Question

Origin is the centre of the square with one of its vertices at $(3,4),$ then the other vertices are

• A. $(-3,4),(-3,-4),(3,-4)$
• B. $(-4,3),(-3,-4),(4,-3)$
• C. $(-4,3),(-4,-3),(3,-4)$
• D. $(3,4),(-4,-3),(4,-3)$

Answer 1

The correct option is B $(-4,3),(-3,-4),(4,-3)$

We know that the diagonals of a square bisect each other and intersect at the centre. It is given that the centre is the origin.

Now let the vertex diagonally opposite to $A=(3,4)$ be $C=(x,y)$.

Hence $(0,0)$ is the midpoint of that diagonal connecting $(x,y)$ and $(3,4)$.

Thus application of section formula gives us

$(0,0)=\frac{x+3}{2},\frac{y+4}{2}$

$x=-3$ $y=-4$.

Hence $C(x,y)=(-3,-4)$.

Now let the third vertex be $B=(h,k)$

Application of Pythagoras theorem gives us $A{B}^2+B{C}^2=A{C}^2$

Since it is a square, $AB=BC$.

Substitution in the Pythagoras theorem gives us .

$AB=BC=\frac{AC}{\sqrt{2}}$ ...(i)

Now, using distance formula,

$AB=BC$ implies

$\sqrt{(h-3{)}^2+(k-4{)}^2}=\sqrt{(h+3{)}^2+(k+4{)}^2}$

$(h-3{)}^2+(k-4{)}^2=(h+3{)}^2+(k+4{)}^2$

$12h+16k=0$

$3h=-4k$ ...(ii)

And

$AB=\frac{AC}{\sqrt{2}}$ from (i).

Hence application of distance formula gives us

$\sqrt{(h-3{)}^2+(k-4{)}^2}=\frac{10}{\sqrt{2}}$

Now $h=\frac{-4k}{3}$ from (ii)

Hence

$(-\frac{4k}{3}-3{)}^2+(k-4{)}^2=50$

$(4k+9{)}^2+9(k-4{)}^2=450$

$16k^2+72k+81+9k^2-72k+144=450$

$25k^2=450-144-81$

$25k^2=225$

$5k=\pm 15$

$k=\pm 3$

Hence $h=\mp 4$

Thus the other two vertices are

$(4,-3)$ and $(-4,3)$

Therefore the remaining three vertices are

$(4,-3),(-3,4),(-3,-4)$.