Maths / Coordinate Geometry / Distance Formula

QUESTION

The opposite vertices of a square are (0, 1) and (4, 3). Find the coordinates of the other two vertices.

EXPLANATION
Explain TypeExplanation Content
Text

The opposite vertices of a square are A(0, 1) and C(4, 3). Let the coordinates of the other two vertices be B(x,y) and D (m,n)

$\large AB = \sqrt{(x- 0)^2+(y-1)^2}= \sqrt{x^2+(y-1)^2}$

$\large BC = \sqrt{(x- 4)^2+(y-3)^2}$

As these are sides of a square they are equal

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

squaring both sides we get

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

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

$\large (x + (x- 4))(x - (x- 4)) =((y-3)+(y-1))((y-3)-(y-1))$

$\large (2x - 4)(4) =(2y-4)(-2)$

$\large 8x - 16 =-4y+8 \Rightarrow 8x+4y-24=0\Rightarrow 2x+y-6=0$

$\dpi{120} \large y= 6-2x$                                                           (1)

As ABC is a rt triangle

$\dpi{120} \large AC^2 = AB^2+ BC^2$

$\large AC = \sqrt{(4- 0)^2+(3-1)^2}= \sqrt{16+4}=\sqrt{20}$

$\large 20 = x^2 +(y-1)^2 + (4-x)^2+(3-y)^2$

$\large 20 = x^2 +y^2 +1 -2y + x^2 +16-8x+y^2 +9-6y$

$\large 20 = 2x^2 +2y^2 +26 -8y -8x$

$\large 2x^2 +2y^2 +26 -8y -8x -20 = 0$

$\large 2x^2 +2y^2 +6 -8y -8x = 0$

$\large x^2 +y^2 +3 -4y -4x = 0$

Putting the value of y from Eq 1

$\large x^2 +(6-2x)^2 +3 -4(6-2x) -4x = 0$

$\large x^2 +36+4x^2- 24x +3 -24+8x -4x = 0$

$\large 5x^2 +15-20x= 0$

$\large x^2 -4x+3 = 0$

$\dpi{120} \large x^2 -3x-x+3=0$

$\dpi{120} \large x(x -3) -1(x -3)=0$

$\dpi{120} \large (x -3) (x-1)=0$

either x = 3 or x =1

If x= 3 then y = 6-2x = 6-2(3) = 6-6 = 0

if x= 1 then y = 6-2x = 6-2(1) = 6-2 = 4

So the coordinates are (3,0) and ( 1, 4)



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