A line passes through #(5 ,6 )# and #(7 ,3 )#. A second line passes through #(2 ,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 #(5    ,6   )# and #(7    ,3   )#.  A second line passes through #(2    ,8  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Jayden Jackson · Accepted Answer

#color(blue)((1,19/2)#

First we need to find the equation of the line that passes through the points #(5,6) and (7,3)#

Using point slope form of a line:

#(y_2-y_1)=m(x_2-x_1)#

Where #m# is the gradient.

#m=(y_2-y_1)/(x_2-x_1)=(6-3)/(5-7)=-3/2#

#y-3=-3/2(x-7)#

#y=-3/2x+27/2#

The line passing through point #(2,8)# has to be parallel to the line we just found. If two lines are parallel, then they have the same gradients.

Using point slope form of a line:

#y-8=-3/2(x-2)#

#y=-3/2x+11#

To find another point that this line passes through, we just plug in a value for #x# and calculate the corresponding value of #y#

#x=1#

#y=-3/2(1)+11=19/2#

So #(1,19/2)# in another point.

Mason Adams · Answer

undefined To find a point that the second line may pass through if it's parallel to the first line passing through (5,6) and (7,3), we can use the same slope as the first line.

The slope of the first line is given by:
$$ m = \frac{{y_2 - y_1}}{{x_2 - x_1}} $$

Substituting the given points (5,6) and (7,3) into the equation, we get:
$$ m = \frac{{3 - 6}}{{7 - 5}} = \frac{{-3}}{{2}} $$

Now, for the second line to be parallel, it must have the same slope. So, using the point-slope form of a line equation:
$$ y - y_1 = m(x - x_1) $$
$$ y - 8 = -\frac{3}{2}(x - 2) $$

Expanding and rearranging, we get:
$$ y - 8 = -\frac{3}{2}x + 3 $$
$$ y = -\frac{3}{2}x + 11 $$

To find another point, we can choose any value for $ x $ and solve for $ y $. Let's take $ x = 0 $:
$$ y = -\frac{3}{2}(0) + 11 $$
$$ y = 11 $$

So, another point on the second line could be (0,11).