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Section 9.1 Properties of Lines

Subsection 1. Find the slope of a line

We can tell whether two lines are parallel, perpendicular, or neither by comparing their slopes.

Subsubsection Examples

Example 9.1.

Find the slope of the line \(~6x-8y = 9\)

Solution

The easiest way to find the slope of this line is to put its equation into slope-intercept form by solving for \(y\text{.}\)

\begin{align*} -8y \amp = -6x+9\\ y \amp = \dfrac{-6x+9}{-8} = \dfrac{3}{4}x - \dfrac{9}{8} \end{align*}

The slope of the line is \(\dfrac{3}{4}\text{.}\)

Example 9.2.

Find the slope of the line whose intercepts are \((112,0)\) and \((0,140)\text{.}\)

Solution

We use the slope formula.

\begin{equation*} m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{140-0}{0-112} = \dfrac{140}{-112} = \dfrac{-5}{4} \end{equation*}

Subsubsection Exercises

Find the slope of the line \(~2.8x+3.6y = 1.2\)

Answer
\(\dfrac{7}{9}\)

Find the slope of the line \(~6x=72\)

Answer
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Find the slope of the line that passes through \((15,-6)\) and \((10,-3)\text{.}\)

Answer
\(\dfrac{-3}{5}\)

Find the slope of a line that describes a 35% grade (or incline).

Answer
\(\dfrac{7}{20}\)

Subsection 2. Use the point-slope formula

Recall the point-slope formula for finding the equation of a line:

\begin{equation*} y-y_1 = m(x-x_1) \end{equation*}

Subsubsection Examples

Example 9.7.

Find the \(y\)-intercept of the line of slope \(\dfrac{-2}{3}\) that passes through \((-8,9)\text{.}\)

Solution

We first use the point-slope formula to find the equation of the line.

\begin{align*} y-9 \amp = \dfrac{-2}{3}(x+8) \amp \amp \blert{\text{Multiply by 3 to clear the fraction.}}\\ 3y-27 \amp = -2x-16 \amp \amp \blert{\text{Add 27 to both sides.}}\\ 3y \amp = -2x+11 \amp \amp \blert{\text{Divide both sides by 3.}}\\ y \amp = \dfrac{-2}{3}x + \dfrac{11}{3} \end{align*}

The \(y\)-intercept of the line is \(\left(0, \dfrac{11}{3}\right)\text{.}\)

Example 9.8.

Find the equation of the line with \(x\)-intercept \((2,0)\) that passes through \((-3,-3)\text{.}\)

Solution

We first compute the slope, using the two points given.

\begin{equation*} m = \dfrac{0+3}{2+3} = \dfrac{3}{5} \end{equation*}

Now we can use the point-slope formula.

\begin{align*} y-0 \amp = \dfrac{3}{5}(x-2) \amp \amp \blert{\text{Multiply by 5 to clear the fraction.}}\\ 5y \amp = 3x-6 \amp \amp \blert{\text{Divide both sides by 5.}}\\ y \amp = \dfrac{3}{5}x - \dfrac{6}{5} \end{align*}

Subsubsection Exercises

Find the equation of the line that has slope \(\dfrac{-1}{3}\) and passes through \((-4,-6)\text{.}\)

Answer
\(y = \dfrac{1}{3}x=\dfrac{22}{3}\)

Find the equation of the line that has slope \(0\) and passes through \((25,64)\text{.}\)

Answer
\(y=64\)

Find the \(y\)-intercept of the line that passes through \((-5,49)\) and has slope \(\dfrac{-4}{5}\text{.}\)

Answer
\((0,0)\)

Find the \(y\)-intercept of the line that has slope \(\dfrac{-7}{3}\) and passes through \((-6,0)\text{.}\)

Answer
\((0,-14)\)

Subsection 3. Use properties of geometric figures.

Analytic geometry uses algebra to help solve geometric problems.

Subsubsection Example

Example 9.13.

The figure shows isosceles triangle \(ABC\) and its altitude \(\overline{AP}\text{.}\) (Recall that the altitude of a triangle is the segment perpendicular to the base that passes through the opposite vertex.) For this problem, we'll use the following property:

In an isoceles triangle, the altitude bisects the base.

isosceles triangle
  1. Find the equation of the line that includes \(\overline{AP}\text{.}\)
  2. Find the coordinates of point \(P\text{.}\)
  3. Find the length of the segment \(\overline{AP}\text{.}\)
Solution

  1. Because \(\overline{AP}\) is perpendicular to \(\overline{BC}\text{,}\) we can find its slope. The coordinates of \(B\) and \(C\) are \((6,0)\) and \((2,-6)\text{,}\) so the slope of \(\overline{BC}\) is

    \begin{equation*} \dfrac{-6-0}{-2-6} = \dfrac{3}{4} \end{equation*}

    The slope of \(\overline{AP}\) is the negative reciprocal of \(\dfrac{3}{4}\text{,}\) or \(\dfrac{-4}{3}\text{.}\)

    Now we can use the point-slope formula with \(m=\dfrac{-4}{3}\) and the coordinates of \(A(-4,5)\) to calculate the equation of the line.

    \begin{align*} \dfrac{-4}{3} \amp = \dfrac{y-5}{x+4} \amp \amp \blert{\text{Cross-multiply.}}\\ 3(y-5) \amp = -4(x+4) \amp \amp \blert{\text{Apply the distributive law.}}\\ 3y-15 \amp = -4x-16 \amp \amp \blert{\text{Add 15 to both sides.}}\\ 3y \amp = -4x-1 \amp \amp \blert{\text{Divide both sides by 3.}}\\ y \amp = \dfrac{-4}{3}x - \dfrac{1}{3} \end{align*}

  2. Because the altitude bisects the base, point \(P\) is the midpoint of \(\overline{BC}\text{.}\) The coordinates of \(B\) and \(C\) are \((6,0)\) and \((2,-6)\text{,}\) so we use the midpoint formula to find the coordinates of \(P\text{.}\)

    \begin{align*} x \amp = \dfrac{-2+6}{2} = 2\\ y \amp = \dfrac{-6+0}{2} = -3 \end{align*}

    The coordinates of \(P\) are \((2,-3)\text{.}\)

  3. The coordinates of \(P\) are \((2,-3)\text{,}\) and the coordinates of \(A\) are \((-4,5)\text{.}\) We use the distance formula to find the length of \(\overline{AP}\text{.}\)

    \begin{equation*} \overline{AP} = \sqrt{(-4-2)^2+(5+3)^2} = \sqrt{36+64} = 10 \end{equation*}

Subsubsection Exercises

The line \(y=-2x+9\) is tangent at point \(Q\) to a circle with center \(C(-4,2)\text{.}\) For this problem, use the following property:

The tangent to a circle is perpendicular to the radius through the point of tangency.

circle and tangent line
  1. Find the equation of the line through \(C\) and \(Q\text{.}\)
  2. Find the coordinates of \(Q\text{.}\)
  3. Find the radius of the circle, \(r=CQ\text{.}\)
  4. Find the equation of the circle.
Answer
  1. \(\displaystyle y=\dfrac{1}{2}x=4\)
  2. \(\displaystyle (2,5)\)
  3. \(\displaystyle 3\sqrt{5}\)
  4. \(\displaystyle (x+4)^2+(y-2)^2=45\)

In the right triangle shown, \(\overline{ST}\) is parallel to the shorter leg, and \(h=\dfrac{9}{10}r\text{.}\) For this problem, use the following property:

A line parallel to the base of a triangle cuts off a similar triangle. (Recall that the sides of similar triangles are proportional.)

triangle cut parallel to base
  1. Write an expression for \(r\) in terms of \(h\text{.}\)
  2. Find an expression for \(b\) in terms of \(h\text{.}\)
  3. If \(OS=\dfrac{2}{3}h\) find an expression for \(ST\) in terms of \(h\text{.}\)
Answer
  1. \(\displaystyle r=\dfrac{10}{9}h\)
  2. \(\displaystyle b=\dfrac{\sqrt{19}}{9}h\)
  3. \(\displaystyle ST = \dfrac{2\sqrt{19}}{27}h\)

The quadrilateral \(ABCD\) has vertices \(~A(-4,-1),~B(-2,5),~C(8,5)~\) and \(~D(6,-1).\)

triangle cut parallel to base
  1. Show that \(ABCD\) is a parallelogram (its opposite sides are parallel).
  2. Find equations for the lines through the diagonals, \(\overline{AC}\) and \(\overline{BD}\text{.}\)
  3. Find the intersection of the diagonals, \(P\text{.}\)
  4. Find the lengths of \(\overline{AP}\) and \(\overline{PC}\text{,}\) and the lengths of \(\overline{BP}\) and \(\overline{PD}\text{.}\)

  5. This example illustrates the following property of parallelograms:

    The diagonals of a parallelogram each other.

Answer
  1. \(\displaystyle m_{\overline{AB}}=m_{\overline{CD}}=\dfrac{3}{2}; ~m_{\overline{AD}}=m_{\overline{BC}}=0\)
  2. \(\displaystyle y=\dfrac{1}{2}x+1; ~y=\dfrac{-3}{4}x+\dfrac{7}{2}\)
  3. \(\displaystyle (2,2)\)
  4. \(\displaystyle AP=PC=3\sqrt{5}; ~BP=PD=5\)
  5. bisect