f.3 Coordinate Geometry

2011-04-13 5:38 am

1.it is given that A(a,0),B(0,3),and C(3,40 are three vertices of square ABCD.If A lies on the positive x-axis,find the value of a

2.Three points A(-3,4),B(5,-2) and P(3,K) are given.If the slope of AB is equal to the sum of the slopes of AP and BP,find the value of K

回答 (3)

用戶7639

2011-04-13 6:17 am

✔ 最佳答案

圖片參考：http://i.imgur.com/p98TF.gif

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pingshek

2011-04-13 5:55 pm

1.
Is C (3, 4) or (3, 40) ?

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[匿名]

2011-04-13 6:10 am

Answer:

1. the gradient/ slope of BC is (4-3)/(3-0) = 1/3
as it is a square, the slope of AB/AC x the slope of BC = -1
the slope of AB/AC= -3

AB: y-3 = -3 (x-0)
substitue x as 0 because point A is lying on the x-axis
that gives you -3x=-3 , x=1

AC: y-4 = -3 (x-3)
again substitute x as 0
This results in -4= -3x+9, x=13/3

There are two outcomes.!?

However, because ABCD is a square, which means the four sides are equal in length.

length of BC= ((3-0)^2+(4-3)^2)^(1/2) = 10^(1/2)

length of AC= ((4-13/3)^2+(3-0)^2)^(1/2) =9.1111111111^ (1/2)

length of AB= ((3-0)^2+(0-1)^2)^(1/2)= 10^(1/2)

Because BC=AB so the coordinates of A are (1,0), a=1

2. Slope of AB = (-2-4)/(5-(-3)) = -3/4

Slope of AP= (k-4)/((3-(-3))= (k-4)/6

Slope of BP= (k-(-2))/(3-5)= -(K+2)/2

Slope of AB = Slope of AP + Slope of BP

-3/4=(k-4)/6-(k+2)/2 multiply the equation by 12
-9=2(k-2)-6(k+2)
-9 = 2k-4-6k-12
4k= -11
k= -11/4

The value of K is -11/4

參考: me

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