A line passes through #(8 ,2 )# and #(6 ,7 )#. A second line passes through #(3 ,8 )#. 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 #(8    ,2   )# and #(6   ,7   )#.  A second line passes through #(3    ,8  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Caroline Aucoin · Accepted Answer

Infinite points given by #(a,(31-5a)/2)#

As the first line passes through #(8,2)# and #(6,7)# its slope is

#(7-2)/(6-8)=-5/2#

Let the other point be #(a,b)# and as line joining #(a,b)# and #(3,8)# is parallel to first line, it will too have a slope of #-5/2#

Hence #(b-8)/(a-3)=-5/2#

or #2b-16=-5a+15#

i.e. #2b=-5a+31#

or #b=(31-5a)/2#

Hence the point #(a,(31-5a)/2)# is the other point.

Observe that by changing various values of #a#, we can have many points and hence therecould be infinite points. For example if #a=-3#, the point is #(-3,23)#.

graph{((x-8)^2+(y-2)^2-0.02)((x-6)^2+(y-7)^2-0.02)((x-3)^2+(y-8)^2-0.02)(5x+2y-31)(5x+2y-44)=0 [-6.38, 13.62, 0, 10]}

Carson Arnold · Answer

undefined To find a point that the second line may pass through if it is parallel to the first line, we need to use the concept that parallel lines have the same slope.

The slope of the first line passing through (8, 2) and (6, 7) is calculated as:

$ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 2}{6 - 8} = \frac{5}{-2} = -\frac{5}{2} $

Since the second line is parallel to the first, it will have the same slope.

Now, we can use the slope and the given point (3, 8) to find the equation of the second line and then find another point on it. Using point-slope form:

$ y - y_1 = m(x - x_1) $

Substituting the values, we get:

$ y - 8 = -\frac{5}{2}(x - 3) $

Expanding and simplifying:

$ y - 8 = -\frac{5}{2}x + \frac{15}{2} $

$ y = -\frac{5}{2}x + \frac{31}{2} $

Now, we can choose any value for $ x $ and find the corresponding $ y $ to get another point on the line. Let's choose $ x = 0 $:

$ y = -\frac{5}{2}(0) + \frac{31}{2} $

$ y = \frac{31}{2} $

So, another point on the second line is (0, 15.5).