Show that the origin and p(7,4) are equidistant from the l: y= -4/3+20/3

Answer 1

I have seen this same question several times in the last two days,
each time it contained the same typo, so it must be the same poster

line equation should be
y= (-4/3)x+20/3
.. btw, it was answered by oobleck, who pointed out what I call
method 2 below.

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easiest way:
show that the midpoint of the segment joining (0,0) and (7,4) lies on the line
midpoint = ( (0+7)/2 , (0+4)/2) = (7/2, 2)

line: y = (-4/3)x + 20/3
for (7/2 , 2)
LS = 7/2
RS = (-4/3)(7/2) + 20/3 = 2 = LS
Yes, the two stated points are equidistant

method 2:
for ax + by + c = 0 , the shortest distance to the point (p,q) is
| ap + bq + c | / √(a^2 + b^2)

so change y = (-4/3)x + 20/3 to
3y = -4x + 20
4x + 3y - 20 = 0
for the given line:
distance from (0,0) = |0 + 0 - 20|/√25 = 4
distance from (7,4) = |28 + 12 - 20|/√25 = 4