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Formula of a line passing through (-5,-2) and x-7y=- 21

2007-01-23 18:01:16 · 6 answers · asked by Rubyx 2 in Science & Mathematics Mathematics

6 answers

maybe you mean it is parallel to the line given and passes through that point? if so then:
you would find the slope of the line given by getting it to y=mx+b form
7y=x+21
y=1/7 x+3
so the slope =1/7
then use pt slope form (y-ysub1)=m(x-xsub1)
to get (y+2)=1/7(x+5)
there you go (if its supposed to be perpendicular you would swap the 1/7 out for its opp reciprocal, -7)

2007-01-23 18:12:27 · answer #1 · answered by ☞danbighands☜ 3 · 2 0

i guess your problem is find the equation/formula of a line passing through (-5,-2) and is parallel to x-7y= -21.

if its the case...
find slope of the given line, that is y = x/7 + 3 hence, slope is 1/7
then substitute -5 for x and -2 for y in y+mx + b to find the y-intercept,as shown here,

-2 = (1/7)(-5) + b
b = 5/7 - 2 = - 9/7

so the equation is: y = x/7 - 9/7 or
7y = x - 9.
or x - 7y = 9.

i hope this helps!

2007-01-24 02:19:28 · answer #2 · answered by 13angus13 3 · 0 0

Line parallel to x - 7y = -21 and passing through (-5,-2).

Slope of given line is m = 1/ 7
so equation of line will be
(y + 2 )= 1/7 (x + 5 )

If u need a line perpendicular to the given line x - 7y = -21 and passing through (-5,-2).

then use slope m = -7
Thus equation of line will be
(y+2 ) = -7*(x+5)

// Remember Equation of line passing through point( a, b) and having slope M is
(y - b) = M*(x - a)

2007-01-24 02:26:13 · answer #3 · answered by @rrsu 4 · 1 0

You have a serious problem.

The formula that you give already specifies a line. Unfortunately, the point that you give isn't on the line. You can show that by substitution:

-5 -7(-2) = -21
-5 +14 = 9 != -21

2007-01-24 02:05:10 · answer #4 · answered by tony1athome 5 · 0 1

I got your problem :)
It is simply to find equation of a line passing through point (-5,-2) and line
x-7y =-21
take the line first
We can solve this by fistly finding any point passing through line x-7y =-21
Put y =1
So x = -14

So we have got another point as A(-14,1) along with B(-5,-2)

So now we can find equation of the required line!!!
Hence,
y-y1 = m(x-x1)
m= (y2-y1)/(x2-x1)
= (-2-1)/(-5+14)
= -(1/3)

So equation is
y-1 = (-1/3) (x+14)
So,
3y - 3 = -x-14
So,
3y + x +11 =0
Is the required equation... :)

2007-01-24 03:04:51 · answer #5 · answered by Amit B 2 · 0 0

how about if you close that book turn off your computer and go to sleep now ..
And for tomorrow just try to sit next to the smartest kids in the class..

2007-01-24 02:15:57 · answer #6 · answered by Anonymous · 0 2

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