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

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

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Answer 1 · 2016-03-24T04:16:45

A point that makes $Q_{1} Q_{2}$ $Q_1Q_2$ // $P_{1} P_{2}$ $P_1P_2$
$Q_{2} (0, \frac{3}{2})$ $Q_2(0, 3/2)$

Explanation:

Given:
1) Line $p$ $p$ given by points $P_{1} (5, 0) and P_{2} (7, 3)$ $P_1(5,0) and P_2(7,3)$ , $p \Rightarrow ¯ ¯¯¯¯¯¯ ¯ P_{1} P_{2}$ $p => bar(P_1P_2)$
2) Line $q$ $q$ given by points $Q_{1} (5, 0) and Q_{2} (x, y)$ $Q_1(5,0) and Q_2(x,y)$ , $q \Rightarrow ¯ ¯¯¯¯¯¯¯ ¯ Q_{1} Q_{2}$ $q => bar(Q_1Q_2)$
Required $Q_{2} (x, y)$ $Q_2(x,y)$ ?
We are going to use the following Definition and Principles:
a) Equation of line $y_{p} = m_{p} x + b_{p}$ $y_p=m_px+b_p$ ,
b) It's perpendicular line passing through $Q_{i}$ $Q_i$ and given by: $y_{p e r} = - \frac{1}{m} x + b_{p e r}$ $y_(per)= -1/mx+b_(per)$
c) Equation of line $y_{q} = m_{q} x + b_{q}$ $y_q=m_qx+b_q$
d) Now given c) any point $Q_{2} (x, y)$ $Q_2(x,y)$ on $y_{q}$ $y_q$ will yield:
So let's get started:

$\Rightarrow (a) y_{p} = m_{p} x + b_{p}$ $color(red)(=>(a) y_p= m_px+b_p)$
$m_{p} = \frac{3 - 0}{7 - 5} = \frac{3}{2}$ $m_p=(3-0)/(7-5)=3/2$ ; $y_{p} = \frac{3}{2} x + b_{p}$ $y_p= 3/2x+b_p$ insert $P_{1} (5, 0)$ $P_1(5,0)$ and solve for $b = - \frac{15}{2}$ $b=-15/2$
$y_{p} = \frac{3}{2} x - \frac{15}{2}$ $y_p= 3/2x-15/2$

$\Rightarrow (b) y_{p e r} = - \frac{1}{m} x + b_{p e r}$ $color(blue)(=>(b) y_(per)= -1/mx+b_(per))$
$m_{p e r} = - \frac{1}{m_{p}} = - \frac{2}{3}$ $m_(per)= -1/m_p=-2/3$
$y_{p e r} = - \frac{2}{3} + b_{p e r}$ $y_(per)= -2/3+b_(per)$ insert $Q_{1} (3, 6)$ $Q_1(3,6)$ and solve for $b_{p e r} = 4$ $b_(per)=4$
$y_{p e r} = - \frac{2}{3} + 4$ $y_(per)= -2/3+4$

$\Rightarrow (c) y_{q} = m_{q} x + b_{q}$ $color(green)(=>(c) y_(q)= m_qx+b_(q))$
$m_{q} = m_{P} = \frac{3}{2}$ $m_(q)= m_(P)=3/2$
$y_{q} = \frac{3}{2} + b_{q}$ $y_(q)= 3/2+b_(q)$ insert $Q_{1} (3, 6)$ $Q_1(3,6)$ and solve for $b_{q} = \frac{3}{2}$ $b_(q)=3/2$
$y_{q} = \frac{3}{2} x + \frac{3}{2}$ $y_(q)= 3/2x+3/2$

$\Rightarrow (d) f \in d Q_{2} (x, y) using y_{q} = \frac{3}{2} x + \frac{3}{2}$ $color(magenta)(=>(d) find Q_2(x,y) " using "y_(q)= 3/2x+3/2$
Last pick any value for x and find y, you have your points:
Let's try $x = 0; y_{q} = \frac{3}{2}$ $x=0; y_q=3/2$
#Q_2(0, 3/2)

To show this two lines are parallel find the vectors:
$- -- \to P_{1} P_{2} = \to p \Rightarrow ((7 - 5, 3 - 0) = 2 i + 3 j + 0 k$ $vec(P_1P_2)=vec(p)=>((7-5, 3-0) = 2i +3j+0k$
$- -- \to Q_{1} Q_{2} = \to q \Rightarrow ((3 - 0, 6 - \frac{3}{2}) = 3 i + \frac{9}{2} j + 0 k$ $vec(Q_1Q_2) = vec(q)=>((3-0, 6-3/2) = 3i +9/2j+0k$
$vec(P_1P_2)xxvec(Q_1Q_2) =det[(i,j,k),(2,3,0),(3,9/2,0)]$
$vec(P_1P_2)xxvec(Q_1Q_2) =k[2*9/3-9] = 0$
This confirm that li $p and q$ are parallel to one another and you are done.

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

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Explanation:

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

1 Answer

Explanation:

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