Consider the line whose equation is 4x – 8y = 8. Write the equation of the line, in point-slope form, that is perpendicular to the line and passes through the point (1, 2).
Point slope form is defined as y - y1 = m(x - x1)
Find the slope which is y2-y1/x2-x1
1/2
Plug your slope back into the equation
y=y1=1/2(x-x1)
I find it easier to use standard form which is y = mx+b
m = slope
@satellite73 help i am confused
We will start over... have you found y?
2?
Rewrite the equation in standard form as y =
2=1/2x-1?
No... it should be in the form y = mx+b
y=1/2x-1
Here is how u want to approach this. I can't tell you are close, just little mistakes here and there I sounds like.
That is what I was trying to explain. Thanks Demo
this is the question right Write the equation of the line, in point-slope form, that is perpendicular to the line and passes through the point (1, 2).?
yss
did you get the answer, or do you still need help?
still need help
ok but you did most of it already you found the slope of \[4x – 8y = 8.\] is \(m=\frac{1}{2}\)
the perpendicular line has a slope that is the "negative reciprocal" i.e. flip it and change the sign the negative reciprocal of \(\frac{1}{2}\) is \(-2\)
now the question really becomes find the equation of the line with slope \(-2\) through \((1, 2)\)
for that you use the "point slope" formula \[y-y_1=m(x-x_1)\] which in our case is \[y-2=-2(x-1)\]
to put it in the "slope intercept" form, solve for \(y\) the steps are always the same first multiply out on the right and get \[y-2=-2x+2\] then add \(2\) to finish with \[y=-2x+4\] and that is done
ok thanks i know get it
good not too hard you got another one to try?
i have a differnt one on how to graph a eqation
like the last one?
um like the last one we graphed kinda x+3y<6 that is a greater than or equal sign
\[x+3y\leq 6\] i.e. "less than or equal to" right?
yes
ok lets do this the easy way
if \(x=0\) you get \(3y=6\) or \(y=2\) so we know \((0,2)\) is on the graph, where the line hits the \(y\) axis
if \(y=0\) you get \(x=6\) so you know \((6,0)\) is on the graph, where the line crosses the \(x\) axis lets plot those two points only
|dw:1399526681104:dw|
now since it is \(x+3y\leq 6\) the graph is not just a line but the whole region BELOW the line
shade in all stuff underneath
ok so thats it
yup
k thanks
yw
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