Verified answer

M = (m,0)

N = (0,n)

then the slopes are all consistent

-n/m = (8-n)/12

we also know that the distances are equal

squares of the distances:

Case I: the y-intercept is closer to (12,8) thus m is negative, n is positive.

m^2 + n^2 = 12^2 + (8-n)^2

we have

m = 12n/(n-8)

substituting in

m^2 + n^2 = 12^2 + (8-n)^2

we have m = -12, n = 4 .

Case II. The x-intercept is closer to (12,8) . Thus m is positive and n is negative.

Again, m = 12n/(n-8) .

m^2 + n^2 = (12-m)^2 + (8)^2

eventually ...

m = 6

n = -8

thus two possible scenarios:

A(12,8) , N(0,4) , M(-12, 0) , thus B(-24, -4)

or

A(12,8) , M(6,0) , N(0,-8) , thus B(-6, -16) .

Since it is given that M and N lie on coordinated axes

let coordinates of M = (0,y1) and N = (x1,0)

coordinates of A = (12,8) , let B = (x2,y2)

Because M and N trisect AB, M is the mid point of AN.

so x - coordinate of M = (12+x1) /2

y-cordinate of M = (8+0) /2 = 4

so (12+x1)/2 = 0 and y1 = 4

x1 = -12

so coordinates of M = (0,4)

coordinates of N = (-12,0)

now N is the mid point of BM

therefore (a + 0) /2 = -12 : a = -24

similarly (b + 4) /2 = 0 : b = -4

so possible coordinates of B = (-24, -4)

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Not enough info. You need to know where either M or N is