How do you find an equation of a line perpendicular to y=x at the point (5,5) in standard form?

put the original equation in the form

y = m x + b
in this case
y = 1 x + 0
so m, the original slope, = 1
Now m' = -1/m
so our new slope = -1/1 = -1
so our new function looks like
y = -1 x + b
we need b
use the point
5 = -5 + b
b = 10
so
y = -x + 10

The slope of y = x is 1, so the slope of a perpendicular line must be -1. (This is becasue the product of the slopes must be -1).

The equation can therefore be written
(y-5)/(x-5) = -1
y-5 = -x +5
y = -x +10

To find an equation of a line that is perpendicular to another line, we need to use the fact that the slopes of perpendicular lines are negative reciprocals of each other.

The given line has a slope of 1, since it is in the form y = x. The negative reciprocal of 1 is -1/1, which simplifies to -1.

To find the equation of the line perpendicular to y = x at the point (5,5), we can use the point-slope form of a linear equation:

y - y1 = m(x - x1)

where (x1, y1) represent the coordinates of the given point and m represents the slope.

Plugging in the values, we have:

y - 5 = -1(x - 5)

Expanding this equation gives us:

y - 5 = -x + 5

Moving the terms around to get the equation in standard form gives us:

x + y = 10

So, the equation of the line perpendicular to y = x at the point (5,5) in standard form is x + y = 10.