If two opposite vertices of a square are(3,4),(1,-1)find the coordinates of the other two vertices
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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
Now, the other diagonal BD is gonna be perpendicular to AC at its midpoint E.
It is clear to see 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 its equation.
So, m’=-1/m (since BD is perpendicular to AC)
=> m’=-1/5/2= -2/5
Also, c=y-m’x= 3/2-(-2/5)*2 = 23/10
So, eqn 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, so 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 have:
y = 1/2, 5/2
So the remaining two points of the square are (9/2, 1/2) & (-1/2, 5/2)
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