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Which is the distance between the point with the coordinates (1,2) and the line with the equation 2x-3y=-2?

  • math -

    distance from (h,k) to ax+by+c=0 is

    |ah+bk+c|/√(a^2+b^2)

    so, you have

    |1*2 - 2*3 + 2|/√(4+9) = 2/√13

    or, if you can't recall the formula, you can work it out by realizing that the shortest distance will be along a perpendicular line through the point.

    2x-3y = -2 has slope 2/3
    The perpendicular thus has slope -3/2
    Now you have a point and a slope, so the equation of the perpendicular line through (1,2) is

    y-2 = -3/2 (x-1)
    y = -3/2 x + 7/2

    The perpendicular intersects at (17/13,20/13)

    The distance from (1,2) to (17/13,20/13) is √(16/169+36/169) = √52/13 = 2/√13

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