SOLUTION: Find the distances of the point (4,-2) from the line 2x+3y-6=0​

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Question 1174470: Find the distances of the point (4,-2) from the line 2x+3y-6=0​
Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.

There is a remarkable formula to calculate the distance from a given point to a given straight line in a coordinate plane.


    Let the straight line in a coordinate plane is defined in terms of its linear equation 

         a*x + b*y + c = 0,

    where "a", "b" and "c" are real numbers, and let P = (x%5B0%5D,y%5B0%5D) be the point in the coordinate plane. 

    Then the distance from the point P to the straight line is equal to

        d = abs%28a%2Ax%5B0%5D+%2B+b%2Ay%5B0%5D+%2B+c%29%2Fsqrt%28a%5E2+%2B+b%5E2%29.


Regarding this formula, see the lesson
    The distance from a point to a straight line in a coordinate plane
in this site.


Your straight line is 2x + 3y - 6 = 0.   The point is  (4,-2)


Substitute the given data  a= 2, b= 3, c= -6,  x%5B0%5D = 4,  y%5B0%5D= -2  into the formula to get the distance under the question


    abs%282%2A4+%2B+3%2A%28-2%29+-+6%29%2Fsqrt%282%5E2+%2B+3%5E2%29 = abs%28-4%29%2Fsqrt%2813%29 = 4%2Fsqrt%2813%29 = %284%2Asqrt%2813%29%29%2F13 = 1.1094  (approximately).


Answer.  The distance is  %284%2Asqrt%2813%29%29%2F13 = 1.1094  (approximately).

Solved.