Answer:
This distance is approx. 12.8.
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.)