Quiz Matematika - Universityku

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Bayangan garis y = 2x + √2 setelah dirotasikan sejauh 45° berlawanan jarum jam adalah …

Quiz Matematika

Jawaban Terkonfirmasi

Asumsi titik rotasinya berada di pusat koordinat (0,0)
Rotasi 45 derajat berlawanan arah jarum jam:
$left[begin{array}{cc}x'\y'end{array}right]=left[begin{array}{cc}cos 45&-sin 45\sin 45&cos45end{array}right]left[begin{array}{cc}x\yend{array}right] \\ left[begin{array}{cc}x'\y'end{array}right]=left[begin{array}{cc}frac{1}{2}sqrt{2}&-frac{1}{2}sqrt{2}\frac{1}{2}sqrt{2}&frac{1}{2}sqrt{2}end{array}right]left[begin{array}{cc}x\yend{array}right]$

$left[begin{array}{cc}x'\y'end{array}right]=left[begin{array}{cc}frac{1}{2}sqrt{2}(x-y)\frac{1}{2}sqrt{2}(x+y)end{array}right] \\ $Maka:$ \ x-y=x'sqrt{2} \ x+y=y'sqrt{2} \ --$Jumlah$-- \ 2x=sqrt{2}(x'+y') \ x=frac{1}{2}sqrt{2}(x'+y') \ $Dan,$ \ y=frac{1}{2}sqrt{2}(y'-x')$

Substitusikan masing-masing:
$y=2x+sqrt{2} \ frac{1}{2}sqrt{2}(y'-x')=2timesfrac{1}{2}sqrt{2}(x'+y')+sqrt{2} \ frac{1}{2}sqrt{2}(y'-x')=sqrt{2}(x'+y')+sqrt{2} \ $Kalikan $sqrt{2} \ y'-x'=2(x'+y')+2 \ y'-x'=2x'+2y'+2 \ y'-2y'=2x'+x'+2 \ -y'=3x'+2 \ boxed{y'=-3x'-2}$

Atau dapat dihilangkan tanda aksennya.
y = -3x-2
3x + y + 2 = 0