A line passes through #(5 ,6 )# and #(2 ,8 )#. A second line passes through #(7 ,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 #(5    ,6   )# and #(2    ,8   )#.  A second line passes through #(7    ,1  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Autumn Armstrong · Accepted Answer

One specific point would be (4,3)

Slope of the line passing through (5,6) and (2,8) would be #(8-6)/(2-5)= -2/3# Slope of the line parallel to this one would also be #-2/3#. If the line passing through (7,1) also passes through a point #(x, y)#, then its slope can also be expressed as #(y -1)/(x -7)# Then #-2/3 = (y -1)/(x -7)#. Solve this eq. for y. #y= (17-2x)/3#. Now assign any value to x and calculate the corresponding y. Thus there would be infinite number of points satisfying the equation. A specific point can be obtained by assigning the value x=4 for which y would be 3. this point would thus be (4,3).

Savannah Aguilar · Answer

undefined If the second line is parallel to the first line, it will have the same slope. To find another point on the second line, we can use the slope-intercept form of the equation for a line:

$$ y = mx + b $$

Where $ m $ is the slope of the line.

Given that the first line passes through the points $(5, 6)$ and $(2, 8)$, we can find the slope using the formula:

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

$$ m = \frac{{8 - 6}}{{2 - 5}} = \frac{{2}}{{-3}} = -\frac{{2}}{{3}} $$

Since the second line is parallel to the first line, it will also have a slope of $-\frac{{2}}{{3}}$.

Given that the second line passes through the point $(7, 1)$, we can use the point-slope form of the equation for a line to find another point on the second line:

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

Substituting $ m = -\frac{{2}}{{3}} $ and $(x_1, y_1) = (7, 1)$:

$$ y - 1 = -\frac{{2}}{{3}}(x - 7) $$

$$ y - 1 = -\frac{{2}}{{3}}x + \frac{{14}}{{3}} $$

$$ y = -\frac{{2}}{{3}}x + \frac{{14}}{{3}} + 1 $$

$$ y = -\frac{{2}}{{3}}x + \frac{{17}}{{3}} $$

So, one other point that the second line may pass through if it is parallel to the first line is $(x, y) = (0, \frac{{17}}{{3}})$.