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?

1 Answer
Mar 24, 2016

Answer:

A point that makes #Q_1Q_2# // #P_1P_2#
#Q_2(0, 3/2)#

Explanation:

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

#color(red)(=>(a) y_p= m_px+b_p)#
#m_p=(3-0)/(7-5)=3/2#; #y_p= 3/2x+b_p# insert #P_1(5,0)# and solve for #b=-15/2#
#y_p= 3/2x-15/2#

#color(blue)(=>(b) y_(per)= -1/mx+b_(per))#
#m_(per)= -1/m_p=-2/3#
#y_(per)= -2/3+b_(per)# insert #Q_1(3,6)# and solve for #b_(per)=4#
#y_(per)= -2/3+4#

#color(green)(=>(c) y_(q)= m_qx+b_(q))#
#m_(q)= m_(P)=3/2#
#y_(q)= 3/2+b_(q)# insert #Q_1(3,6)# and solve for #b_(q)=3/2#
#y_(q)= 3/2x+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_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.