Question 862337: Hi, I'm having trouble with this problem:
Write the equation of the line that contains the indicated pair of points. Express the final equation in standard form. (-5,6),(2,-2).
I got using the formula m=y2-y1/x2-x1
-2-6/2-5= slope 8/3
formula: y2-y1=m(x2-x1)
y-6=8/3--multiply by 3 to get rid of denominator
=-58=8x-3y
Thank you!
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you started good but ended bad.
let's do it again so you can see where you went wrong.
the formula for the slope is:
m = (y2-y1) / (x2-x1)
you have 2 points.
assign one of them to (x1,y1) and assign the other to (x2,y2)
it doesn't matter which you assign to which. you'll get the same answer either way.
so we'll assign:
(x1,y1) = (-5,6)
(x2,y2) = (2,-2)
the slope is calculated as follows:
m = (y2 - y1) / (x2 - x1) = (-2-6) / (2--5) = (-8) / (2+5) = -8/7
your slope turns out to be -8/7
now you apply that slope to the point slope form of the equation.
the point slope form of the equation is:
y = y1 = m(x - x1)
you take one of your points.
any one of them will do.
you replace x1,y1 in this equation with their respective values from that point.
we'll choose (2,-2) as the point.
this means we are now assigning the point (2,-2) to (x1,y1) in the point slope form of the equation.
the fact that this was the point (x2,y2) in the slope equation is not relevant.
this is a new equation and we can assign any point on the line to (x1,y1) in this new equation, even if it was the point (x2,y2) in the slope equation.
the assignment of points to (x1,y1) or (x2,y2) is relevant only to the equation in which they are used.
the only requirement is that the points chosen have to be unique points that are on the line.
so we assigned the point (2,-2) to (x1,y1) in the point slope form of the equation of y-y1 = m(x-x1) to get:
y--2 = m(x-2)
this becomes:
y+2 = m(x-2) because y--2 is equal to y+2 (minus of a minus is a plus).
you know that m = -8/7 from the slope equation we worked on before.
the equation now becomes:
y+2 = -8/7 * (x-2)
simplify to get:
y + 2 = -8/7 * x - 2 * (-8/7)
simplify this to get:
y + 2 = -8/7 * x + 16/7
subtract 2 from both sides of this equation to get:
y = -8/7 * x + 16/7 - 2
this is equivalent to:
y = -8/7 * x + 16/7 - 14/7
simplify to get
y = -8/7 * x + 2/7
this can also be written as:
y = (-8/7)x + (2/7)
if this equation is good, then both points should be solutions to this equation.
we'll try (x,y) = (-5,6) to test this out.
this time we are replacing x with -5 and y with 6 to evaluate the equation.
equation of:
y = (-8/7)x + (2/7) becomes:
6 = (-8/7)(-5) + (2/7) which becomes:
6 = 40/7 + 2/7 which becomes:
6 = 42/7 which becomes:
6 = 6
the equation is true when x = -5 and y = 6.
it will also be true when x = 2 and y = -2.
you can do that for yourself.
the graph of the equation is shown below:
horizontal lines were drawn at y = 6 and y = -2
if you trace a vertical line down from the intersection of those horizontal lines with the line of the equation, you will see that x = -5 and x = 2 respectively when y = 6 and y = -2.
the easiest way to trace a vertical line is to take a rectangular piece of paper and slide the horizontal edge of the paper along the bottom of your screen until the vertical edge of your paper lines up with the intersection points. trace down or up the vertical edge of the paper until you see the number that's on the x-axis.
|
|
|
