IF one of the vertices of a square is (3,2) and one of the diagonalls is along the line 3x+4y+8=0, then find the centre of the square and other vertices.

Explanation

We have square ABCD, where coordinates of A are (3,2).
Clearly, point (3,2) does not satisfy the line 3x+4y+8=0.
Thus, diagonal BD is along this line.
M(h,k) is centre of the square, which is the foot of perpendicular from vertex A on the diagonal BD.
`therefore (h-3)/(3) = (k-2)/(4) = -((3(3)+4(2)+8))/(3^(2)+4^(2))`
`rArr (h-3)/(3) = (k-2)/(4) = -1`
`rArr (h,k) -= (0,-2),`
which is the centre of the square.
M is midpoint of AC.
`therefore C -= (-3,-6)`
`" Now," AM = (|3(3)+4(2)+8|)/(sqrt(3^(2) + 4^(2))) = 1`
`therefore MB=MD=1`
`" Slope of BD " = -(3)/(4) = tan theta, " where" theta " is inclination of BD with x-axis."`
Therefore , coordinates of B and D are
`(0+-1*"cos" theta, -2+-1*"sin" theta)`
`-=(0+-1(-(4)/(5)),-2+-1 ((3)/(5)))`
`-=(underset(+)(-)(4)/(5),-2+- (3)/(5))`
`-=(-(4)/(5),-(7)/(5)), ((4)/(5),-(13)/(5))`