Question

find the coordinate of any 3 points which are equidistant from A(-1 8) and B(3 4)

Answer 1

Answer:

C(1,6), D(2,7) and E(3,8)

Step-by-step explanation:

The equation of the line passing through the points A and B is given by

y = mx+c

where m = $\frac{4-8}{3-(-1)}$ = -1

and c = y - mx = 4-(-1)(3) = 7

So we have equation y = -x+7.

The middle point on the line y= -x+7 is given by

C($\frac{-1+3}{2}$,$\frac{8+4}{2}$) = C(1,6)

A line perpendicular to the above line has slope

m2 = $\frac{-1}{m}$ = 1

So the equation of the line perpendicular to y = -x+7 and passing through C(1,6) is given by

y-y1 = m2(x-x1)

y-6 = 1(x-1)

y = x+5

According to a theorem, all the points lying on the line y = x+5 would be equidistant from the points A and B.

So with x = 1,2,3 we have y = 6,7,8

Hence C(1,6), D(2,7) and E(3,8) are equidistant from A and B, which can also be verified using distance formula.