1.8 Coordinate Geometry
1.8.1 Learning Objectives
By the end of this section, you should be able to:
- plot points and describe regions in the coordinate plane;
- calculate distances and midpoints;
- graph equations and find intercepts;
- write and interpret equations of circles; and
- test graphs for symmetry.
1.8.2 Points, Distance, and Midpoint
The coordinate plane uses a horizontal \(x\)-axis and vertical \(y\)-axis. Their intersection is the origin \((0,0)\). A point \((x,y)\) is located by moving \(x\) units horizontally and \(y\) units vertically.
For \(A(x_1,y_1)\) and \(B(x_2,y_2)\), the distance formula is
\[ d(A,B)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \]
and the midpoint is
\[ M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right). \]
Example 1.38 Distance and Midpoint
For \(A(-2,5)\) and \(B(4,-3)\),
\[ d=\sqrt{6^2+(-8)^2}=10, \]
\[ M=\left(\frac{-2+4}{2},\frac{5-3}{2}\right)=(1,1). \]
Exercise 1.44 Identify the quadrant containing \((-4,6)\).
Show answer
Answer: II
Exercise 1.45 Find the distance from \((1,2)\) to \((4,6)\).
Show answer
Answer: \(5\)
Exercise 1.46 Find the midpoint of \((-5,3)\) and \((7,-1)\).
Show answer
Answer: \((1,1)\)
1.8.3 Graphs and Intercepts
The graph of an equation in \(x\) and \(y\) consists of all points whose coordinates satisfy the equation.
- To find \(x\)-intercepts, set \(y=0\).
- To find \(y\)-intercepts, set \(x=0\).
Example 1.39 Intercepts
For \(y=x^2-4\), setting \(y=0\) gives \(x=\pm2\). Setting \(x=0\) gives \(y=-4\).
The intercepts are \(\boxed{(-2,0),(2,0),(0,-4)}\).
Exercise 1.47 Find both intercepts of \(2x+3y=12\).
Show answer
Answer: \((6,0)\) and \((0,4)\)
1.8.4 Circles
A circle with centre \((h,k)\) and radius \(r\) has equation
\[ (x-h)^2+(y-k)^2=r^2. \]
Example 1.40 Circle from Centre and Radius
A circle with centre \((3,-2)\) and radius \(5\) has equation
\[ \boxed{(x-3)^2+(y+2)^2=25}. \]
Exercise 1.48 State the centre and radius of \((x+1)^2+(y-4)^2=9\).
Show answer
Answer: centre \((-1,4)\), radius \(3\)
An expanded circle equation can be returned to standard form by completing the square in both \(x\) and \(y\).
Example 1.41 Circle from Expanded Form
Rewrite \(x^2+y^2-6x+4y-12=0\).
\[ (x^2-6x)+(y^2+4y)=12 \]
\[ (x-3)^2+(y+2)^2=25. \]
The centre is \((3,-2)\) and the radius is \(5\).
1.8.5 Symmetry
Symmetry can be tested algebraically:
| Symmetry | Substitution that leaves the equation unchanged |
|---|---|
| \(x\)-axis | Replace \(y\) by \(-y\) |
| \(y\)-axis | Replace \(x\) by \(-x\) |
| Origin | Replace \((x,y)\) by \((-x,-y)\) |
Example 1.42 Symmetry
The equation \(y=x^4-2x^2\) is unchanged when \(x\) is replaced by \(-x\), so its graph is symmetric about the \(y\)-axis.
1.8.6 Conceptual Takeaways
- Ordered pairs connect algebraic equations with geometric locations.
- Distance comes from the Pythagorean theorem.
- Midpoint coordinates are averages.
- Intercepts are found by setting one coordinate equal to zero.
- Standard form reveals a circle’s centre and radius.
- Symmetry can reduce the amount of graphing needed.
1.8.7 Skills You Should Be Able to Do
- Plot points and describe plane regions.
- Apply distance and midpoint formulas.
- Determine whether a point satisfies an equation.
- Find intercepts and sketch basic graphs.
- Write and interpret circle equations.
- Complete the square for circles.
- Test graphs for symmetry.
1.8.8 Practice Problems with Solutions
Find the distance between \((-3,-1)\) and \((5,5)\).
Show Solution
\[ d=\sqrt{8^2+6^2}=\boxed{10}. \]
Find the midpoint of \((8,-6)\) and \((-2,4)\).
Show Solution
\[ M=\left(\frac{8-2}{2},\frac{-6+4}{2}\right)=\boxed{(3,-1)}. \]
Determine whether \((2,7)\) lies on \(y=x^2+3\).
Show Solution
Since \(2^2+3=7\), the point \(\boxed{\text{does lie on the graph}}\).
Find the intercepts of \(y=2x^2-8\).
Show Solution
Setting \(y=0\): \[ 2x^2-8=0\Longrightarrow x=\pm2. \] Setting \(x=0\) gives \(y=-8\). The intercepts are \[ \boxed{(-2,0),(2,0),(0,-8)}. \]
Write the equation of the circle with centre \((-4,1)\) and radius \(6\).
Show Solution
\[ \boxed{(x+4)^2+(y-1)^2=36}. \]
Find the centre and radius of \(x^2+y^2+8x-2y-8=0\).
Show Solution
\[ (x^2+8x)+(y^2-2y)=8 \] \[ (x+4)^2+(y-1)^2=25. \] The centre is \(\boxed{(-4,1)}\) and the radius is \(\boxed{5}\).
Test \(x^2+y^2=16\) for symmetry.
Show Solution
Replacing \(x\), \(y\), or both by their opposites leaves the equation unchanged. The circle is symmetric about the \(x\)-axis, \(y\)-axis, and origin.