急F4 Equations of Straight Line

27a) The slope ofL1 = -(5/-1)=5The y-intercept of L1 =-(10/-1)=10 b)Let the coordinates of the intersect point of L1 and L2be (0,y)Sub (0,y) into equation of L1-y+10=0y=10∴the coordinates of the intersect point of L1 and L2 is (0,10)∵L2⊥L1∴slope of L2 =-1/5∴The equation of L2 isy-10=(-1/5)(x)5y-50=-xx+5y-50=0 28a)Let the coordinates of P and Q are (0,y) and (x,0)respectivelySub (0,y) into -2x+5y=105y=10y=2∴P(0,2)Sub (x,0) into -2x+5y=10-2x=10x=-5∴Q(-5,0) b)∵QR=9∴The coordinates of R =(9-5,0)=(4,0)The equation of l2 is y-0=[(0-2)/(4-0)](x-4)2y=-x+4x+2y-4=0
29a)Let the coordinates of A and B be (x,0) and (0,y)respectivelySub (x,0) into L5x-60=05x=60x=12∴A(12,0) Sub (0,y) into L-12y-60=0-12y=60y=-5∴B(0,-5) b)The perimeter of triangle OAB=√[(12^2)+(5^2)]+12+5=13+17=30

2011-09-09 18:45:15 補充：
The perimeter of triangle OAB


={√[(12^2)+(5^2)]}+12+5


=13+17


=30