i have a stupid math ?...?

Question

What is the equation of the line containing the given point and perpendicular to the given line?

(0,2); 8x+5y=7

Answer 1

first, there is NO stupid math question...! and people are !

Answer 2

the equation of a line can be determined if either 2 points are given or any 1point and the slope of the line are given. here in ur question,
1 point is given and slope can be found out.

first we find the slope of the line 8x+5y=7 by reducing it into y = mx +c,
8x+5y =7
or 5y = - 8x +7
or y = -8/5 x + 7/5
here m = -8/5
this is the slope of the line 8x +5y =7.

let m1 be the slope of the line perpendicular to the given line.
then to be perpendicular for these 2 lines, following conditions should be satisfied
m * m1 = -1
or m1 = -1/m
or m1 = -1(-8/5)
or m1 = 5/8

now we have to determine the equation of a line which has slope 5/8 and which passes through (0,2)
general relation is
y-y1 = m1(x-x1)
or y - 2 = 5/8 (x - 0)
or y -2 = 5/8 x
or 8y - 16 = 5x
or 5x -8y+16 = 0
this is the required equation

Answer 3

the equation is: Y= -8x/5 +2

8x+5y=7
5y= -8x+7
y= -8x/5 + 7/5
y = -8x/5 + b
then substutute the points (0,2)
2= -8/5(0) +b
2= 0+b
+2=b
Y = -8/5x +2

Answer 4

First, you need to set it up to slope-intercept form:
y=mx+b

For the 8x+5y=7 problem, you need to subtract 8x on both sides:
5y=-8x+7

Divide 5 on both sides:
y=-8/5x+7/5

Since m=-8/5, the inverse for that would be 5/8

Next, you want to use the point slope form:
y-y1=m(x-x1)

y-2=5/8(x-0)
y-2=5/8x
y=5/8x+2

Now, you should be able to take both problems and graph them to see if they intersect.

I hope this helps!

Answer 5

to be perpendicular, slopes must be negative recipricals
8x+5y=7
5y=-8x+7
y=-1.6x+1.4 slope is -1.6
neg recip will be -(1/-1.6)=.625
y=.625x+b substitute 0,2 for x,y
2=.625*0+b
b=2

equation is
y=.625x+2

Answer 6

5y=7-8x
y=7/5-8/5x
y-(2)=5/8(x-(0))
y-2=5/8x
y=5/8x+2