lingkaran

Persamaan lingkaran

Persamaan lingkaran yang mempunyai diameter ruas garis AB, dengan A(x_{1},y_{1}) dan B(x_{2},y_{2}) adalah (x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0

Bukti:

misal pusat lingkaran adalah P, maka titik P ditengah ruas garis AB, P\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right), karena ruas garis AB adalah diamter, maka 2r=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}.sehingga r^{2}=\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{4}.

Persamaan lingkaran dengan pusat P dan jari-jari r adalah:

(x-\frac{x_{1}+x_{2}}{2})^{2}+(y-\frac{y_{1}+y_{2}}{2})^{2}=r^{2}

(x-\frac{x_{1}+x_{2}}{2})^{2}+(y-\frac{y_{1}+y_{2}}{2})^{2}=\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{4}

x^{2}-(x_{1}+x_{2})x+(\frac{x_{1}+x_{2}}{2})^{2}+y^{2}-(y_{1}+y_{2})y+(\frac {y_{1}+y_{2}}{2})^{2}= \frac{(x_{1}-x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(x_{1}-x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}-(\frac {x_{1}+x_{2}}{2})^{2}-(\frac {y_{1}+y_{2}}{2})^{2}

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(x_{1}-x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}-\frac {(x_{1}+x_{2})^{2}}{4}-\frac {(y_{1}+y_{2})^{2}}{4}

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(x_{1}-x_{2})^{2}}{4}-\frac {(x_{1}+x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}-\frac {(y_{1}+y_{2})^{2}}{4}

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y = \frac{(x_{1}-x_{2}+x_{1}+x_{2})(x_{1}-x_{2}-x_{1}-x_{2})}{4}+\frac{(y_{1}-y_{2}+y_{1}+y_{2})(y_{1}-y_{2}-y_{1}-y_{2})}{4}

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(2x_{1})(-2x_{2})}{4}+\frac{(2y_{1})(-2y_{2})}{4}

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y=(-x_{1}x_{2})+(-y_{1}.y_{2})

x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y+(x_{1}x_{2})+(y_{1}.y_{2})=0

x^{2}-(x_{1}+x_{2})x+(x_{1}x_{2})+y^{2}-(y_{1}+y_{2})y+(y_{1}.y_{2})=0

(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0

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Soal kombinatorik (1)

Berikut ini soal-soal dan pembahasan kombinatorik yang dapat digunakan untuk menambah wawasan tentang konsep -konsep kombinatorik.
1. Buktikan bahwa _{n+1}C_{r}= _{n}C_{r-1}+_{n}C_{r}

Bukti :

kita gunakan rumus kombinasi _{n}C_{r}= \frac{n!}{(n-r)!.r!}
Sehingga :

_{n}C_{r-1}= \frac{n!}{(n-(r-1))!.(r-1)!}

_{n}C_{r-1}= \frac{n!}{(n+1-r)!.(r-1)!}

_{n}C_{r-1}= \frac{n!.r}{(n+1-r)(n-r)!.r.(r-1)!}

_{n}C_{r-1}= \frac{r}{(n+1-r)}.\frac{n!}{(n-r)!.r!}

sehingga :
_{n}C_{r-1}+_{n}C_{r}= \frac{r}{(n+1-r)}.\frac{n!}{(n-r)!.r!}+\frac{n!}{(n-r)!.r!}

_{n}C_{r-1}+_{n}C_{r}= \left(\frac{r}{(n+1-r)}+1\right).\frac{n!}{(n-r)!.r!}

_{n}C_{r-1}+_{n}C_{r}= \left(\frac{r+n+1-r}{(n+1-r)}\right).\frac{n!}{(n-r)!.r!}

_{n}C_{r-1}+_{n}C_{r}= \left(\frac{n+1}{(n+1-r)}\right).\frac{n!}{(n-r)!.r!}

_{n}C_{r-1}+_{n}C_{r}= \frac{(n+1).n!}{(n+1-r).(n-r)!.r!}

_{n}C_{r-1}+_{n}C_{r}= \frac{(n+1)!}{(n+1-r)!.r!}

_{n}C_{r-1}+_{n}C_{r}=_{n+1}C_{r}

2. Buktikan bahwa _{n}C_{r+1}= \frac{n-r}{r+1}._{n}C_{r}

Bukti :
_{n}C_{r+1}=\frac{n!}{(n-(r+1))!.(r+1)!}

=\frac{n!}{((n-1)-r)!.(r+1).r!}

=\frac{n!}{((n-1)-r)!.(r+1).r!}.\frac{n-r}{n-r}

= \frac{(n-r).n!}{(r+1).(n-r).((n-1)-r)!.r!}

=\frac{n-r}{r+1}.\frac{n!}{(n-r).((n-1)-r)!.r!}

=\frac{n-r}{r+1}.\frac{n!}{(n-r)!.r!}

=\frac{n-r}{r+1}._{n}C_{r}

Jadi terbukti bahwa _{n}C_{r+1}= \frac{n-r}{r+1}._{n}C_{r}