Iknow it's messy but can someone let me know if this was right? it's asking for the distance from point (-3,-7) to line 5x+4y=39

This distance is approx. 12.8.

Step-by-step explanation:

Distance from point (-3,-7) to line 5x+4y=39?

Keep in mind that this distance MUST be perpendicular to the line 5x + 4y = 39.  Find the slope of a line perpendicular to this 5x + 4y = 39.  

Solving 5x + 4y = 39 results in 4y = -5x + 39. or y = (-5/4)x + 39/4.  The slope of a line perpendicular to y = (-5/4)x + 39/4 is the negative reciprocal of -5/4, and is 4/5.

Find the equation of this perpendicular line.  It passes through (-3,-7) and has 4/5 as its slope.  Its equation is y-[-7] = (4/5)(x-[-3]), or y+7=(4/5)x + 12/5.  Thus, y = -7 + (4/5)x + 12/5.

We must find the point of intersection of this line with the given 5x + 4y = 39.   Substitute y = -7 + (4/5)x + 12/5 for y in 5x + 4y = 39:

5x + 4(-7 + (4/5)x + 12/5= 39.  Simplifying, we get:

5x - 28 + 4 [(4/5)x + 12/5] = 39.  Further, 5x - 28 + (16/5)x + 48/5 = 39.

Multiply each term by 5 to eliminate fractions:

25x - 140 + 16x + 48 = 195.  Then 41x - 92 = 195, or 41x = 287.  Dividing both sides by 41, we get x = 7.  If x = 7, then 5x + 4y = 39 becomes 35 + 4y = 39, and so y = 1.  The two lines intersect at (7,1).

The distance between the given point (-3,-7) and (7,1) is found as follows:

d = √{ (7-[-3])^2 + (1-[-7])^2 }

  = √{100 +64) = √164 = 12.8 (approx.)

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step-by-step explanation:

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