Question

The linear equation y=mx+b contains the point (-5, 5) and is perpendicular to the line 5x+25y=11.

Answer 1

The linear equation y=mx+b containing the point (-5, 5) and perpendicular to the line 5x+25y=11 is y = -5/11 x + 30/11

The formula for finding the equation of a line in point-slope form is expressed as:

y-y0 = m(x-x0) where:

m is the slope

(x0, y0) is any point on the line

Given the following

(x0, y0) = (-5, 5)

Get the slope of the given line 5x+25y=11.

Rewrite in the form y = mx+b

25y = -5x + 11

y = -5/11 x + 11/25

Compare with the original equation

mx = -5/11 x

m = -5/11

Get the required equation

y - 5 = -5/11(x-(-5))

11(y-5) = -5(x+5)

11y-55 = -5x - 25

11y+5x = -25 + 55

11y+5x = 30

11y = -5x + 30

y = -5/11 x + 30/11

Hence the linear equation y=mx+b containing the point (-5, 5) and perpendicular to the line 5x+25y=11 is y = -5/11 x + 30/11

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