = \sqrt{(9 - 1)^2 + (-10 + 2)^2} = \sqrt{128} = 8\sqrt{2} \)
\( AC = \sqrt{(9 + 3)^2 + (-10 - 2)^2} = \sqrt{288} = 12\sqrt{2} \)
Since \( AB + BC = 4\sqrt{2} + 8\sqrt{2} = 12\sqrt{2} = AC \), the points A, B and C are collinear.

Example 4. Find a point on the X-axis which is equidistant from the points (5, 4) and (−2, 3).
Answer:
Let the required point be P(x, 0), since its ordinate is zero. Let A(5, 4) and B(−2, 3). Since AP = BP, we have \( AP^2 = BP^2 \):
\( (x - 5)^2 + (0 - 4)^2 = (x + 2)^2 + (0 - 3)^2 \)
\( x^2 - 10x + 25 + 16 = x^2 + 4x + 4 + 9 \Rightarrow -14x = -28 \Rightarrow x = 2 \)
The required point is (2, 0).

Example 5. The vertices of a triangle are (−2, 0), (2, 3) and (1, −3). Is the triangle equilateral, isosceles or scalene?
Answer:
Let A(−2, 0), B(2, 3) and C(1, −3).
\( AB = \sqrt{(2 + 2)^2 + (3 - 0)^2} = 5 \), \( BC = \sqrt{(1 - 2)^2 + (-3 - 3)^2} = \sqrt{37} \), \( AC = \sqrt{(1 + 2)^2 + (-3 - 0)^2} = 3\sqrt{2} \).
Clearly AB ≠ BC ≠ AC. Therefore ABC is a scalene triangle.

Example 6. The length of a line segment is 10. If one end is at (2, −3) and the abscissa of the second end is 10, show that its ordinate is either 3 or −9.
Answer:
Let A(2, −3) and let the second end be B(10, y). Then \( AB = \sqrt{(10 - 2)^2 + (y + 3)^2} = 10 \).
\( 64 + y^2 + 6y + 9 = 100 \Rightarrow y^2 + 6y - 27 = 0 \Rightarrow (y + 9)(y - 3) = 0 \)
Therefore y = −9 or y = 3.

Example 7. Show that the points (−2, 5), (3, −4) and (7, 10) are the vertices of a right triangle.
Answer:
Let A(−2, 5), B(3, −4) and C(7, 10). Then
\( AB^2 = (3 + 2)^2 + (-4 - 5)^2 = 106 \), \( BC^2 = (7 - 3)^2 + (10 + 4)^2 = 212 \), \( AC^2 = (7 + 2)^2 + (10 - 5)^2 = 106 \).
We see that \( AB^2 + AC^2 = 106 + 106 = 212 = BC^2 \), so \( \angle A = 90^\circ \). Thus ABC is a right triangle, right-angled at A.

Example 8. If the distance of P(x, y) from A(5, 1) and B(−1, 5) are equal and \( ax = by \), find a and b.
Answer:
Since \( AP = BP \), \( AP^2 = BP^2 \):
\( (x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2 \)
\( -10x + 25 - 2y + 1 = 2x + 1 - 10y + 25 \Rightarrow -12x + 8y = 0 \Rightarrow 3x = 2y \)
Comparing with \( ax = by \) (in the form 3x = 2y), a = 3 and b = 2.

Check Your Progress

Check Your Progress 1. Find the distance between the points (−2, 7) and (10, 12).
Answer:
\( \sqrt{(10 + 2)^2 + (12 - 7)^2} = \sqrt{144 + 25} = \) 13.

Check Your Progress 2. The distance between (6, 2) and (0, x) is 10. Find the value of x.
Answer:
\( 36 + (x - 2)^2 = 100 \Rightarrow (x - 2)^2 = 64 \Rightarrow x - 2 = \pm 8 \). So x = 10 or x = −6.

Check Your Progress 3. What points on the x-axis are equidistant from (7, 6) and (−3, 4)?
Answer:
Let P(x, 0). \( (x - 7)^2 + 36 = (x + 3)^2 + 16 \Rightarrow -14x + 85 = 6x + 25 \Rightarrow x = 3 \). The point is (3, 0).

Check Your Progress 4. Show that (2, −2), (−1, 2) and (3, 5) are the vertices of an isosceles triangle.
Answer:
Let A(2, −2), B(−1, 2), C(3, 5). \( AB = \sqrt{9 + 16} = 5 \), \( BC = \sqrt{16 + 9} = 5 \), \( AC = \sqrt{1 + 49} = 5\sqrt{2} \). Since AB = BC, the triangle is isosceles (in fact right-angled at B, as \( AB^2 + BC^2 = 50 = AC^2 \)).

Check Your Progress 5. Show that the points (3, 7), (6, 5) and (15, −1) lie on a straight line.
Answer:
Let A(3, 7), B(6, 5), C(15, −1). \( AB = \sqrt{9 + 4} = \sqrt{13} \), \( BC = \sqrt{81 + 36} = 3\sqrt{13} \), \( AC = \sqrt{144 + 64} = 4\sqrt{13} \). Since \( AB + BC = 4\sqrt{13} = AC \), the points are collinear.