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

Violet Aldrich · Accepted Answer

#(0, 3) and (3/2, 0)#, for instance

#A = (4, 7), B = (7, 1), C = (1,1)# #frac{Delta y}{Delta x} = frac{1-7}{7-4} = frac{-6}{3} = -2# #AB: y = - 2x + 15# #CD: y = -2x + b, y(1) = 1 => 1 = -2 + b => b =3# #x = 0 => y = 3# #y = 0 => x = 3/2#

Aaliyah Anderson · Answer

undefined To find a point that the second line may pass through if it is parallel to the first line, we can use the slope of the first line and the point given for the second line.

The slope of the first line can be calculated using the formula: 
$$ m = \frac{{y2 - y1}}{{x2 - x1}} $$
Where $(x1, y1)$ and $(x2, y2)$ are the coordinates of the two points on the first line.

Using the given points $(4, 7)$ and $(7, 1)$, the slope of the first line is:
$$ m = \frac{{1 - 7}}{{7 - 4}} = \frac{{-6}}{{3}} = -2 $$

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

Now, using the point $(1, 1)$ on the second line and the slope $m = -2$, we can find the equation of the second line in point-slope form:
$$ y - y1 = m(x - x1) $$
$$ y - 1 = -2(x - 1) $$
$$ y - 1 = -2x + 2 $$
$$ y = -2x + 3 $$

To find another point on this line, we can choose any value for $x$ and then solve for $y$. Let's choose $x = 2$:
$$ y = -2(2) + 3 = -4 + 3 = -1 $$

So, when $x = 2$, $y = -1$. Therefore, the point $(2, -1)$ lies on the second line and is parallel to the first line passing through the points $(4, 7)$ and $(7, 1)$.