Given a line ABCD in which AB = BC = CD, B= (0, 3) and C= (1, 8). Find the co-ordinates of A and D.

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Solution

Equation of line BC is
y=5x+3
BC = $\sqrt{(0 - 1)^{2} + (3 - 8)^{2}}$ = $\sqrt{26}$
let the coordinates of A is x1, y1
then from distance formula AB=BC
We had
$x 1^{2} + (y 1 - 3)^{2}$ = 26
solving this with equation of line we get x1=-1 and y1=-2
(-1, -2)
similarly
for point D let co-ordinated are x2, y2
$(x 2 - 1)^{2} + (y 2 - 8)^{2}$ =26
substituting y2 from equation of line
we get
$(x 2 - 1)^{2} = 1$
fro this x2 =2
and from equation of line y2=13
(2,13)

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