Question

Question ⤵

☞Two opposite angular points of a square ABCD are A(-1, 2) and C(3,-2). Find the coordinates of the remaining angular points of the square.
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☞Full Explanation Solution.

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Answer 1

Required Answer is in attachment!

• Formula used in solution:

$\star\;{\bf {Distance\: Formula}}\implies \sf\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}$

$\star\;{\bf {Pythagoras\: theorem}}\sf \implies\:H^2=B^2+P^2$

$\sf Here\begin{cases}\sf H^2\: refers\:to\:square\:of\: hypotenuse.\\\sf P^2 \: refers\:to\: square\:of\: perpendicular.\\\sf B^2\: refers\:to\: square\:of\:base.\end{cases}$

• Identity used:

$\star\;{\bf {(A+B)^2=}}\sf A^2+B^2+2AB$

$\star\;{\bf {(A-B)^2=}}\sf A^2+B^2-2AB$

 

Answer 2

We know

All sides of Square are Equal

⟼ AB = BC

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

$\sf\pink{squaring \: both \: side... }$

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

$\sf\red{x {}^{2} } + 1 + 2x + \sf\red{y {}^{2} + 4 } - 4y = \sf\red{x {}^{2} } + 9 - 6x + \sf\red{y {}^{2} + 4 } + 4y \\ 2x - 4y + 1 = - 6x + 4y + 9 \\ 2x + 6x = 4y + 4y + 9 - 1$

${\bold{\green{\underline{\underline{Each \: Terms \: Divide \: by \: 8}}}}}$

$\;\large{\boxed{\bf{\purple{x = y + 1}}}} - - (i)$

${\bold{\pink{\underline{\underline{sum \: of \: square \: of \: other \: two \: sides}}}}}$

⟼ AC ² = AB² + BC²

[ 3- (-1) ] ² + (-2 -2) ² = x² + 1 + 2x + y +4 - 4y + x² + 9 - 6x + y² + 4 + 4y

$4 {}^{2} + ( - 4) {}^{2} = 2x { }^{2} - 4x + 18 \\ 16 + 16 = 2x {}^{2} + 2y {}^{2} - 4x + 18 \\ 2x {}^{2} + 2y {}^{2} - 4x + 18 - 32 = 0$

$2x {}^{2} + 2y {}^{2} - 4x - 14 = 0 \\ \;\large{\boxed{\bf{\red{x {}^{2} + y {}^{2} - 2x - 7 = 0 }}}} - (ii)$

$\sf\colorbox{green}{ putting \: the \: value \: of \: x \: see(eq.i)}$

$(y + 1) {}^{2} + y {}^{2} - 2(y + 1) - 7 = 0 \\ y {}^{2} + 1 + \sf\red{2y} + y {}^{2} - \sf\red{2y} - 2 - 7 = 0 \\⇒ 2y {}^{2} - 8 = 0 \\ ⇒2(y {}^{2} - 4) = 0 \\⇒ y {}^{2} - 4 = 0 \\⇒ y {}^{2} = 4 \\⇒ y = ± \sqrt{4} \\ ⇒y = ± \: 2$

When y = 2

$x = y + 1 \\ = 2 + 1 \\ = 3$

When y = - 2

$x = y + 1\\ = - 2 + 1 \\ = - 1$

Therefore, Other Angular Points are

$(3,2) \: and \: ( - 1 ,- 2)$

$\sf\blue{hope \: this \: helps \: you!! \: }$

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@Mahor2111

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I had posted the solution in morning itself, but they reported my answers. without any Valid Reason __i don't know why pro Thala Report my All answers :(

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