Question

Let the opposite angular points of a square be (3, 4) and (1, –1). Find the coordinates of the remaining angular points

Answer 1

Concept:
Application:

Let ABCD be a square and let A ( 3,4) and c ( 1,-1) be the given angular points.

Let B ( x,y) be the unknown vertex.

Then AB = BC

$\Rightarrow A{B}^2=B{C}^2$

$\Rightarrow (x-3{)}^2+(y-4{)}^2=(x-1{)}^2+(y+1{)}^2$

$\Rightarrow 4x+10y-23=0$

$\Rightarrow x=\frac{23-10y}{4}\dots \dots \left(1\right)$

In right- angled triangle ABC , we have

$A{B}^2+B{C}^2=A{C}^2$

$\Rightarrow (x-3{)}^2+(y-4{)}^2+(x-1{)}^2+(y+1{)}^2$

$=(3-1{)}^2+(4+1{)}^2$

$\Rightarrow x^2+y^2-4x-3y-1=0\dots \dots \left(2\right)$

Substituting the value of x from (1) and (2),

we get

${\left(\frac{23-10y}{4}\right)}^2+y^2-(23-10y)-3y-1=0$

$\Rightarrow 4y^2-12y+5=0\Rightarrow (2y-1)(2y-5)=0$

$\Rightarrow y=\frac{1}{2}$ or $\frac{5}{2}$

Putting $y=\frac{1}{2}$ and $y=\frac{5}{2}$ respectively in (1) , we get

$x=\frac{9}{2}$ and $\frac{-1}{2}$ respectively.

Hence , the required vertices of the square are $(\frac{9}{2},\frac{1}{2})$ and $(-\frac{1}{2},\frac{5}{2})$