CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Two opposite vertices of a square are $$\left( {3,4} \right)$$ and $$\left( {1, - 1} \right)$$ Find the coordinates of the other vertices.


Solution

Let $$PQRS$$ be a square.

$$P(3,4)$$ and $$R(1,-1)$$

Let co-ordinates of a point $$Q(x,y)$$

$$PQ=QR$$

Squaring on both sides,

$$PQ^2=QR^2$$

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

Solving this we get,

$$\Rightarrow$$  $$x^2-3x+9+y^2-8y+16=x^2-2x+1+y^2+2y+1$$

$$\Rightarrow$$  $$4x+10y=23$$        ---- ( 1 )

$$\Rightarrow$$  $$x=\dfrac{23-10y}{4}$$        ----- ( 2 )

Length of the hypotenuse $$=\sqrt{2}\times side$$

$$\Rightarrow$$  $$PR=\sqrt{2}PQ$$

$$\Rightarrow$$  $$PR^2=2PQ^2$$

$$\Rightarrow$$  $$(3-1)^2+(4+1)^2=2[(x-3)^2+(y-4)^2]$$

$$\Rightarrow$$  $$4+25=2[x^2-6x+9+y^2-8y+16]$$

$$\Rightarrow$$  $$29=2[x^2-6x+y^2-8y+25]$$

$$\Rightarrow$$  $$29=2x^2-12x+2y^2-16y+50$$

$$\Rightarrow$$  $$29=2\left(\dfrac{23-10y}{4}\right)^2-12\left(\dfrac{23-10y}{4}\right)+2y^2-16y+50$$            [ From ( 2 ) ]

$$\Rightarrow$$  $$29=\dfrac{2}{16}(529-460y+100y^2)-69+30y+2y^2-16y+50$$

$$\Rightarrow$$  $$29=66.125-57.5y+12.5y^2-69+30y+2y^2-16y+50$$

$$\Rightarrow$$  $$29=47.125-43.5y+14.5y^2$$

$$\Rightarrow$$  $$14.5y^2-43.5y+18.125=0$$

By soving we get,

$$y=\dfrac{5}{2}$$ and $$y=\dfrac{1}{2}$$

Substituting value of $$y$$ in equation ( 1 ) we get,

$$\Rightarrow$$  $$x=\dfrac{-1}{2}$$ and $$x=\dfrac{9}{2}$$

Vertices of a square are $$\left(\dfrac{9}{2},\dfrac{1}{2}\right)$$ and $$\left(\dfrac{-1}{2},\dfrac{5}{2}\right)$$




Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image