What is the equation of a line that is perpendicular to y-3=-1(x-2) and passes through the point (4,-1)

The equation of a line that is perpendicular to y - 3 = -1(x - 2) and passes through the point (4, -1) is y = x - 5

Solution:

Given that we have to find the equation of a line that is perpendicular to y - 3 = -1(x - 2) and passes through the point (4, -1)

The slope intercept form is given as:

y = mx + c eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Given that the line that passes through the point  ( 4, -1 )  and is perpendicular to the line y - 3 = -1(x - 2)

Given line is perpendicular to y - 3 = -1(x - 2)

y - 3 = -1(x - 2)

y - 3 = -x + 2

y = -x + 2 + 3

y = -x + 5

On comparing the above equation with eqn 1, we get slope of line , m = -1

We know that product of slope of a line and slope of line perpendicular to it is -1

-1 x slope of line perpendicular to it = -1

slope of line perpendicular to it = 1

Given point is (4, -1)

Now we have to find the equation of line passing through (4, -1) with slope m = 1

Substitute m = 1 and (x, y) = (4, -1) in eqn 1

-1 = 1(4) + c

-1 = 4 + c

c = -5

Substitute c = -5 and m = 1 in eqn 1 to get required equation of line

y = 1x - 5

y = x - 5

Thus the equation of line is found

step-by-step explanation:

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step-by-step explanation: