SOLUTION: Hello, I'm Alex. Could I please ask your assistance on these two Homeschool American School Problems because I don't have a teacher here to help? (1) What is the equation of th

Question 63653This question is from textbook
: Hello, I'm Alex. Could I please ask your assistance on these two Homeschool American School Problems because I don't have a teacher here to help?
(1) What is the equation of the perpendicular bisector of the line between points (2,2) and (6, 6)?
(2) What is the (a)directrix and the (b)focus of the parabola y = x^2 - 5x + 4?
This question is from textbook

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Hello, I'm Alex. Could I please ask your assistance on these two Homeschool American School Problems because I don't have a teacher here to help?
(1) What is the equation of the perpendicular bisector of the line between points A(2,2)=(X1,Y1) and B(6, 6)=(X2,Y2)?
PERPENDICULAR BISECTOR OF AB GOES THROUGH MIDPOINT OF AB.....SAY C AND IS
PERPENDICULAR TO AB.
COORDINATES OF C = [(X1+X2)/2,(Y1+Y2)/2] = [(2+6)/2,(2+6)/2=(4,4)= (X',Y')
SLOPE OF AB = (Y2-Y1)/(X2-X1)= (6-2)/(6-2)=4/4=1
HENCE SLOPE OF ITS PERPENDICULAR = -1/1 = -1 = M [PRODUCT OF SLOPES OF 2 PERPENDICULAR LINES =-1]
HENCE EQN. OF PERPENDICULAR BISECTOR IS
Y-Y'=M(X-X')
Y-4=-1(X-4)=-X+4
Y=-X+8
(2) What is the (a)directrix and the (b)focus of the parabola y = x^2 - 5x + 4?
Y = [X^2-2*X*(5/2)+(5/2)^2]-(5/2)^2+4
(X-2.5)^2 = (Y+2.25)
COMPARING WITH STD. EQN. OF PARABOLA
(X-H)^2 = 4A(Y-K)
4A=1......A=1/4=0.25
VERTEX = (H,K)......= (2.5,-2.25)
AXIS = X-H=0.......X-2.5=0
FOCUS = [H,K+A].....[2.5,-2.25+0.25]=(2.5,-2)
DIRECTRIX = Y-K+A=0......Y+2.25+0.25=0.......Y+2.5=0
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