In die diagram is die vergelyking van lyn $AF$ $y = -x - 11$ - NSC Mathematics - Question 3 - 2023 - Paper 2
Question 3
In die diagram is die vergelyking van lyn $AF$ $y = -x - 11$. $B$, 'n punt op die $x$-as, is die middelpunt van die reglyn wat $A(-1; t)$ en $C$ verbind. Die inklin... show full transcript
Worked Solution & Example Answer:In die diagram is die vergelyking van lyn $AF$ $y = -x - 11$ - NSC Mathematics - Question 3 - 2023 - Paper 2
Step 1
Waarde van $t$
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Om die waarde van t te bereken, gebruik ons die punt A(−1;t) en substitueer dit in die vergelyking van die lyn y=−x−11 sodat:
t=−(−1)−11
t = 1 - 11 = -10.
Die waarde van t is -10.
Step 2
Grootte van $ ext{α}$
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Die grootte van extα kan bereken word deur die tangens van die helling te neem. Gegewe dat die referensie lyn van A tot F het ons:
an ext{α} = rac{1}{1} = 1.
Daarom, extα=an−1(1)=45exto.
Grootte van extα is 135° (aangesien dit in die tweede kwadrant is).
Step 3
Gradiënt van $AC$ tot die naaste heelgetal
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Die gradiënt van die lyn AC kan bereken word met:
an(63,43exto)=mACmAC=2.
Die gradiënt van AC tot die naaste heelgetal is 2.
Step 4
Bepaal die vergelyking van $AC$ in die vorm $y = mx + k$
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Die lynvergelyking kan geskryf word as y−y1=m(x−x1). Hier is A(−1;−10) en m=2:
y+10=2(x+1)y+10=2x+2y=2x−8.
Die vergelyking van lyn AC is y=2x−8.
Step 5
Koördinate van $C$
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Die koördinate van C is:
ewline y_C = 10.$$
Dus die koördinates van $C$ is $(9; 10)$.
Step 6
Grootte $F ext{ÉD}$
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Die grootte van die hoek FextEˊD kan bepaal word deur die alternatiewe hoeke te gebruik:
extEAB=63,43extoextEextFextEˊD=63,43exto
En FextEˊD is dus 108,43exto.
Step 7
Bepaal die vergelyking van $G$ in sirkel
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Die vergelyking van die sirkel met middelpunt G kan geskryf word as:
(x−xG)2+(y−yG)2=r2.
Hier, ons moet die koördinates van G bepaal. Aangesien G op die x-as is, is die koördinates (−11;19). Dus:
