<PRE><B>Question 870657</B>
A triangle is formed by the points A(-1,3), B(5,7 and C(0,8)
(a) show that angle ACB is a right angle
(b) Find the coordinates of the point where the line through B parallel to AC cuts the x-axis. 
Answer by distance formula 
(A)
AB=square root of (5+1)^2+ (7-3)^2 =36 +16 =SQUARE ROOT OF 52
AC=SQUARE ROOT OF (0+1)^2 +(8-3)^2 = 1+ 25 = SQUARE ROOT OF 26
BC =SQUARE ROOT OF (0-5)^2 +(8-7)^2 = 25 + 1= SQUARE ROOT OF 26
BY USING PYTHAGORAS THEOREM FOR RIGHT TRIANGLE 
 (SQUARE ROOT OF 52)^2 =(SQUARE ROOT OF 26)^2 + ( SQUARE ROOT OF 26)^2
52 = 26 +26
52 = 52 PYTHAGORAS THEOREM SATISFIED THEREFORE WE CAN SAY ANGLE  ACB  IS RIGHT ANGLE  PROVED 
(B)
SLOP OF AC =8-3/0+1 = 5/1
SLOPE OF LINE PASSING THROUGH B AND INTERSECT X AXIS LET COORDINATE AT X AXIS IS (X,0)
SLOPE OF LINE = 0-7/X-5 =-7/X-5
SLOPE OF PARALLEL LINES ARE ALWAYS EQUAL
-7/X-5 =5/1
-7 =5(X-5)
-7= 5X-25
-7+25 =5X
18=5X
X=18/5= 3.6
COORDINATES ARE (3.6,0) AT X AXIS 

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