# Analytic geometry:perpendicular to y=7x+1 and passing at a diistance square root of 2 from (4,-2)

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Perpendicular to line y=7x+1 (rearranging it) -> 7x-y=1

for a line to be perpendicular just interchange the constant of the 2 variables and change the sign of y

x+7y+P=0

now we have to solve for P using the formula of distance from a line to a point

d=| (Ax_{1} + By_{1} + P)/(sqrt(A^2 + B^2)) | (absolute value)

where:

d=sqrt2

x_{1}=4

y_{1}=-2

A=1

B=7

sqrt2=| { (1*4) + (7*-2) + P }/(sqrt50) |

solving for P gives you P=20 so the line is x+7y+20=0

for a line to be perpendicular just interchange the constant of the 2 variables and change the sign of y

x+7y+P=0

now we have to solve for P using the formula of distance from a line to a point

d=| (Ax_{1} + By_{1} + P)/(sqrt(A^2 + B^2)) | (absolute value)

where:

d=sqrt2

x_{1}=4

y_{1}=-2

A=1

B=7

sqrt2=| { (1*4) + (7*-2) + P }/(sqrt50) |

solving for P gives you P=20 so the line is x+7y+20=0