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What is the standard form of an equation that passes through the point (-1,5) and has a slope of 1/2?

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2 Answers

Hi Mikk;
The point-slope formula is...
(y-y1)=m(x-x1)
m=slope
(y-5)=(1/2)(x--1)
Subtracting a negative number is the same as adding a positive number.
(y-5)=(1/2)(x+1)
y-5=(1/2)x+(1/2)
Let's add 5 to both sides...
5+y-5=(1/2)x+(1/2)+5
y=(1/2)x+(5 1/2)
The equation is now in slope-intercept form of...
y=mx+b, b is the y-intercept, the value of y when x=0.
 
The standard form of the equation would be...
Ax+By=C, neither A nor B equal zero, and A is greater than zero.
y=(1/2)x+(5 1/2)
Let's subtract (1/2)x from both sides...
-(1/2)x+y=(1/2)x+(5 1/2)-(1/2)x
(-1/2)x+y=5 1/2
A is not greater than zero.  To render it such, we will multiply both sides by -1...
-1[(-1/2)x+y)=-1(5 1/2)
(1/2)x-y=-5 1/2
There is a nice way to access an equation called the point-slope form of a line.  This form looks like :  
                                    y - y1 = m (x-x1)
 
All you do is plug the slope and your point into the equation to get your line... like this:
 
                                    y - 5 = ½ (x - -1)                        and then simplify...
 
 ANSWER:                 y - 5 = ½ (x + 1)