Question

In the standard $(x,y)$ coordinate plane above, $3$ of the vertices of a rectangle are shown. Find the $4^{th}$ vertex of the rectangle.

• A. $(3,-7)$
• B. $(4,-8)$
• C. $(5,-1)$
• D. $(8,-3)$
• E. $(9,-3)$

Answer 1

The correct option is A $(3,-7)$

We know the vertices of A(2,1), B(-1,-1),D(6,-5)

There are two ways to solve

i)graphically:when all the points are plotted ,we get the point (3,-7)

ii)when we solve,

The distance of BC=distance of DC.

$\sqrt{(x+1{)}^2+(y+1{)}^2}=\sqrt{(x-6{)}^2+(y+5{)}^2}$

Squaring,$(x+1{)}^2+(y+1{)}^2=(x-6{)}^2+(y+5{)}^2$

Solving ,we get the equation

$14x-8y+14=0⟹(x,y)=(3,-7)$