Question

1. Persamaan garis yang melalui titik (-5, 1) dan tegak lurus 2x + y = 10

2. Persamaan garis yang melalui titik (4, -3) dan (-2,7)

3. Nilai x yang merupakan penyelesaian dari 2x-3y = 13 dan x + 5y = -13

4. Himpunan penyelesaian dari sistem persamaan 5x - 7 = 8 dan 2x - 3y = - 7

Answer 1

1) m1 = -2/1 = -2
m2 tegak lurus m1
m2 = -1/m1 = ½

y - 1 = ½(x + 5)

2y - 2 = x + 5

2y - x - 2 - 5 = 0

x - 2y + 7 = 0

2) y1 = -3 , y2 = 7
x1 = 4 , x2 = -2

(y-y1)/(y2-y1) = (x-x1)/(x2-x1)

(y+3)/(7+3) = (x-4)/(-2-4)

(y+3)/10 = (x-4)/-6

-6(y + 3) = 10(x - 4)

-6y - 18 = 10x - 40

-10x -6y -18 + 40 = 0

10x + 6y - 22 = 0

3)
2x - 3y = 13 (×5) 10x - 15y = 65
x + 5y = -13 (×3) 3x + 15y = -39
~~~~~~~~~~~~~~~~~~~~~~~~~~~ +
13x = 26

x = 26/13

x = 2

4)
5x-7y=8 (×2) 10x - 14y = 16
2x-3y=-7 (×5) 10x - 15y = -35
~~~~~~~~~~~~~~~~~~~~~~~~ -
y = 51

5x - 7(51) = 8
5x = 8+357
5x = 365
x = 365/5
x = 73

hp = { 73, 51 }

Answer 2

1. m¹ = 2x + y = 10. m = -a/b
a = 2. = -2/1 = -2
b = 1

m¹ = -1/m¹ = ½

y - y¹ = m(x-x¹)
y - (-5) = ½(x - 1)
y + 5 = ½x - ½
y = ½x - ½ - 5
y = ½x - (-9/10)
y = ½x + 9/10

2. y - y¹/y²-y¹ = x - x¹/x²-x¹
y - (-3)/7 - (-3) = x - 4/-2-4
(y+3)/10 = (x-4)/-6

-6(y+3) = 10(x-4)
-6y-18 = 10x-40
-6y = 10x-40+18
-6y = 10x - 22 : -6
y = 10x/-6 - 22/-6
y = -5x/3 + 11/3