Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!

Algebra ->  Algebra  -> College  -> Linear Algebra -> Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!     (Log On)

   

Question 167433: find the lope, the slope point form, slope intercept form, and standard form of the equation through the points (-5,6)(3,10) for slope intercept and standard form: find the lope, the slope point form, slope intercept form, and standard form of the equation through the points (-5,6)(3,10) for slope intercept and standard form
Answer by jojo14344(1030) About Me  (Show Source):
You can put this solution on YOUR website!

y+9=(-2/5)(x-3), EQN 1
It follows Point Slope Form Eqn: y=mx+b
Continuing,
y=(-2/5)x+(6/5)-9
y=(-2/5)x+(6-45)/5
y=(-2/5)x+(-39/5) ---> highlight(slope=(-2/5))
In standard form:ax+by=c, then EQN 1 becomes:
y+9=-2/5(x-3) ---> y+(2/5)x=(6/5)-9
(2/5)x+y=(6-45)/5=-39/5, multiply eqn by "5"
highlight(2x+5y=-39) --------> Standard Equation
.
Using Point-Slope Form for points (-5,6)(3,10):
m=(y[2]-y[1])/(x[2]-x[1])
m=(10-6)/(3-(-5))=4/(3+5)=4/8
highlight(m=1/2), Slope
Thru points (-5,6), standard line eqn:
Via y=mx+b
6=(1/2)(-5)+b
6+(5/2)=b --> b=(12+5)/2 ---> b=17/2, y-intercept
Therefore it follows,
y=(1/2)x+(17/2) ---> (-1/2)x+y=17/2, Multiply by "2" the eqn:
highlight(-x+2y=17), STANDARD FORM
We'll see the graph:
drawing(300,300,-20,8,-15,15,grid(1),graph(300,300,-20,8,-15,15,(-2/5)x-(39/5),(1/2)x+(17/2)),circle(-5,6,.20),circle(3,10,.20))--- See Green Line -x+2y=17 passes thru points (-5,6)(3,10)
Thank you,
Jojo