Question 341095: Find an equation in standard form that has a slope of 1/3 and goes through (1,6).
Please help
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A line and a point example.
Write in standard form the eqation of a line that satisfies the given conditions. Perpendicular to 9x+3y=36, through (1,2)
-----------------
Find the slope of the line. Do that by putting the equation in slope-intercept form, y = mx + b. That means solve for y.
9x+3y = 36
3y= - 9x + 36
y = -3x + 12
The slope, m = -3
------------------
The slope of lines parallel is the same.
The slope of lines perpendicular is the negative inverse, m = +1/3
----------------
Use y = mx + b and the point (1,2) to find b.
2 = (1/3)*1 + b
b = 5/3
The equation is y = (1/3)x + 5/3 (slope-intercept form)
x - 3y = -5 (standard form)
------------------------
For further assistance, or to check your work, email me via the thank you note, or at Moral Loophole@aol.com
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Find an equation in standard form that
has a slope of 1/3 and goes through (1,6).
---------------------
Form: y = mx + b
You are told m = 1/3 and y = 6 when x = 1.
Solve for "b":
6 = (1/3)1 + b
b = 17/3
-----
y = (1/3)x + (17/3)
======================
Cheers,
Stan H.
==========
|
|
|
