# Algebra

posted by delmore

Given a line containing the points(1,4), (2,7) and (3,10) determine that slope-intercept form of the equation, provide one additional point on this line, and graph the funtion.

Start by putting the first point into point-slope form: y-y1=m(x-x1), so y-4=m(x-1). Slope, or m, is y2-y1 over x2-x1. With the first two points, slope=7-4 over 2-1, or 3 over 1, which is 3. The equation is then y-4=3(x-1). After doing basic algebra, the slope-intercept form of the equation is y=3x+1.

Using the slope of 3, add three to the y-value of the last point, (3,10), and 1 to the x-value; you are basically doing the slope equation backwards here. Your new point could be (4,13).

To graph the function, just plot the points you were given; in fact, you only need 2 of them, and then draw your line.

i would say (4,13) or (0,1) depending which direction you want to go.
the way i see it is like this.

for the point (1,4) to get to the next point (2,7) they are going up 1 number on the x coordinate point and 3 on the y coordinate point.I know that theres a formula for this but i can't remember what it is but i just know that you line is a diagnol line.

I just set up two equations and solved them simultaneously.
y = mx + b
==============
substitute points 1,4 and 2,7
eqn 1:   4=m(1)+b
eqn 2:   7=m(2)+b
solve for m and b.
m=3 and b=1
from work done by the two previous posters, both 4,13 and 0,1 are on the line as well as the third point given in the problem of 3,10. And y = 3x+1 is the slope intercept form.
I just thought I would give an alternative method of solving the problem.

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