A line passes through #(6 ,2 )# and #(2 ,1 )#. A second line passes through #(3 ,2 )#. What is one other point that the second line may pass through if it is parallel to the first line?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
#(1,3/2)# or graph{y = x/4 + 1/2 [-10, 10, -5, 5]} any point on the line #y = x/4 + 5/4#
First, find the gradient of the first line from the two points given.
gradient #=(rise)/(run)#
gradient #=(y1-y2)/(x1-x2)#
#(6,2) = (x1,y1)# and #(2,1) = (x2,y2)#
gradient #=(2-1)/(6-2)#
gradient #=(1)/(4)#
This is the first line: (# y = x/4 + 1/2 #) - the equation was found by substituting a coordinate into #y = x/4 + c# to find c. (but you don't need to find the equation, this graph is just for explanation purposes)
graph{y = x/4 + 1/2 [-10, 10, -5, 5]}
The second line is parallel, meaning it has the same gradient, which is #1/4#. We also have one point #(3,2)#.
Substitute the point #(3,2)# and gradient of #1/4#in to #y = mx + c# where #m=#gradient to find the value of c.
#2 = 1/4(3) +c#
#c =5/4#
Therefore the equation for the second line is # y = x/4 + 5/4 #, as shown:
graph{y = x/4 + 5/4 [-10, 10, -5, 5]}
Now, just choose any random x-value and find the corresponding y-value to get a point on that line.
