Algebra help!?

Question

Passing through (5,-9) and perpendicular to the line whose equation is x+7y-12=0? I am not sure how to answer this and I have tons of these problems....can someone please help? Thank you in advance!

Answer 1

to get equation of straight line you need just to know
1) slope of line
2) point on the line
for your problem
slope of given line represented by x+7y-12=0
slope -a/b ( eq is in form ax +by +c = 0 ) a is coeff. of x , b is coeff of y
so slope of this line = -1/7
slope of line perpendicular to it = ( - reciprocal)
so slope of required line = 7
( 5 , -9) slope = 7
equation is formed like that :
(y-y1) = m(x-x1)
such that : m = slope , ( x1,y1) given point
( y -(-9) ) = 7 ( x - 5 )
( y +9 ) = 7x-35
equation is y = 7x - 44
or in general form 7x - y - 44 = 0
hope that helppppsss

Answer 2

First, put your equation into y = mx + b form:

y = -x/7 + 12/7

This gives you the slope of this line: -1/7. A line perpendicular to this line would have a slope of 7 (negative reciprocal):

y = 7x + b

Now, plug in the point in question (5, -9) and solve for b:

-9 = 7(5) + b
-9 = 35 + b
b = -44

So, your final equation would be:

y = 7x - 44

Answer 3

write the slope form eq for the line passing thru (5,-9) as y-y1=m(x-x1) where x1, y1 are the first points of x and y co-ordinates mis the slope.=> y+9=m(x-5) -------- (1)
The 2nd eq is x+7y-12=0 ---------- (2) slope= -(x cordinate)/(y coordinate) = -1/7
If (1) & (2) are perpendicular, the products of their slopes must be -1.So slope for (1) would be 7. put this in
(1) =>7x-y-44=0.

Answer 4

For increasing cubic binomials the final formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Answer 5

you just need to find that line i.e (5,-9) in the qudratic plane ,then you can easily solve this.