Write the equation of the line that is parallel to the line 4x - 3y = -12 and passes through the point (-3, 4).
Don't know how to do this one!
Write the equation of the line that is parallel to the line y = -3x + 12 and passes through the point (-1, 6). Need help with this one too
equation of line parallel to given line has the same slope
I would just start by changing the format to y=mx+b
I don't understand.
\[\Large \bf y_{}-y_{1}=m(x_{}-x_{1}) ~~\\~~m: Slope\] \[\Large \bf (x_{1},y_{1})=(-3,4)\]
would it be y = 4x - 4
I'm so confused.
the first question what is the slope ?
rvc's format works too but y= 4x-4 is wrong First you subtract 4x from both sides and then divide everything by -3 for (y=mx+b)
The logic behind these types of problems is that whatever you do on one side of the equal sign you do on the other
yeah
oh whoops I forgot to divide everything. y= 4/3x + 8?
8 or 3?
4x - 3y = -12 subtract 4x from both sides -3y = -4x - 12 divide everything by -3 on both sides y = (4/3)x + (-12/-3) so... y= (4/3)x + 4 That is the original formula. To get the parallel line you rewrite it as y=(4/3)x + b Plug in the provided coordinate (-3,4)
\[4x-3y=12-->3y=4x-12-->y=\frac{4}{3}x-\frac{12}{3}\]
Okay I got it for both, thank you both! I wish I could give two medals.
os is lagging :(
It really always lags a lot.
All the best!
One way to do these problems is to recognize that the slope is determined by the coefficients of \(x\) and \(y\), and a parallel line has the same slope. Therefore, all you need to do is plug in your point in the original left hand side and find the new constant term: \[4x-3y=-12\]passing through point \((-3,4)\): \[4(-3)-3(4) = -12-12 = -24\]so the new equation is simply \[4x-3y=-24\] and if you solve that for \(y\) to put it in \(y = mx+b\) form (slope-intercept), you get \[-3y=-24-4x\]\[y = \frac{4}{3}x+8\]
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