Question 215585
What is the y intercept of the points (1, 202) and (4, 382)?


Step 1.  We need to find an equation in slope intercept form given as y=mx+b where m is the slope and b is the y-intercept at point (0,b).


Step 2.  The slope of the line m is given as


{{{ m=(y2-y1)/(x2-x1)}}}


where for our example is x1=1, y1=202, x2=4 and y2=382 (think of {{{slope=rise/run}}}).  You can choose the points the other way around but be consistent with the x and y coordinates.  You will get the same result.


Step 3.  Substituting the above values in the slope equation gives


{{{m=(382-202)/(4-1)}}}


{{{m=180/3}}}


{{{m=60}}}


Step 4.  The slope is calculated as 60 or m=60.


Step 5.  Now use the slope equation of step 1 and choose one of the given points.  I'll choose point (1,202).   Letting y=y2 and x=x2 and substituting m=60 in the slope equation given as,


{{{ m=(y2-y1)/(x2-x1)}}}



{{{ 60=(y-202)/(x-1)}}}


{{{ 60=(y-202)/(x-1)}}}


Step 6.  Multiply both sides of equation by x-1 to get rid of denomination found on the right side of the equation



{{{ 60(x-1)=(x-1)(y-202)/(x-1)}}}



{{{ 60x-60=y-202}}}



Step 7.  Now simplify and put the above equation into slope-intercept form.


{{{60x-60=y-202}}}


Add 202 to both sides of the equation


{{{60x-60+202=y-202+202}}}


{{{60x+142=y}}}


{{{y=60x+142}}}   ANSWER in slope-intercept form.  m=60 and y-intercept=142


Step 8.  See if the other point (4, 382) or x=4 and y=382 satisfies this equation


{{{y=60x+142}}}


{{{382=60*4+142}}}


{{382=382}}}  So the point (4,382) satisfies the equation and is on the line.  In other words, you can use the other point to check your work.


Note: above equation can be also be transform into standard form as


{{{-60x+y=142}}}


See graph below to check the above steps.


*[invoke describe_linear_equation -60, 1, 142]


I hope the above steps were helpful. 

 
And good luck in your studies!


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


Respectfully,
Dr J