Hi,
I need help with this coursework question please.
y - -5 = -1/3 (x-1)
I know I need to minus 5 from each side but the minus one-third is confusing me when multiplying out the brackets.
Would I have to times each side by -3?
Hi,
I need help with this coursework question please.
y - -5 = -1/3 (x-1)
I know I need to minus 5 from each side but the minus one-third is confusing me when multiplying out the brackets.
Would I have to times each side by -3?
Hi mrszone,
I'm not really clear what you are trying to accomplish. It looks to me like you have an equation of a line in point-slope form. A point on the line is (1, -5) and the slope is $\displaystyle -\frac{1}{3}$
Simplifying, you get:
$\displaystyle y+5=-\frac{1}{3}(x-1)$
What else do you want to do?
Do you need to convert it to another form like standard or slope-intercept form?
A "perpedicular bisector" is a line which is perpendicular to a given segment, and cuts that segment at its midpoint. If the equation you posted represents the line which contains the segment, we can find the perpendicular slope for the bisector.
But until we know what the segment is that you're supposed to bisect, I see no way to proceed. Sorry!
This is a fuller description of what I have so far:
Find the equation of the perpendicular bisector of AB when
A=(3,1) B=(-1,-11)
This is what I have worked out so far:
-1 - 3 = -4 which is the run
-11 - 1 = -12 which is the rise
-12 / -4 = 3 which is the slope
the negative reciprocal is -1/3
the midpoint is (1/2(3+-1)), (1/2(1+-11))
which gives you a midpoint of 1,-5
this leaves an equation of y - -5 = -1/3(x-1)
y + 5 = -1/3(x-1)
all the exercise examples show the final equation after multiplying out the brackets and this is what I am trying to do.
So, all you want to do is put this equation in another form, right?
$\displaystyle y+5=-\frac{1}{3}(x-1)$
$\displaystyle y=-\frac{1}{3}x+\frac{1}{3}-5$
$\displaystyle y=-\frac{1}{3}x-\frac{14}{3}$
The above is in 'Slope-intercept form'. To convert to Standard form,
$\displaystyle y=-\frac{1}{3}x-\frac{14}{3}$
$\displaystyle 3y=-x-14$
$\displaystyle x+3y=-14$