CBSE Class 10 Mathematics Coordinate Geometry Notes Set 09

Coordinates of a Point

Two perpendicular number lines meet at the origin O. The horizontal one is the x-axis and the vertical one is the y-axis. The position of a point P is given by an ordered pair (x, y): x (the abscissa) is its distance from the y-axis and y (the ordinate) is its distance from the x-axis, each with a sign.
1. x is positive to the right of the y-axis and negative to its left.
2. y is positive above the x-axis and negative below it.
3. The axes divide the plane into four quadrants. Signs of (x, y) are (+, +) in the first, (-, +) in the second, (-, -) in the third and (+, -) in the fourth quadrant.
4. A point on the x-axis has y = 0 and a point on the y-axis has x = 0.

Polar Coordinates

A point can also be fixed by its distance r from the origin (the pole) and the angle \( \theta \) that the line joining it to the pole makes with the positive x-axis, measured anticlockwise. The pair is written (r, \( \theta \)).
1. Polar to Cartesian: \( x = r\cos\theta,\; y = r\sin\theta \).
2. Cartesian to polar: \( r = \sqrt{x^2 + y^2},\; \tan\theta = \dfrac{y}{x} \), choosing the quadrant from the signs of x and y.

Distance Formula

The distance between P(x₁, y₁) and Q(x₂, y₂) follows from Pythagoras' theorem on the right triangle whose legs are the horizontal and vertical differences:
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
The distance of (x, y) from the origin is \( \sqrt{x^2 + y^2} \).

Section Formula

The point dividing the segment joining (x₁, y₁) and (x₂, y₂) internally in the ratio m : n is
\[ \left( \frac{m x_2 + n x_1}{m + n},\; \frac{m y_2 + n y_1}{m + n} \right) \]
For external division in the ratio m : n replace n by -n: \( \left( \dfrac{m x_2 - n x_1}{m - n},\; \dfrac{m y_2 - n y_1}{m - n} \right) \). The midpoint is the case m = n = 1.

Area of a Triangle

The area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is
\[ \text{Area} = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
1. Three points are collinear exactly when this area is zero.
2. The area of a quadrilateral is found by splitting it along a diagonal into two triangles.

Locus

A locus is the path of a point that moves under a given geometrical condition. To find its equation take a general point (x, y) on it and translate the condition into an equation in x and y.
1. Points at a fixed distance from a fixed point form a circle: \( (x - a)^2 + (y - b)^2 = r^2 \).
2. Points equidistant from two fixed points form the perpendicular bisector of the segment joining them.
3. The line x = k is the locus of points at distance |k| from the y-axis, and y = k that of points at distance |k| from the x-axis.

Straight Lines

1. Slope of the line through two points: \( m = \dfrac{y_2 - y_1}{x_2 - x_1} \).
2. Slope form: \( y = mx + c \). Point-slope form: \( y - y_1 = m(x - x_1) \). Intercept form: \( \dfrac{x}{a} + \dfrac{y}{b} = 1 \).
3. Two lines are parallel when their slopes are equal, so a line parallel to \( ax + by + c = 0 \) has the form \( ax + by + k = 0 \).
4. Two lines are perpendicular when the product of their slopes is -1, so a line perpendicular to \( ax + by + c = 0 \) has the form \( bx - ay + k = 0 \).
5. The angle between lines of slopes m₁ and m₂ is given by \( \tan\theta = \left| \dfrac{m_1 - m_2}{1 + m_1 m_2} \right| \).

Worked Examples

Example 1. Find the area of the triangle with vertices (1, -3), (3, 1), (1, 4).
Area = \( \tfrac12 |1(1 - 4) + 3(4 + 3) + 1(-3 - 1)| = \tfrac12 |-3 + 21 - 4| = 7 \) square units.

Example 2. The point (0, -3) divides AB in the ratio 3 : 2 and A is (-6, -9). Find B(h, k).
\( \dfrac{3h + 2(-6)}{5} = 0 \Rightarrow h = 4 \) and \( \dfrac{3k + 2(-9)}{5} = -3 \Rightarrow k = 1 \). So B = (4, 1).

Example 3. A point moves so that its distance from (0, a) equals its distance from (0, -a). Find its locus.
\( x^2 + (y - a)^2 = x^2 + (y + a)^2 \Rightarrow 4ay = 0 \Rightarrow y = 0 \). The locus is the x-axis.

Example 4. Find the line through (-5, 4) perpendicular to 4x + 3y + 5 = 0.
A perpendicular line has the form 3x - 4y + k = 0. Substituting (-5, 4): -15 - 16 + k = 0, so k = 31. The line is 3x - 4y + 31 = 0.