Distance Between Two Points
1. Formula:
\[
d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}
\]
2. Worked Example:
Find the distance between points \(P_1(1, 2)\) and \(P_2(4, 6)\).
Solution:
\[
d = \sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
So, the distance is \(5\) units.
Slope of a Line Joining Two Points
1. Formula:
\[
m = \frac{y_2 – y_1}{x_2 – x_1}
\]
2. Worked Example:
Find the slope of the line joining points \(P_1(2, 3)\) and \(P_2(5, 11)\).
Solution:
\[
m = \frac{11 – 3}{5 – 2} = \frac{8}{3}
\]
So, the slope is \(\frac{8}{3}\).
Equation of a Line Through Two Points
1. Formula:
\[
y – y_1 = \frac{y_2 – y_1}{x_2 – x_1}(x – x_1)
\]
2. Worked Example:
Find the equation of the line passing through \(P_1(1, 2)\) and \(P_2(3, 4)\).
Solution:
\[
y – 2 = \frac{4 – 2}{3 – 1}(x – 1)
\]
Simplifying:
\[
y – 2 = \frac{2}{2}(x – 1) \quad \Rightarrow \quad y – 2 = x – 1
\]
\[
y = x + 1
\]
So, the equation of the line is \(y = x + 1\).
General Equation of a Line
1. Formula:
\[
Ax + By + C = 0
\]
2. Worked Example:
Find the general equation of the line with slope \(m = 2\) and passing through the point \((1, 3)\).
Solution:
Using the point-slope form:
\[
y – 3 = 2(x – 1)
\]
Simplifying:
\[
y – 3 = 2x – 2 \quad \Rightarrow \quad y = 2x + 1
\]
The general equation is:
\[
2x – y + 1 = 0
\]
Distance from a Point to a Line
1. Formula:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
2. Worked Example:
Find the distance from the point \((3, 4)\) to the line \(2x – y + 1 = 0\).
Solution
\[
d = \frac{|2(3) – (4) + 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|6 – 4 + 1|}{\sqrt{4 + 1}} = \frac{|3|}{\sqrt{5}} = \frac{3}{\sqrt{5}} \approx 1.34
\]
So, the distance is approximately \(1.34\) units.
