Diketahui ruas garis AB dengan titik A(-5,4) dan titik B(3,6). Jika titik C terletak pada A B → \overrightarrow{AB} A B <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> sedemikian sehingga A C → : C B → = 3 : 2 \overrightarrow{AC}\ :\ \overrightarrow{CB}\ =\ 3\ :\ 2 A C <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> : C B <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 3 : 2 . Tentukanlah koordinat titik C!

( − 1 5 , 26 5 ) \left(-\frac{1}{5},\frac{26}{5}\right) ( − 5 1 , 5 2 6 )

( 1 5 , 26 5 ) \left(\frac{1}{5},\frac{26}{5}\right) ( 5 1 , 5 2 6 )

( 26 5 , 1 5 ) \left(\frac{26}{5},\frac{1}{5}\right) ( 5 2 6 , 5 1 )

( 26 5 , − 1 5 ) \left(\frac{26}{5},-\frac{1}{5}\right) ( 5 2 6 , − 5 1 )

Jika diketahui Titik F terletak pada DE sedemikian sehingga D F → : F E → = 1 : 3 \overrightarrow{DF}\ :\ \overrightarrow{FE}\ =\ 1\ :\ 3 D F <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> : F E <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 1 : 3 , dan ruas garis DE dengan titik D (1,2) dan E(2,3). Maka nilai dari koordinat titik F adalah...

( 5 4 , 9 4 ) \left(\frac{5}{4},\frac{9}{4}\right) ( 4 5 , 4 9 )

( 9 4 , 5 4 ) \left(\frac{9}{4},\frac{5}{4}\right) ( 4 9 , 4 5 )

( 4 5 , 4 9 ) \left(\frac{4}{5},\frac{4}{9}\right) ( 5 4 , 9 4 )

( 4 9 , 4 5 ) \left(\frac{4}{9},\frac{4}{5}\right) ( 9 4 , 5 4 )

Diketahui segitiga ABC dengan koordinat A(2, 3, 4), B(5, 0, 1), dan C(4, 2, 5). Titik P membagi AB sehingga A P → : A B → = 2 : 3 \overrightarrow{AP\ }\ :\ \overrightarrow{AB}=2:3 A P <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> : A B <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 2 : 3 . Panjang vektor PC adalah

3 2 3\sqrt{2} 3 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

10 \sqrt{10} 1 0 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

15 \sqrt{15} 1 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

20 \sqrt{20} 2 0 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

9 2 9\sqrt{2} 9 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Kolinear

Authored by yayah shulhiyyah

Mathematics

10th Grade

Used 28+ times

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MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Diketahui ruas garis AB dengan titik A(-5,4) dan titik B(3,6). Jika titik C terletak pada $\overrightarrow{AB}$ sedemikian sehingga $\overrightarrow{AC}\ :\ \overrightarrow{CB}\ =\ 3\ :\ 2$ . Tentukanlah koordinat titik C!

$\left(\frac{1}{5},\frac{26}{5}\right)$

$\left(\frac{26}{5},\frac{1}{5}\right)$

$\left(-\frac{1}{5},\frac{26}{5}\right)$

$\left(\frac{26}{5},-\frac{1}{5}\right)$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Jika diketahui Titik F terletak pada DE sedemikian sehingga $\overrightarrow{DF}\ :\ \overrightarrow{FE}\ =\ 1\ :\ 3$ , dan ruas garis DE dengan titik D (1,2) dan E(2,3). Maka nilai dari koordinat titik F adalah...

$\left(\frac{5}{4},\frac{9}{4}\right)$

$\left(\frac{9}{4},\frac{5}{4}\right)$

$\left(\frac{4}{5},\frac{4}{9}\right)$

$\left(\frac{4}{9},\frac{4}{5}\right)$

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Titik A (3, 2, -1) B (1 , -2, 1) C (7, p - 1, -5) segaris. Tentukan nilai p...

-13

-11

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Diketahui segitiga ABC dengan koordinat A(2, 3, 4), B(5, 0, 1), dan C(4, 2, 5). Titik P membagi AB sehingga $\overrightarrow{AP\ }\ :\ \overrightarrow{AB}=2:3$ . Panjang vektor PC adalah

$\sqrt{10}$

$\sqrt{15}$

$\sqrt{20}$

$3\sqrt{2}$

$9\sqrt{2}$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Diketahui A(1,2,3), B(3,3,1), dan C(7,5,−3), Jika A,B, dan C segaris (kolinear), maka $\overrightarrow{AB\ }:\ \overrightarrow{BC}=....$ adalah ⋯⋅

1 : 2

2 : 1

2 : 5

5 : 7

7 : 5

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Nama lain dari kolinear ialah

Vektor posisi

Vektor kesamaan

Vektor negatif

Vektor segaris

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Diketahui: A(3,-1,4) , B(3,-4,6). Titik P membagi AB sehingga AP : PB = 3 : 2 ,maka vektor P= ...

(3, -2, 2)

(3, 2, 2)

(5, -2 , 3)

(5, 2 , 3)

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Kolinear

Diketahui ruas garis AB dengan titik A(-5,4) dan titik B(3,6). Jika titik C terletak pada $\overrightarrow{AB}$ sedemikian sehingga $\overrightarrow{AC}\ :\ \overrightarrow{CB}\ =\ 3\ :\ 2$ . Tentukanlah koordinat titik C!

Jika diketahui Titik F terletak pada DE sedemikian sehingga $\overrightarrow{DF}\ :\ \overrightarrow{FE}\ =\ 1\ :\ 3$ , dan ruas garis DE dengan titik D (1,2) dan E(2,3). Maka nilai dari koordinat titik F adalah...

Titik A (3, 2, -1) B (1 , -2, 1) C (7, p - 1, -5) segaris. Tentukan nilai p...

Diketahui segitiga ABC dengan koordinat A(2, 3, 4), B(5, 0, 1), dan C(4, 2, 5). Titik P membagi AB sehingga $\overrightarrow{AP\ }\ :\ \overrightarrow{AB}=2:3$ . Panjang vektor PC adalah

Diketahui A(1,2,3), B(3,3,1), dan C(7,5,−3), Jika A,B, dan C segaris (kolinear), maka $\overrightarrow{AB\ }:\ \overrightarrow{BC}=....$ adalah ⋯⋅

Nama lain dari kolinear ialah

Diketahui: A(3,-1,4) , B(3,-4,6). Titik P membagi AB sehingga AP : PB = 3 : 2 ,maka vektor P= ...

Diketahui: A(3,-1,4) , B(3,-4,6), C(-1,5,4). Titik P membagi AB sehingga AP : PB = 3 : 2 ,maka vektor PC= ...

Agar a = (x, 4, 7) dan b = (6, y, 14) segaris, nilai 2x - y = ...

Diberikan titik A(3, -5, -4), B(6,-1,3), dan C(12,n,m). Jika A, B, dan C segaris, nilai (m - n)= ...

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Discover more resources for Mathematics

Kolinear

Diketahui ruas garis AB dengan titik A(-5,4) dan titik B(3,6). Jika titik C terletak pada AB→\overrightarrow{AB}AB sedemikian sehingga AC→ : CB→ = 3 : 2\overrightarrow{AC}\ :\ \overrightarrow{CB}\ =\ 3\ :\ 2AC : CB = 3 : 2 . Tentukanlah koordinat titik C!

Jika diketahui Titik F terletak pada DE sedemikian sehingga DF→ : FE→ = 1 : 3\overrightarrow{DF}\ :\ \overrightarrow{FE}\ =\ 1\ :\ 3DF : FE = 1 : 3 , dan ruas garis DE dengan titik D (1,2) dan E(2,3). Maka nilai dari koordinat titik F adalah...

Titik A (3, 2, -1) B (1 , -2, 1) C (7, p - 1, -5) segaris. Tentukan nilai p...

Diketahui segitiga ABC dengan koordinat A(2, 3, 4), B(5, 0, 1), dan C(4, 2, 5). Titik P membagi AB sehingga AP → : AB→=2:3\overrightarrow{AP\ }\ :\ \overrightarrow{AB}=2:3AP : AB=2:3 . Panjang vektor PC adalah

Diketahui A(1,2,3), B(3,3,1), dan C(7,5,−3), Jika A,B, dan C segaris (kolinear), maka AB →: BC→=....\overrightarrow{AB\ }:\ \overrightarrow{BC}=....AB : BC=.... adalah ⋯⋅

Nama lain dari kolinear ialah

Diketahui: A(3,-1,4) , B(3,-4,6). Titik P membagi AB sehingga AP : PB = 3 : 2 ,maka vektor P= ...

Diketahui: A(3,-1,4) , B(3,-4,6), C(-1,5,4). Titik P membagi AB sehingga AP : PB = 3 : 2 ,maka vektor PC= ...

Agar a = (x, 4, 7) dan b = (6, y, 14) segaris, nilai 2x - y = ...

Diberikan titik A(3, -5, -4), B(6,-1,3), dan C(12,n,m). Jika A, B, dan C segaris, nilai (m - n)= ...

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Popular Resources on Wayground

Discover more resources for Mathematics

Diketahui ruas garis AB dengan titik A(-5,4) dan titik B(3,6). Jika titik C terletak pada $\overrightarrow{AB}$ sedemikian sehingga $\overrightarrow{AC}\ :\ \overrightarrow{CB}\ =\ 3\ :\ 2$ . Tentukanlah koordinat titik C!

Jika diketahui Titik F terletak pada DE sedemikian sehingga $\overrightarrow{DF}\ :\ \overrightarrow{FE}\ =\ 1\ :\ 3$ , dan ruas garis DE dengan titik D (1,2) dan E(2,3). Maka nilai dari koordinat titik F adalah...

Diketahui segitiga ABC dengan koordinat A(2, 3, 4), B(5, 0, 1), dan C(4, 2, 5). Titik P membagi AB sehingga $\overrightarrow{AP\ }\ :\ \overrightarrow{AB}=2:3$ . Panjang vektor PC adalah

Diketahui A(1,2,3), B(3,3,1), dan C(7,5,−3), Jika A,B, dan C segaris (kolinear), maka $\overrightarrow{AB\ }:\ \overrightarrow{BC}=....$ adalah ⋯⋅