Question

Find the distance between each pair of parallel lines with the given equation Y = -3/2x + 4 and y=-3/2x -1 (System)Instead of using the parallel distance formula can you do it similar in the picture shown..

Answer 1

To solve the exercise we can first find the equation of a line perpendicular to the given lines.

Two lines are parallel if their lines are equal.

`\begin{gathered} m_1=m_2 \\ \text{ Where }m_1\text{ is the slope of the first line and }m_2\text{ is the slope of the second line} \end{gathered}`

And two lines are perpendicular if their respective slopes have the following relationship:

`m_1=-(1)/(m_2)`

Then, the slope of the line perpendicular to the given lines is:

`\begin{gathered} m_1=-(3)/(2) \\ -(3)/(2)=-(1)/(m_2) \\ \text{ Multiply by -1 from both sides of the equation} \\ -1\cdot-(3)/(2)=-1\cdot-(1)/(m_2) \\ (3)/(2)=(1)/(m_2) \\ \text{ Apply the cross prduct} \\ 3\cdot m_2=1\cdot2 \\ 3m_2=2 \\ \text{ Divide by 3 from both sides of the equation} \\ (3m_2)/(3)=(2)/(3) \\ m_2=(2)/(3) \end{gathered}`

Now, let us find a point through which the second line passes:

`\begin{gathered} x=0 \\ y=-(3)/(2)x-1\Rightarrow\text{ Second line} \\ \text{ Replace the value of x} \\ y=-(3)/(2)(0)-1 \\ y=0-1 \\ y=-1 \\ \text{ Then, this line passes through the point (0,-1)} \end{gathered}`

Now, using the point-slope formula we can find the equation of the line perpendicular to the given parallel lines:

`\begin{gathered} $y-y_1=m(x-x_1)$\Rightarrow\text{ Point-slope formula} \\ \text{ Where m is the slope of the line and }(x_1,y_1)\text{ is a point through the line passes} \end{gathered}`
`\begin{gathered} m=(2)/(3) \\ (x_1,y_1)=(0,1) \\ y-y_1=m(x-x_1) \\ y-(-1)=(2)/(3)(x-0) \\ y-(-1)=(2)/(3)(x) \\ y+1=(2)/(3)x \\ \text{ Subtract -1 from both sides of the equation} \\ y+1-1=(2)/(3)x-1 \\ y+1-1=(2)/(3)x-1 \\ $$\boldsymbol{y=(2)/(3)x-1}$$ \end{gathered}`

Now, we can find the point where the first line and the line that is perpendicular to the two given lines meet:

`\begin{gathered} y=-(3)/(2)x+4\Rightarrow\text{ First line } \\ y=(2)/(3)x-1\Rightarrow\text{ Perpendicular line} \\ \text{ We equalize and solve for x} \\ -(3)/(2)x+4=(2)/(3)x-1 \\ \text{ Subtract 4 from both sides of the equation} \\ -(3)/(2)x+4-4=(2)/(3)x-1-4 \\ -(3)/(2)x=(2)/(3)x-5 \\ \text{ Subtract }(2)/(3)x\text{ from both sides of the equation} \\ -(3)/(2)x-(2)/(3)x=(2)/(3)x-5-(2)/(3)x \\ (-(3)/(2)-(2)/(3))x=-5 \\ ((-3\cdot3-2\cdot2)/(2\cdot3))x=-5 \\ ((-9-4)/(6))x=-5 \\ (-13)/(6)x=-5 \\ \text{Multiply by 6 from both sides of the equation} \\ 6\cdot(-13)/(6)x=-5\cdot6 \\ -13x=-30 \\ \text{ Divide by -13 from both sides of the equation} \\ (-13x)/(-13)=(-30)/(-13) \\ x=(30)/(13) \\ x\approx2.3\Rightarrow\approx\text{ it reads`

Now, we can plug the value of x into the equation of the first line to find its respective y-coordinate: