peraamaan garis y=a.x+b melalui titik p(√2,5) dan q(3√2,7) maka nilai a dan b adalah?

p([tex] \sqrt{2} [/tex], 5) ----> [tex] \sqrt{2} [/tex]a + b = 5
q(3[tex] \sqrt{2} [/tex], 7) ---> 3[tex] \sqrt{2} [/tex]a + b = 7

eliminasikan hasilnya :
[tex]-2 \sqrt{2} [/tex]a = -2
a = [tex] \frac{-2}{-2 \sqrt{2} } [/tex]
a = 1/2[tex] \sqrt{2} [/tex]

[tex] \sqrt{2} [/tex]a + b = 5
[tex] \sqrt{2}(1/2 \sqrt{2}) [/tex] + b = 5
1 + b = 5
     b = 5 - 1
     b = 4

5 = av2 + b
7 = 3av2 + b
-2 = - 2av2 ========> a=v2/2 maka b= 5-1 = 4
jadi nilai a= v2/2 dan b = 4