Coordinate Geometry: The Line

Revision

In your previous course you will have coved some of the basics of the line such as

  • Plot a point such as \(p(3,2)\)
  • Find the distance between two points
  • Find the midpoint of a line
  • Find the distance between two points

Distance between 2 points

The formula for the distance between two points \(p(x1, y1)\) and \(p(x2, y2)\) is:

Length of \(\left [ ab \right ]\) is \(\sqrt{(x_{2} -x_{1})^2 +(y_{2}-y_{1})^2}\)

This is because the horizontal distance of the line is given as \(x2-x1\) while the vertical distance of the line is \((y2-y1)\). Pythagorus tells us that the length of the hypotenuse of a right angled triangle is:

\[(hypotenuse)^2 = (base)^2 + (height)^2\]\[hypotenuse = \sqrt{(base)^2 + (height)^2}\]

Midpoint of a Line Segment

The midpoint, \(m\), of a line segment joining \(a\left(x_{1},y_{1}\right)\) and \(b\left(x_{2},y_{2}\right)\) is

\[\left(\frac{x_{1} +x_{2}}{2},\frac{y_{1} +y_{2}}{2}\right)\]

Example

If \(a\left(-1,3\right)\) and \(b\left(5,7\right)\) are 2 points in the plane, find (i) \(\mid ab \mid\) and (ii) the midpoint of \(\left[ab\right]\).

Solution

\((i)\) The points are \(a\left(-1,3\right)\) which corresponds to \(a\left(x_{1},y_{1}\right)\) and \(b\left(5,7\right)\) which corresponds to \(b\left(x_{2},y_{2}\right)\)

\[ x_{1} = -1 \\y_{1} = 3\\x_{2} = 5\\y_{2} = 7\]
\[\begin{eqnarray} \mid ab \mid & = & \sqrt{(x_{2} -x_{1})^2 +(y_{2}-y_{1})^2} \\ & = & \sqrt{(5-(-1))^2 +(7-3)^2} \\ & = & \sqrt{36 +16} \\ & = & \sqrt{52} \end{eqnarray}\]

\((ii)\) Midpoint of \(\left[ab\right]\) :

\[\begin{split}& = & \left(\frac{x_{1} +x_{2}}{2},\frac{y_{1} +y_{2}}{2}\right) \\ & = & \left(\frac{-1 + 5 }{2},\frac{3 + 7}{2}\right)\\ & = & (2,5)\end{split}\]

Extras

\[(a + b)^2 = a^2 + 2ab + b^2\]\[(a - b)^2 = a^2 - 2ab + b^2\]\[x^2=y^2\]
\[\cos (2\theta) = \cos^2 \theta - \sin^2 \theta\]
\[\begin{split}(a + b)^2 &= (a + b)(a + b) \\ &= a^2 + 2ab + b^2\end{split}\]
\[\begin{eqnarray} y & = & ax^2 + bx + c \\ f(x) & = & x^2 + 2xy + y^2 \end{eqnarray}\]
\[\begin{eqnarray} \sqrt{ (k-0)^2 +( -2k + 2)^2} &=& \sqrt{(k +2)^2 +(-2k +6)^2} \\ \Rightarrow k^2 + 4k^2 -8k +4 &=& k^2 +4k +4 +4k^2 -24k +36 \\ \Rightarrow -8k +4 &=& 40 -20k \\ \Rightarrow 12k &=& 36 \\ \Rightarrow k &=& 3 \\ \end{eqnarray}\]
(1)\[e^{i\pi} + 1 = 0\]