The Equations of Parallel or Perpendicular Lines

Two parallel lines have the same slope (see The Relative Position of Two Lines).

1. Determine the value of the slope of the parallel line, i.e., the value of its parameter |m|. The slope is the same for the equation sought.

2. In the equation |y=mx+b|, replace the parameter |m| with the slope determined in step 1.

3. In the same equation, replace |x| and |y| with the coordinates |(x,y)| of the given point.

4. Isolate parameter |b| to find the value of the y-intercept.

5. Write the straight line equation in the form |y=mx+b| with the values of parameters |m| and |b|.

What is the equation of the line that is parallel to the line |y = 3x + 4| and passes through point |(2,1)|?

1. Since the lines are parallel, they have the same slope. The value of the parameter |m| in |y=3x+4| is |3|.

2. Replace parameter |m| of the equation |y=mx+b| with |3|. ||y = 3x + b||

3. Using the given point |(2,1)|, replace |y| with |1| and |x| with |2|. ||\begin{align} y &= 3x + b \\ 1 &= 3(2) + b \\ 1 &= 6 + b \end{align}||

4. Isolate the parameter |b|. ||\begin{align} 1 &= 6 + b \\ 1-6 &= b \\ -5 &= b \end{align}||

5. Write the equation in its functional form with the parameters |m=3| and |b=-5|. ||y = 3x -5||

Two perpendicular lines have slopes with a product equal to -1 (see The Relative Position of Two Lines).

1. Find the value of the slope perpendicular to the straight line by applying the following formula: |m_{1}\times m_{2}=-1| where |m_1| is the slope of the given perpendicular line and |m_2| is the slope of the line for which we are looking for the equation.

2. In the equation |y=mx+b|, replace the parameter |m| with the slope determined in step 1.

3. In this same equation, replace |x| and |y| with coordinates |(x,y)| of the given point.

4. Isolate parameter |b| to find the value of the y-intercept.

5. Write the line’s equation in the form |y=mx+b|, with the values of the parameters |m| and |b|.

What is the equation of a line perpendicular to the line given by |y=\dfrac{3}{2}x+7| and passing through point |(3,4)|?

1. Find the value of the perpendicular slope by applying the formula: ||\begin{align} m_{1}\times m_{2} &= -1 \\ \dfrac{3}{2}\times m_{2} &= -1 \\ \Rightarrow\ m_2 &=-1\div\dfrac{3}{2} \\ &= -1\times \dfrac{2}{3} \\ &=\dfrac{-2}{3} \end{align}||

2. Replace the parameter |m| in the equation |y=mx+b| with |\dfrac{-2}{3}.| ||y=-\dfrac{2}{3}x+b||

3. Using the known point, replace |y| with |4| and |x| with |3|. ||\begin{align} 4 &= -\dfrac{2}{3}(3)+b \\ 4 &= -2 + b \end{align}||

4. Isolate the parameter |b|. ||6 = b||

5. Write the equation in its functional form with the parameters |m=-\dfrac{2}{3}| and |b=6|. ||y=-\dfrac{2}{3}x+6||