find an equation of line perpendicular to line x-7y=9 and point(5,4)?

Answer 1

We will put this in the general form of a line y = mx + b

x - 7y = 9

7y = x - 9

y = (1/7)x - 9/7

We have m = 1/7. This is the slope of the line. A line perpendicular to this will have a slope -1/m = -7

We can construct a line through (5,4) with this slope by the following formula

y - y1 = m(x - x1)

y - 4 = -7(x - 5)

y - 4 = -7x + 35

y = -7x + 39

Answer 2

Lets get the equation in slope intercept form

x - 7y = 9
-7y = -x + 9
y = 1/7x - 9/7

Line perpendicular with have negative receiprical slope

So the slope will be y = -7y + b

So we have a point on the new line - substitute

4 = -7(5) + b
4 = -35 + b
39 = b

So the line that is perpendicular will be :
y = -7x + 39

Answer 3

the slope of this line is:
Ax + By = C

slope = -A/B = -1/-7 = 1/7

the slope of the line perpendicular is the negative reciprocal:
-7

the point is (5,4)

using point-slope form:
y-4 = -7 ( x-5)

if you wanted y-intercept form:
y = -7x +39

Answer 4

x-7y=9
y=(x-9)/7

Perpendicular will be opposite of the slope:

y = - 7x +n

As it crosses in point (5,4):

4= -7*5 + n
39=n

y = - 7x + 39

Answer 5

change the eq into slope intercept form
y = 1/7x -9

so eq of line perpendicular to that has a slope of -7

m1*m2 = -1

(y-4)/(x-5)=-7
y-4=-7x +35
y = -7x + 39

Answer 6

First find m1, the slope of this equation.
Then m1*m2= -1, and find m2
Once found m2, then use y=mx + b, where m is m2, and sub point (5,4) to find b,
once found b, together with m2 your get
y=mx + b

Answer 7

y=1/7 +33/7

Answer 8

y=-x/7+9 is simplified, so the line perpendicular is y=7/x, as for point (5,4) idk what u want.

Answer 9

u need more detail in your question