Coordinate geometry connects algebra and geometry by representing points with ordered pairs \((x,y)\) on the Cartesian plane.
Horizontal axis is \(x\)-axis, vertical axis is \(y\)-axis, and intersection is origin \((0,0)\).
Quadrants:
Ordered pair \((a,b)\) means move \(a\) units along \(x\), then \(b\) units along \(y\).
Example: \((-3,2)\) lies in Quadrant II.
For \(A(x_1,y_1)\), \(B(x_2,y_2)\):
\[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \]
This comes from Pythagoras' theorem on horizontal and vertical differences.
\[ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right). \]
Example: Midpoint of \((2,5)\) and \((8,-1)\) is \((5,2)\).
For points \(A(x_1,y_1)\), \(B(x_2,y_2)\) with \(x_2\ne x_1\):
\[ m=\frac{y_2-y_1}{x_2-x_1}. \]
Slope-intercept form:
\[ y=mx+c \]
where \(m\) is slope and \(c\) is \(y\)-intercept.
Point-slope form:
\[ y-y_1=m(x-x_1). \]
Between \(A(x_1,y_1)\) and \(B(x_2,y_2)\), horizontal change is \(\Delta x=x_2-x_1\) and vertical change is \(\Delta y=y_2-y_1\). These are the legs of a right triangle, and segment \(AB\) is its hypotenuse.
\[ AB^2=(\Delta x)^2+(\Delta y)^2 \]
\[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]
A midpoint lies exactly halfway in both horizontal and vertical directions, so each coordinate is an average of the endpoints:
\[ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \]
Slope measures steepness: rise divided by run. For two points, rise is \(y_2-y_1\) and run is \(x_2-x_1\), giving:
\[ m=\frac{y_2-y_1}{x_2-x_1} \]
This is why vertical lines (\(x_2=x_1\)) have undefined slope: the denominator becomes zero.
Coordinate geometry was formalized by Rene Descartes and Pierre de Fermat. Their work connected algebraic equations to geometric curves, creating a major foundation for modern mathematics.
Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\).
\[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]
A midpoint splits a segment into two equal parts in both directions. So the \(x\)-coordinate is halfway between \(x_1\) and \(x_2\), and the \(y\)-coordinate is halfway between \(y_1\) and \(y_2\).
\[ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \]
Slope is rise over run. For points \(A(x_1,y_1)\) and \(B(x_2,y_2)\):
\[ m=\frac{y_2-y_1}{x_2-x_1} \]
Rearranging the same idea for a generic point \((x,y)\) on the line through \((x_1,y_1)\) gives point-slope form:
\[ y-y_1=m(x-x_1) \]
Expanding gives slope-intercept form \(y=mx+c\), where \(c\) is the \(y\)-intercept.
Answer: Quadrant III.
Answer: \(\sqrt{6^2+8^2}=\sqrt{100}=10\).
Answer: \(\left(3,2\right)\).
Answer: \(m=\frac{7-(-1)}{6-2}=\frac84=2\).
Answer: \(y-4=3(x-1)\Rightarrow y=3x+1\).
Answer: Pythagoras gives \(AB^2\), so we take the positive square root to get the actual distance \(AB\).
Answer: For a vertical line, \(\Delta x=0\), so \(m=\frac{\Delta y}{\Delta x}\) would require division by zero.