Question 550800


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			
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