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What is the standard form of an equation that is through Origin and (6,5) ?

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1 Answer

Hello, Jaxon -

To develop the equation in standard form, it is necessary to (1) solve for the slope, m, in the Slope-Intercept form of the linear equation, and (2) using a given point, i.e., (6,5), to solve for "b", the y-intercept, as follows:

Utilize your "y=mx+b" slope-intercept form
[from the standard form of the linear equation in two unknowns (x,y), as in AX + BY = C]

Hence, solving for "m" first: m=(Y2-Y1)/(X2-X1) for X2,Y2) as (6,5) respectively, and through the Origin, (0,0) as (X1,Y1), and thusly, m=(5-0)/(6-0) = 5/6.
• So, now, with m=5/6, substitute (6,5) as (X,Y) in the general Y=mX+b to get
• 5= [(5/6) x 6] + b, and
• b= 0,
• So y=mX+ b simplifies to
• y=(5/6)X+ 0,
• and,
• Therefore, Y=(5/6)X, through the origin, where the y-intercept is b= 0.
Answer: 5X - 6Y = 0 where A=5, B = -6, and C = 0 in the Standard Form of the Straight Line (linear equation in two unknowns (X,Y).
Questions? Thanks for this opportunity: hope it arrives to help you in a timely manner, Jaxon: I would be glad to tutor you in your Algebra remotely.
Thanks again,

David / "Mr. Mac."
Wyzant Algebra Tutor
Wichita, KS 67213