Jarak dua titik
| d | = √ {(x1-x2)²+(y1-y2)²}
*
a. y = 2x² + 3x + 5 dan y = 11 - x
titik potong keduanya di tentukan dengan y - y = 0
2x² + 3x + 5 - (11 - x ) = 0
2x² + 3x + x + 5 - 11 = 0
2x² + 4x - 6= 0
2(x²+ 2x - 3) =0
2(x + 3)(x - 1)= 0
x= - 3 atau x = 1
y = 11- x
titik potong :(x,y)
x = - 3 , y = 14 --> (-3, 14)
x= 1 , y = 10 --> (1, 10)
Jarak segmen garis d = √(-3-1)² + (14 -10)²
d = √(16 + 16)
d = 4√2
b . y = x² + 2x + 1 dan y = - x² +2x + 9
y1 - y2 = 0
x² + 2x+ 1 - ( - x² +2x + 9) = 0
x² + 2x + 1 + x² -2x - 9= 0
2x² -8= 0
x² = 4
x = 4 atau x = - 4
y = x² + 2x + 1
x= 4 , y= 25
x = -4 , y = 9
jarak segmen garis d = √{(4 + 4)²+(25-9)²
d = √(64 + 256)
d = √320= 8√5
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Verified answer
Jarak dua titik
| d | = √ {(x1-x2)²+(y1-y2)²}
*
a. y = 2x² + 3x + 5 dan y = 11 - x
titik potong keduanya di tentukan dengan y - y = 0
2x² + 3x + 5 - (11 - x ) = 0
2x² + 3x + x + 5 - 11 = 0
2x² + 4x - 6= 0
2(x²+ 2x - 3) =0
2(x + 3)(x - 1)= 0
x= - 3 atau x = 1
y = 11- x
titik potong :(x,y)
x = - 3 , y = 14 --> (-3, 14)
x= 1 , y = 10 --> (1, 10)
Jarak segmen garis d = √(-3-1)² + (14 -10)²
d = √(16 + 16)
d = 4√2
*
b . y = x² + 2x + 1 dan y = - x² +2x + 9
y1 - y2 = 0
x² + 2x+ 1 - ( - x² +2x + 9) = 0
x² + 2x + 1 + x² -2x - 9= 0
2x² -8= 0
x² = 4
x = 4 atau x = - 4
*
y = x² + 2x + 1
x= 4 , y= 25
x = -4 , y = 9
*
jarak segmen garis d = √{(4 + 4)²+(25-9)²
d = √(64 + 256)
d = √320= 8√5