Passing through (5,-6) perpendicular whose equation is x-7y=10 what is

the point slope and the slope intercept form?

Find the slope of the other line:
x-7y = 10
x-7y-x = 10-x
-7y = 10-x
-7y = -x+10
-7y/-7 = (-x+10)/-7
y = (1/7)x-(10/7)
This other line has a slope of 1/7. So the line you have to come up with has a slope that is the negative reciprocal of that. That line's slope will be -7.

Plug in -7 for "m" (the slope) and plug in 5 for "x" and -6 for "y" (from the given point). Solve for "b" (the y-intercept):
y = mx+b
(-6) = (-7)(5)+b
-6 = -35+b
-6+35 = -35+b+35
29 = b

Answer: y=-7x+29

Given equation,
x-7y = 10, ===> -7y = -x +10,
OR,
y = x/7 -10/7, ........................................… [1]
SO, slope, m = 1/7
Slope perpendicular to [1], = -7
Hence,
line passing through (5,-6) perpendicular to x-7y=10, is,

y+6 = -7(x-5) = -7x +35,
OR,
7x +y - 29 = 0 >============================< ANSWER

The normal of x-7y=10 is (1,-7) (since the equation is in the cartesian equation form).

So the vector (point slope) would be:
v = (5, -6) + t(1, -7)

Slope Intercept
Slope = (-7/1) = -7
Point (5, -6)

y=mx+b
-6 = (-7)(5)+b
b = 29

y = -7x+29

point slope: v = (5, -6) + t(1, -7)
slope intercept: y = -7x+29

Hope this helped!
Good Luck!

x - 7y = 10 --> y = (x - 10)/7

Slope of perpendicular = -7

Eqn of perp. line:

y + 6 = -7(x - 5)

y = -7x + 29