help!! And please label step clearly!!!
If the length of the line segment joining the points (2,3) and (k,1-k) is 4, then k=
A.2 B.4 C.0 or 4 D.-2 or 2
A(2, 5) and B(6, -3) are two points. If P is a point lying on the straight line x=y such that AP=PB, then the coordinates of P are
A, (-2, -2) B.(-2,4) C.(1,1) D.(4,1)
[10 points] F.4 General Maths coordinates
2008-02-13 3:10 am
回答 (3)
2008-02-13 3:27 am
2008-02-15 1:50 am
Clear step.
2008-02-13 3:24 am
If the length of the line segment joining the points (2,3) and (k,1-k) is 4, then k=
(2-k)^2+(3-(1-k))^2=4^2
4-4k+k^2+(2+k)^2=16
4-4k+k^2+4+4k+k^2-16=0
2k^2-8=0
k^2-4=0
k^2=4
k=2 or -2
ans: D
設P=(x,y)
AP=PB
(y-5)^2+(x-2)^2=(y+3)^2+(x-6)^2
y^2-10y+25+x^2-4x+4=y^2+6y+9+x^2-12x+36
-10y-4x+29=6y-12x+45
8x+8y-16=0
x+y-2=0----(1)
x=y----(2)
Sub. (2) in (1)
y+y-2=0
2y=2
y=1
sub. y=1 in (2)
x=1
so, the coordinates of P are (1,1)<---ans: C
(2-k)^2+(3-(1-k))^2=4^2
4-4k+k^2+(2+k)^2=16
4-4k+k^2+4+4k+k^2-16=0
2k^2-8=0
k^2-4=0
k^2=4
k=2 or -2
ans: D
設P=(x,y)
AP=PB
(y-5)^2+(x-2)^2=(y+3)^2+(x-6)^2
y^2-10y+25+x^2-4x+4=y^2+6y+9+x^2-12x+36
-10y-4x+29=6y-12x+45
8x+8y-16=0
x+y-2=0----(1)
x=y----(2)
Sub. (2) in (1)
y+y-2=0
2y=2
y=1
sub. y=1 in (2)
x=1
so, the coordinates of P are (1,1)<---ans: C
收錄日期: 2021-04-25 17:19:21
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