Question

If two opposite vertices of a square are (3, 4) and (1,-1), find the coordinates of the other two vertices.​

Answer 1

Answer :-

Let the line joining the points A (3,4) & C (1,-1) be y=mx+c

Then, m=(4+1)/(3-2)=5/2

and c=y-mx= -1 - (5/2)*1= -7/2

So, the eqn is y=(5/2)x - 7/2

And,

And,the other diagonal BD will be perpendicular to AC at its midpoint E.

Its clear that E=((3+1)/2, (4+(-1))/2) = (2, 3/2)

Let the equation of BD be y=m’x+c’

Since E lies on BD, it will satisfy the equation.

So, m’=-1/m (since BD and Ac are perpendicular to each other)

=> m’=-1/5/2= -2/5

Also, c = y-m’x

=> 3/2-(-2/5)*2 = 23/10

So, equation is y=(-2/5)x + 23/10

Now clearly, BE=AC/2.

=> sqrt ( (x–2)^2 + (y–3/2)^2) = 1/2 * sqrt( (3–1)^2 + (4-(-1))^2 )

=> (x–2)^2 + (y–3/2)^2 = 29/4

But since (x,y) lies on BD, therefore

y=(-2/5)x + 23/10

=> (x–2)^2 + ( (-2/5)x + 23/10 -3/2)^2 = 29/4

=> (x–2)^2 + 4/25 * (x–2)^2 = 29/4

=> (x–2)^2 * 29/25 = 29/4

=> (x–2)^2 = 25/4

=> x = 2+-5/2

=> x = 9/2, -1/2

from y = (-2/5)x + 23/10, we get :

y = 1/2, 5/2

And,

The remaining two points of the square are (9/2, 1/2) & (-1/2, 5/2)