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anonymous
 4 years ago
Can you help me with this one: Find a point in first quadrant which belongs to line x+y=8 and which is equally distanced from point (2,8) and line x3y+2=0.
anonymous
 4 years ago
Can you help me with this one: Find a point in first quadrant which belongs to line x+y=8 and which is equally distanced from point (2,8) and line x3y+2=0.

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Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0The equation of the perpendicular to x3y=2=0 through the point (2,8) is y=3x+14. The required point must be on this perpendicular since the distance from a point to a line is on the perpendicular. So equate y=1/3x +2/3 and y =3x+14

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3}x+\frac{2}{3}=3x+14\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0If x = 4, y = 2 and that is the required point.
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