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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 x-3y+2=0.

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  1. Mertsj
    • 4 years ago
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    The equation of the perpendicular to x-3y=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

  2. Mertsj
    • 4 years ago
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    \[\frac{1}{3}x+\frac{2}{3}=-3x+14\]

  3. Mertsj
    • 4 years ago
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    \[x=4\]

  4. Mertsj
    • 4 years ago
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    If x = 4, y = 2 and that is the required point.

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