A line passes through #(5 ,0 )# and #(7 ,9 )#. A second line passes through #(3 ,6 )#. 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    ,0  )# and #(7   ,9   )#.  A second line passes through #(3    ,6  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Noah Atkinson · Accepted Answer

#(2,3/2)# #"establish the equation of the parallel line"# #"the equation of a line in "color(blue)"slope-intercept form "# is. #•color(white)(x)y=mx+b# #"where m is the slope and b the y-intercept"# #"to calculate m use the "color(blue)"gradient formula"# #color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))# #"let "(x_1,y_1)=(5,0)" and "(x_2,y_2)=(7,9)# #rArrm=(9-0)/(7-5)=9/2# #• " parallel lines have equal slopes"# #rArr"slope of parallel line "=9/2# #rArry=9/2x+b color(blue)" is the partial equation"# #"to find b substitute "(3,6)" into the partial equation"# #6=27/2+brArrb=-15/2# #rArry=9/2x-15/2larrcolor(blue)"equation of parallel line"# #"choose any value for x and substitute into the equation"# #x=2toy=9-15/2=3/2# #"another point on the line is "(2,3/2)#

Theodore Abad · Answer

undefined If the second line is parallel to the first line passing through (5,0) and (7,9), it must have the same slope as the first line. The slope of the first line can be found using the formula:

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

Substituting the coordinates (5,0) and (7,9) into the formula, we get:

$m = \frac{{9 - 0}}{{7 - 5}} = \frac{9}{2}$

So, the slope of the second line must also be $\frac{9}{2}$ to be parallel.

Using the point-slope form of a line $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope, and substituting the given point (3,6) and the slope $\frac{9}{2}$, we can find the equation of the second line:

$y - 6 = \frac{9}{2}(x - 3)$

Expanding and rearranging, we get:

$y = \frac{9}{2}x - \frac{27}{2} + 6$

$y = \frac{9}{2}x - \frac{15}{2}$

To find another point on this line, we can choose any value of $x$ and solve for $y$. Let's take $x = 1$ for simplicity:

$y = \frac{9}{2}(1) - \frac{15}{2} = \frac{9}{2} - \frac{15}{2} = -3$

Therefore, another point on the second line is (1, -3).