# A line passes through (5 ,0 ) and (7 ,3 ). A second line passes through (3 ,6 ). What is one other point that the second line may pass through if it is parallel to the first line?

Mar 24, 2016

A point that makes ${Q}_{1} {Q}_{2}$ // ${P}_{1} {P}_{2}$
${Q}_{2} \left(0 , \frac{3}{2}\right)$

#### Explanation:

Given:
1) Line $p$ given by points ${P}_{1} \left(5 , 0\right) \mathmr{and} {P}_{2} \left(7 , 3\right)$, $p \implies \overline{{P}_{1} {P}_{2}}$
2) Line$q$ given by points ${Q}_{1} \left(5 , 0\right) \mathmr{and} {Q}_{2} \left(x , y\right)$, $q \implies \overline{{Q}_{1} {Q}_{2}}$
Required ${Q}_{2} \left(x , y\right)$?
We are going to use the following Definition and Principles:
a) Equation of line ${y}_{p} = {m}_{p} x + {b}_{p}$,
b) It's perpendicular line passing through ${Q}_{i}$ and given by: ${y}_{p e r} = - \frac{1}{m} x + {b}_{p e r}$
c) Equation of line ${y}_{q} = {m}_{q} x + {b}_{q}$
d) Now given c) any point ${Q}_{2} \left(x , y\right)$ on ${y}_{q}$ will yield:
So let's get started:

$\textcolor{red}{\implies \left(a\right) {y}_{p} = {m}_{p} x + {b}_{p}}$
${m}_{p} = \frac{3 - 0}{7 - 5} = \frac{3}{2}$; ${y}_{p} = \frac{3}{2} x + {b}_{p}$ insert ${P}_{1} \left(5 , 0\right)$ and solve for $b = - \frac{15}{2}$
${y}_{p} = \frac{3}{2} x - \frac{15}{2}$

$\textcolor{b l u e}{\implies \left(b\right) {y}_{p e r} = - \frac{1}{m} x + {b}_{p e r}}$
${m}_{p e r} = - \frac{1}{m} _ p = - \frac{2}{3}$
${y}_{p e r} = - \frac{2}{3} + {b}_{p e r}$ insert ${Q}_{1} \left(3 , 6\right)$ and solve for ${b}_{p e r} = 4$
${y}_{p e r} = - \frac{2}{3} + 4$

$\textcolor{g r e e n}{\implies \left(c\right) {y}_{q} = {m}_{q} x + {b}_{q}}$
${m}_{q} = {m}_{P} = \frac{3}{2}$
${y}_{q} = \frac{3}{2} + {b}_{q}$ insert ${Q}_{1} \left(3 , 6\right)$ and solve for ${b}_{q} = \frac{3}{2}$
${y}_{q} = \frac{3}{2} x + \frac{3}{2}$

color(magenta)(=>(d) find Q_2(x,y) " using "y_(q)= 3/2x+3/2
Last pick any value for x and find y, you have your points:
Let's try x=0; y_q=3/2
Q_2(0, 3/2)

To show this two lines are parallel find the vectors:
vec(P_1P_2)=vec(p)=>((7-5, 3-0) = 2i +3j+0k
vec(Q_1Q_2) = vec(q)=>((3-0, 6-3/2) = 3i +9/2j+0k#
$\vec{{P}_{1} {P}_{2}} \times \vec{{Q}_{1} {Q}_{2}} = \det \left[\begin{matrix}i & j & k \\ 2 & 3 & 0 \\ 3 & \frac{9}{2} & 0\end{matrix}\right]$
$\vec{{P}_{1} {P}_{2}} \times \vec{{Q}_{1} {Q}_{2}} = k \left[2 \cdot \frac{9}{3} - 9\right] = 0$
This confirm that li $p \mathmr{and} q$ are parallel to one another and you are done.