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distance formula for point and line: given the point (x,y) and line in the form Ax + By = C the distance is given by: d = |Ax + By + C| / sqrt(A^2 + B^2)
|dw:1433039250805:dw| just to make it a little more clear
ACTUALLY, WAIT, just a small correction! the line should be in the form Ax + By + C = 0 but everything else is correct
Find the equation of the line that is perpendicular to 2x-3y+2=0 and this perpendicular line goes through this point |dw:1433039456627:dw|
mark the point of intersection between these two lines as point P |dw:1433039484436:dw|
|dw:1433039495540:dw| the answer you want is the distance d (use the distance formula to compute that distance)
Thank you guys so much I'm I think I'm going something wrong with the calculation Idk what is it
what are you getting for the equation of the perpendicular line?
13 is a number, not an equation
I'm not even sure
I give up I've no idea how to do it
2x-3y+2=0 2x-3y+2-2=0-2 2x-3y = -2
swap the coefficients on 2x-3y = -2 to get -3x+2y = -2 then make the x coefficient positive to get 3x+2y = -2 3x+2y = -2 is perpendicular to 2x-3y = -2, but we don't know if the perpendicular line goes through (5,6)
Any equation perpendicular to the given line is of this form 3x+2y = C where C is any real number plug in (x,y) = (5,6) 3x+2y = C 3(5)+2(6) = C 27 = C C = 27 so the equation 3x+2y = 27 is perpendicular to the given equation AND it goes through (5,6) |dw:1433040359530:dw|
now solve this system of equations 2x-3y = -2 3x+2y = 27 to find the intersection point P |dw:1433040425688:dw|