Find the equation of the line perpendicular to the line4x+8y=16 and through the midpoint of AB where A(-8,12) and B(6,10) Please go step by step.
can you fist find the midpoint of AB ?
*first
do u know the midpoint formula?
using the midpoint formula, yes (-1, 11)
ok... good... now that we have a point, we need the slope of the line we need...
it needs to be PERPENDICULAR to the line 4x + 8y = 16. can you find the slope of this line?
should be -1/2
yes.... that's the slope of 4x + 8y = 16..... now the line needs to be perpendicular to a slope of -1/2.. what slope would that be?
2/1 or just 2
corrrect... :)
so we have a point, (-1, 11) and the slope, m=2 we can put that info in point-slope form: \(\large y-y_1=m(x-x_1) \)
what form does the line need to be in?
standard form? (like the given line in the original post?)
it is in a different form i think the last answer was x-2y=10 which im not sure what form that is
ok.... stick the info into point-slope form to find the equation of the line we need. after finding that, we can put that in standard form....
ok then its y-11=2(x+1)
yes.... carry out distributive property on the right side to get: y - 11 = 2x + 2 to get it in standard form bring everything with a variable to the left side and move all constants to the right...
ie, move the 2x to the left and move the -11 to the right.
2x+y=13
right?
the answer is x+2y=21 have no clue how the book got that
I'll have to check over the steps
4x+8y=16 4(x + 2y) = 4(4) x + 2y = 4 2y = 4 - x y = (4 - x)/2 y = 2 - x/2 The slope of the original line is definitely -1/2 so the slope of the perpendicular line should be 2.
Looks like your book is wrong.
wow. and I paid for that book too....figures
does it always have to be in standard form? can you run that past me again?
No, there are other forms. Slope Intercept is another form.
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