Msaada wa maswali haya ya Hesabu

sam-Akida

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Write the distance of any point of a line Y=mx + c from the origin as a function of x , hence find the coordinate of the point on the line Y=2x + 5 which is closest to the origin.
 
Refer to C.J Tranter(red cover) topic coodnate geometry also, unaweza kugugooo
 
kuna kiolevel na Ki-Alevel,wataka njia gani?
Ya O level ni hii
Swali linataka tutafute kanuni ya kutafuta umbali kutoka origin kwenda kwenye point yoyote ya kwenye mstari ambao kanuni yake ni y=mx+c
ila inahitajika kanuni hiyo isiwe na y,iwe na x tu.

1)Origin=(0,0)

2)Point yoyote kwenye line ni (x,y)

3)badala ya y weka mx+c ili point iwe (x,mx+c)

4)sasa tunatafuta umbali kutoka (0,0) kwenda (x,mx+c)
formula

5)distance=square root ya [ (x-o)[SUP]2[/SUP]+(mx+c-0)[SUP]2[/SUP]] ,ukisimpilfy unapata

distance=square root ya [ (x)[SUP]2[/SUP] +(mx+c)[SUP]2[/SUP]]

6)ukiangalia hiyo formula haina herufi y hata kidogo ,na ndio maana ya yeye kukuambia distance iwe in terms of x only,ina x tu

ZINGATIA:
hizo m na c usiziangalie saana. Ukihitaji maelezo uni-pm.

KIPENGELE CHA PILI
sasa hivi tumepewa kanuni ambayo m na c vinajulikana na tunaambiwa tutafute umbali toka origin kwenda kwenye hiyo line,

1)kwenye m weka 2,kwenye c weka 5
2)distance=square root ya[(x)[SUP]2[/SUP] +(2x+5)[SUP]2],ukirahisisha unapata
3)distance=square root ya(5xsquare[/SUP]
[SUP]+20x+25) hiyo mbili mbele ya 5x2 ni square
4)ikumbukwe o,o haibadiliki lakini point ya kwenye line inawezakuwa yoyote hivyo umbali unaweza kuwa wowote kutokana na point utakayoitaka wewe
5)swali linataka tuseme ni point gani ya line iko karibu zaidi na origin
6)kanuni hapo ndani ni Quadratic yenye a=5,b=20,c=25
7) value ya x itakayotupa minimum distance ni x=-b/2a (kwa sababu a ni positive)

8)x=-20/(2x5) unapata x=-2.
9)ili kujua value ya y tumia y=2x+5
10)y=2 times -2 +5 upate y=1
11) point yenyewe ni (-2,1)
CONCLUSION
LINE y=2x+5 ina point nyingi sana ila point ambayo iko karibu zaidi na (0,0) ni point (-2,1)
samahani kwa maelezo mareeefu.

[/SUP]
 
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