Question 206576
if the lines are parallel then their slopes have to be equal.
-----
your slope is given by the equation (y2-y1)/(x2-x1)
-----
your first line joins (-2,1) to (6,4)
y2 - y1 = 4 - 1 = 3
x2 - x1 = 6 - (-2) = 8
-----
the slope of your first line is 3/8
-----
your second line joins (-q,5) to (4,q)
for the slopes to be equal, y2-y1 / x2-x1 must equal 3/8
-----
y2-y1 = q-5
x2-x1 = 4-(-q) = q + 4
-----
(q-5)/(q+4) must equal 3/8
-----
if we multiply both sides of this equation by 8 we get:
(8q-40)/(q+4) = 3
-----
if we multiply both sides of this equation by (q+4) we get:
8q-40 = 3q + 12
-----
if we subtract 3q from both sides of this equation and we add 40 to both sides of this equation we get:
5q = 52
-----
if we divide both sides of this equation by 5 we get:
q = (52/5)
this is equivalent to 10.4
-----
our first line passes through the points (-2,1) and (6,4)
our second line passes through the points (-10.4,5) and (4,10.4)
-----
if both lines have the same slope, then the slope of the second line should equal (3/8)
-----
the slope of the second line is y2-y1/x2-x1 = (10.4 - 5)/(4 - (-10.4))
this comes out to be 5.4/14.4
-----
5.4/14.4 is the same ratio as 3/8 so the slopes are the same.
-----
our answer is q = 10.4
----
to graph these equations, we need to get them into the standard form of y = mx + b
-----
m is the slope = 3/8
b is the y intercept which is the value of y when x = 0
-----
to find b, we substitute one of the points in the equation and solve for b.
-----
for the first equation, we will use the point (6,4)
y = mx + b becomes 4 = 6*(3/8) + b
this makes b = 1.75
-----
standard form of our first equation is y = (3/8)*x + 1.75
-----
for the second equation, we will use the point (4,10.4)
y = mx + b becomes 10.4 = 4*(3/8) + b
this makes b = 8.9
-----
standard form of our second equation is y = (3/8)*x + 8.9
-----
graph of both equations looks like the following:
{{{graph(600,600,-12,8,-2,12,(3/8)*x + 1.75,(3/8)*x + 8.9)}}}
-----