The distance between the two points (2, -3) (10, y) is ten units find ‘y’.
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y=3
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Answer:
d=√((x2-x1)²+(y2-y1)²)
10=√((10-2)²+(y+3)²)
squaring both sides
100 =8²+(y+3)²
100-64 =(y+3)²
36=(y+3)²
√(36)=(y+3)
y+3=+6&-6
y=3&-9
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