Jawaban:
Penjelasan dengan langkah-langkah:
Titik A (1,3) dan D(8,4)
M = Y₂-Y₁
X₂-X₁
= (4-3)
(8-1)
= 1
7
Y-y₁=m(X-x₁)
Y-3=1/7(X-1)
Y-3=1/7X-1/7
Y=1/7X-1/7+3
Y=1/7X-1/7+21/7
Y=1/7X+20/7
x (7)
7Y=X+20
-X+7Y=20
-X+7Y-20=0
X-7Y+20=0
Demikian
Persamaan Garis Lurus
[tex]A(1 \: , \: 3) \to ( a_{1},b_1): \: D(8\:,\:4) \to (a_2, b_2)[/tex]
menentukan persamaan garis:
[tex] \frac{y - b_{1}}{ b_{2} - b_{1} } = \frac{x - a_{1} }{ a_{2} - a_{1} } \\ \frac{y - 3}{4 - 3} = \frac{x - 1}{8 - 1} \\ \frac{y - 3}{1} = \frac{x - 1}{7} \\ 1(x - 1) = 7(y - 3) \\ x - 1 = 7y - 21 \\ x - 7y - 1 + 21 = 0 \\ x - 7y + 20 = 0[/tex]
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Jawaban:
[Persamaan Garis Lurus]
Penjelasan dengan langkah-langkah:
Titik A (1,3) dan D(8,4)
M = Y₂-Y₁
X₂-X₁
= (4-3)
(8-1)
= 1
7
Y-y₁=m(X-x₁)
Y-3=1/7(X-1)
Y-3=1/7X-1/7
Y=1/7X-1/7+3
Y=1/7X-1/7+21/7
Y=1/7X+20/7
x (7)
7Y=X+20
-X+7Y=20
-X+7Y-20=0
X-7Y+20=0
Demikian
Verified answer
Penjelasan dengan langkah-langkah:
Persamaan Garis Lurus
[tex]A(1 \: , \: 3) \to ( a_{1},b_1): \: D(8\:,\:4) \to (a_2, b_2)[/tex]
menentukan persamaan garis:
[tex] \frac{y - b_{1}}{ b_{2} - b_{1} } = \frac{x - a_{1} }{ a_{2} - a_{1} } \\ \frac{y - 3}{4 - 3} = \frac{x - 1}{8 - 1} \\ \frac{y - 3}{1} = \frac{x - 1}{7} \\ 1(x - 1) = 7(y - 3) \\ x - 1 = 7y - 21 \\ x - 7y - 1 + 21 = 0 \\ x - 7y + 20 = 0[/tex]