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?

Question

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?

Aaliyah Anderson · Accepted Answer

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

Wyatt Anderson · Answer

undefined If the second line is parallel to the first line passing through points (5, 0) and (7, 3), then the direction of the second line will be the same as that of the first line.

The direction of the first line can be found by calculating the slope using the given points:

$$m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$$
$$m = \frac{{3 - 0}}{{7 - 5}}$$
$$m = \frac{3}{2}$$

Since the second line is parallel to the first line, it will have the same slope. Now, using the given point (3, 6) and the slope $m = \frac{3}{2}$, we can find the equation of the second line using the point-slope form:

$$y - y_1 = m(x - x_1)$$
$$y - 6 = \frac{3}{2}(x - 3)$$
$$y - 6 = \frac{3}{2}x - \frac{9}{2}$$
$$y = \frac{3}{2}x - \frac{9}{2} + 6$$
$$y = \frac{3}{2}x - \frac{9}{2} + \frac{12}{2}$$
$$y = \frac{3}{2}x + \frac{3}{2}$$

So, the equation of the second line parallel to the first line and passing through point (3, 6) is $y = \frac{3}{2}x + \frac{3}{2}$.