Question 550800: 1.What point on x-axis is equidistance from the points(6,7) and (4,-3)?
2.Find the equations of the diagonals of a rectangle whose sides are x+1=0,x-4=0,y+1=0 and y-2=0.
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website!
let Q be the point on the x axis (x,0)
Let P be (6,7)
distance PQ
d^2=((7-0)^2+(6-x)^2
d^2=49+36-12x+x^2
d^2=85-12x+x^2
Let the second point be R (4,-3)
Distance RQ=
d^2=(-3-0)^2+(4-x)^2
d^2=9+16-8x+x^2
D(RQ)=d(PQ)
square
85-12x+x^2=9+16-8x+x^2
60=4x
x=15
point Q(15,0)
CHECK
d^2(RQ)=(-3-0)^2+(4-15)^2
d^2(RQ)=9+121=130
d^2(PQ)=49+81=130
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2.Find the equations of the diagonals of a rectangle whose sides are x+1=0,x-4=0,y+1=0 and y-2=0.
the lines are x=-1, y=-1, x=4, y=2
intersection points of the lines are
(-1,-1),(-1,2),(4,2),(4,-1)
Calculate the slope of the line passing through the points on the diagoinal
(-1,-1) & (4,2)
x1 y1 x2 y2
-1 -1 4 2
slope m = (y2-y1)/(x2-x1)
( 2 - -1 )/( 4 - -1 )
( 3 / 5 )
m= 3/ 5
Plug value of the slope and point ( -1 , -1 ) in
Y = m x + b
-1.00 = - 3/5 + b
b= -1 - - 3/5
b= - 2/ 5
So the equation will be
Y = 3/5 x - 2/5
m.ananth@hotmail.ca
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