Question

Two opposite vertices of a square are (1,-2) and (-5,6). Find the coordinates of the remaining two vertices.​

Answer 1

Answer:

Two opposite vertices of a square are (1,-2) and (-5,6)

Step-by-step explanation:

A=(1,-2)

B=(-5,-2)

C=(-5,6)

D=(1,6)

Answer 2

Answer:

The co ordinates of remaining two verices are (2,5) and (-6,-1)

Step-by-step explanation:

Let the square be ABCD and let the given vertices be A(1,-2) and C(-5,6) and let the other two vertices be B(a,b) and D(c,d) first we find the length of diagonal AC from the distance formula

$AC=\sqrt{(-5-1) {}^{2} +(6-(-2)) {}^{2} } \\ = \sqrt{6 {}^{2} + 8 {}^{2} } = 10$

From properties of square we have

$AB=BC=CD=AD \\ AC=BD$

Therefore AC=BD=10

Also we have

$side = \frac{diagonal}{ \sqrt{2} } \\ = > side = \frac{10}{ \sqrt{2} } = 5 \sqrt{2} \: units$

Therefore,

$AB=BC=CD=AD = 5 \sqrt{2}$

$AB = \sqrt{(a - 1) {}^{2} + (b + 2) {}^{2} } = 5 \sqrt{2} \\ = > (a - 1) {}^{2} + (b + 2) {}^{2} = 50 \\ = > {a}^{2} + {b}^{2} - 2a + 4b - 45 = 0- - > (1)$

$BC= \sqrt{( - 5 - a) {}^{2} + ( 6- b) {}^{2} } = 5 \sqrt{2} \\ = > 50 =( - 5 - a) {}^{2} + ( 6- b) {}^{2} \\ = > {a}^{2} + b {}^{2} + 10a - 12b + 11 = 0 - - > (2)$

subtracting equation 1 from equation 2,we get

$(2) - (1) = > 12a - 16 b+ 56 = 0 \\ = > 3a - 4b + 14 = 0 - - > (3) \\ = > b = \frac{3a + 14}{4}$

Substituting value of b in equation 1 we get

${a}^{2} +( \frac{3a + 14}{4} ) {}^{2} - 2a + 3a + 14 - 45 = 0 \\ = > {a}^{2} + \frac{9 {a}^{2} + 196 + 84a}{16} + a - 31 = 0 \\ = > \frac{25a {}^{2} + 196 + 84a + 16a - 496}{16} = 0 \\ = > 25 {a}^{2} + 100a - 300 = 0 \\ = > {a}^{2} + 4a - 12 = 0 \\ = > a = 2 \\ a = - 6$

$= > b = \frac{3(2) + 14}{4} = 5 \\ b = \frac{3( - 6) + 14}{4} = - 1$

Therefore the vertex can either be (2,5) or (-6,-1)

hence the vertices are (2,5) and (-6,-1)

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