A line passes through #(4 ,3 )# and #(7 ,1 )#. A second line passes through #(1 ,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 #(4    ,3  )# and #(7   ,1   )#.  A second line passes through #(1    ,8  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Aurora Ayala · Accepted Answer

Find the slope of the first line, then apply it to the point of the second line to find #(1+3,8-2)=(4,6)#

We can solve this by finding the slope of the first line, then applying that slope to the point of the second line.

The first line passes through points #(4,3) and (7,1)#. Let's determine the slope.

The equation of slope is #m=(y_2-y_1)/(x_2-x_1)#. Let's plug in our points into this equation:

#m=(1-3)/(7-4)=-2/3#

We can now apply this slope to point #(1,8)#. I'll move 3 to the right on the x axis and 2 down on the y:

#(1+3,8-2)=(4,6)#

Aurora Ayala · Answer

undefined To find a point through which the second line may pass if it is parallel to the first line, we use the slope of the first line, which is determined by the given points (4, 3) and (7, 1). The slope of the first line can be calculated using the formula:

$$m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$$

Once we have the slope of the first line, we can use it to find the equation of the second line. Since the second line is parallel to the first line, it will have the same slope. Then, using the point (1, 8) through which the second line passes, we can find its equation in slope-intercept form $y = mx + b$.

After finding the equation of the second line, we can use it to determine another point on the line by substituting different values of x and solving for y. This will give us various points through which the second line may pass while being parallel to the first line.