Question

Find the point (x,y) on the line y=x that is equidistant from the points (4,-1) and (7,-6)

Answer 1

in other words, we need to find the line that forms thru the midpoint of the line segment created from these points and perpendicular to it; then equate it to y=x

Answer 2

midpoint is add em up and divide by 2

(4,-1)
(7,-6)
------
(11,-7)/2 = (11/2, -7/2)

the slope between them is subtract the points and stack y/x

(7,-6)
-(4,-1)
-------
3,-5 ; slope = y/x = -5/3; our perp slope is flip and negate: 3/5

now we form the equation of the perp line

y = (3/5)x -3(11)/5(2) - 7/2

y = (3/5)x -68/10

and when does this equate with y=x?

y = (3/5)x -68/10 = y = x

(3/5)x -68/10 = x

-68/10 = (2/5)x

-17 = x .... and by default, since y = x, when y = -17 too

(-17,-17)

Answer 3

we can dbl check with pythag thrm

(4,-1) (-17,-17) (7,-6)
17 17 17 17 17 17
---------------------
21,16 0 , 0 24,11

sqrt(21^2+16^2) = sqrt(24^2 + 11^2) ??

http://www.wolframalpha.com/input/?i=sqrt%2821^2%2B16^2%29+%3D+sqrt%2824^2+%2B+11^2%29

Answer 4

goodlooks i get it now i had no idea what the question asked but i get it now thanks

Answer 5

youre welcome :)