有3個步驟又答晒的為最佳
Find unknowns
L1 cuts the x-axis at 5 and the y-axis at 2.
L2 passes through R(1, -2 ) and S(a,3)
L1 ┴ L2
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whether PQ//RS OR PQ┴RS or neither
1 , P(2,5) , Q (5,-1) and M rs = 2
2 , P(k,k) , Q(2k,3k) , R(a-2,3 ) , S(a+2 , 1 )
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Let P(5,2), Q(8,1) and R(7,8) .
1 , Show that 三角形 is right-angled
2 , Find the area of 三角形
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Prove that A(2,3), B(3,6) and C(-2,-9) are collinear (ABC is a straight line)
Find also the point where the line ABC cuts the y-axis
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多條數學題 ( f3 ) 20分
2008-02-25 2:24 am
回答 (1)
2008-02-27 3:14 am
✔ 最佳答案
Slope of L1=(2-0)/(0-5)= -2/5Slope of L2=(-2-3)/(1-a)=5/(a-1)
Slope of L1*Slope of L2= -2/(a-1)= -1 (as L1 ┴ L2)
Therefore, a-1=2, i.e. a=3
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Slope of PQ=(5-(-1))/(2-5)= -2
M rs = 2 =/=Slope of PQ
Therefore, PQ is not parallel to RS
M rs *Slope of PQ= -4=/= -1
Therefore, PQ is not perpendicular to RS
Slope of PQ=(3k-k)/(2k-k)=2
Slope of RS=(3-1)/(a-2-(a+2))=2/(-4)= -1/2
Slope of PQ*Slope of RS= -1
Therefore, PQ┴RS
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Slope of PQ=(2-1)/(5-8)= -1/3
Slope of PR=(2-8)/(5-7)=3
Slope of PQ*Slope of PR= -1
Therefore, PQ┴PR,i.e. Angle QPR=90 degrees
Hence triangle PQR is right-angled at angle QPR
Length of PQ=Square root((8-5)^2+(1-2)^2)=Square root of 10
Length of PR=Square root((7-5)^2+(8-2)^2)=Square root of 40
As triangle PQR is right-angled at angle QPR,
the area of triangle PQR=0.5*Length of PQ*Length of PR=0.5*20=10
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Slope of AB=(6-3)/(3-2)=3
Slope of BC=(-9-6)/(-2-3)=3
Slope of AB=Slope of BC=3
Therefore, AB//BC,i.e.ABC are collinear
Let the equation of the line ABC be y=3x+c
Sub (2,3),3=3*2+c
c= -3
Hence the equation of the line ABC is y=3x-3
Therefore, the y-intercept is 3,i.e.the required point is (0,-3)
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