garis g=2x-3y-6=0 dilatasikan dengan faktor k=3 dan pusat dilatasi p (1,-2) tentukan bayangannya

garis g = 2x - 3y - 6 = 0
A(x, y) Dilatasi [(1, -2), 3] A' (x', y')
[tex] \binom{ {x}^{l} }{ {y}^{l} } = k \binom{x - p}{y - q} + \binom{p}{q} \\ \binom{ {x}^{l} }{ {y}^{l} } = \binom{3 \: \: \: \: \: 0 }{0 \: \: \: \: \: 3} \binom{x - 1}{y + 2} + \binom{1}{ -2 } \\ \binom{ {x}^{l} }{{ y }^{l} } = \binom{3x - 3}{3y + 6} + \binom{1}{ - 2} \\ \binom{ {x}^{l} }{ {y}^{l} } = \binom{3x - 2}{3y + 4} \\ {x} = \frac{( {x}^{l} + 2)}{3} \: \: y = \frac{( {y}^{l} - 4) }{3} [/tex]
substitusi x dan y ke persamaan garis
[tex]2x - 3y - 6 = 0 \\ 2 (\frac{ ({x}^{l} + 2)}{3} ) - 3 ( \frac{( {y}^{l} - 4)}{3} ) - 6 = 0 \\ \frac{2}{3} ( {x}^{l} + 2) - \frac{3}{3} ( {y}^{l} - 4) - 6 = \: 0 [/tex]
semua kali 3 sehingga diperoleh garis bayangannya

[tex]2( {x}^{l} + 2) - 3( {y}^{l} - 4) - 18 = 0 \\ 2 {x}^{l} + 4 - 3 {y}^{l} + 12 - 18 = 0 \\ 2{x}^{l} - 3 {y}^{l} - 2 = 0[/tex]

2x - 3y - 2 = 0