A line passes through #(3 ,6 )# and #(6 ,5 )#. A second line passes through #(4 ,3 )#. 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 #(3    ,6   )# and #(6    ,5   )#.  A second line passes through #(4    ,3  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Sophia Alfred · Accepted Answer

#(1,4)# for instance

The slope of the first equation is #k_1=(Delta y)/(Delta x)=(5-6)/(6-3)=-1/3# #y-6=-1/3(x-3)# => #y=(-x+3)/3+6# => #y=(-x+21)/3# #k_2=k_1=-1/3# #y-3=-1/3(x-4)# => #y=(-x+4)/3+3# => #y=(-x+13)/3# Since the equation of second line is #y=(-x+13)/3# For instance with #x=1# #-> y=(-1+13)/3=4# #-> (1,4)#

Evelyn Arboleda · Answer

undefined If the second line is parallel to the first line, then they have the same slope. The slope of the first line passing through (3, 6) and (6, 5) can be calculated as follows:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{6 - 3} = \frac{-1}{3}$

Since the second line is parallel to the first line, it will also have a slope of -1/3.

Now, we have the slope and one point (4, 3) of the second line. We can use the point-slope form of the equation of a line to find other points on the second line.

Using the slope-intercept form:

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

Substituting the values $m = -\frac{1}{3}$, $x_1 = 4$, and $y_1 = 3$, we get:

$y - 3 = -\frac{1}{3}(x - 4)$

Expanding and simplifying:

$y - 3 = -\frac{1}{3}x + \frac{4}{3}$

$y = -\frac{1}{3}x + \frac{13}{3}$

Now, we can choose any value of x and calculate the corresponding y-coordinate to find other points on the line. For example, let's choose another x-value, say x = 2:

$y = -\frac{1}{3}(2) + \frac{13}{3}$

$y = -\frac{2}{3} + \frac{13}{3}$

$y = \frac{11}{3}$

So, the point (2, 11/3) lies on the second line parallel to the first line passing through (3, 6) and (6, 5).