Question

Show that the points (1, 7), (4, 2), (-1, -1) & (-4, 4) are the vertices of a sqaure.​

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\bf{Given:}}}$

• Point A (1,7)
• Point B (4, 2)
• Point C (-1, -1)
• Point D (-4,4)

$\Large{\underline{\bf{To\:Prove:}}}$

• The points form vertices of a square

$\Large{\underline{\bf{Solution:}}}$

➝ First we have to prove that the distance between the vertices are equal.

➝ By distance formula we know that,

    $\sf Distance=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} }$

➝ Now finding the distance between each of the vertices,

➝ Distance of AB is given by,

    $\sf AB=\sqrt{(4-1)^{2}+(2-7)^{2} }$

    $\sf AB = \sqrt{9+25}$

    AB = √34 units-----(1)

➝ Distance of BC is given by,

    $\sf BC=\sqrt{(-1-4)^{2} +(-1-2)^{2} }$

    $\sf BC=\sqrt{25+9}$

    BC = √34 units----(2)

➝ Distance of CD is given by,

    $\sf CD=\sqrt{(-4+1)^{2} +(4+1)^{2} }$

    $\sf CD=\sqrt{25+9}$

    CD = √34 units--------(3)

➝ Distance of DA is given by,

    $\sf DA=\sqrt{(1+4)^{2}+(7-4)^{2} }$

    $\sf DA=\sqrt{25+9}$

    DA = √34 units----(4)

➝ From equations 1, 2, 3, 4 we have proved that,

   AB = BC = CD = DA = √34 units

➝ Hence ABCD is a rhombus.

➝ Now we have to prove that it is a square.

➝ To prove it is a square, we have to show that diagonals are equal.

➝ Distance of AC is given by,

    $\sf AC=\sqrt{(-1-1)^{2} +(-1-7)^{2} }$

    $\sf AC=\sqrt{4+64}$

   AC = √68 units----(5)

➝ Distance of BD is given by,

    $\sf BD=\sqrt{(-4-4)^{2} +(4-2)^{2} }$

    $\sf BD=\sqrt{64+4}$

  BD = √68 units-----(6)

➝ From equations 5 and 6, AC = BD, that is diagonals are equal.

➝ Hence the rhombus is a square.

➝ Hence proved.

   

Answer 2

$\huge\bold{\underline{Question}}$

Show that the points (1, 7), (4, 2), (-1, -1) & (-4, 4) are the vertices of a sqaure.

$\huge\bold{\underline{Answer}}$

Let, A(1, 7), B(4,2), C(-1,-1) and D(-4,4) are the vertices of a square

We know that,

distance = $\boxed{\bf{\red{\sqrt{(x²-x¹)^{2}+(y²-y¹)^{2}}}}}$

Now,

AB $\sf{\implies \sqrt{(4-1)^{2} + (2-7)^{2}}}$

AB $\sf{\implies \sqrt{(3)^{2} + ( - 5)^{2}}}$

AB $\sf{\implies \sqrt{9 + 25}}$

AB $\sf{\implies \sqrt{34}}$

$\boxed{\sf{\pink{AB\:=\:√34\:units}}}$

BC $\sf{\implies \sqrt{(-1-4)^{2} + (-1-2)^{2}}}$

BC $\sf{\implies \sqrt{(-5)^{2} + (-3)^{2}}}$

BC $\sf{\implies \sqrt{25 + 9}}$

BC $\sf{\implies \sqrt{34}}$

$\boxed{\sf{\pink{BC\:=\:√34\:units}}}$

CD $\sf{\implies \sqrt{(-4+1)^{2} + (4+1)^{2}}}$

CD $\sf{\implies \sqrt{(-3)^{2} + (5)^{2}}}$

CD $\sf{\implies \sqrt{9 + 25}}$

CD $\sf{\implies \sqrt{34}}$

$\boxed{\sf{\pink{CD\:=\:√34\:units}}}$

DA $\sf{\implies \sqrt{(1+4)^{2} + (7-4)^{2}}}$

DA $\sf{\implies \sqrt{(5)^{2} + (3)^{2}}}$

DA $\sf{\implies \sqrt{25 + 9}}$

DA $\sf{\implies \sqrt{34}}$

$\boxed{\sf{\pink{DA\:=\:√34\:units}}}$

Again diagonal ,

AC $\sf{\implies \sqrt{(-1-1)^{2} + (-1-7)^{2}}}$

AC $\sf{\implies \sqrt{(-2)^{2} + (-8)^{2}}}$

AC $\sf{\implies \sqrt{4 + 64}}$

AC $\sf{\implies \sqrt{68}}$

$\boxed{\sf{\pink{AC\:=\:√68\:units}}}$

BD $\sf{\implies \sqrt{(-4-4)^{2} + (4-2)^{2}}}$

BD $\sf{\implies \sqrt{(-8)^{2} + (2)^{2}}}$

BD $\sf{\implies \sqrt{64 + 4}}$

BD $\sf{\implies \sqrt{68}}$

$\boxed{\sf{\pink{BD\:=\:√68\:units}}}$

.°. AB = BC = CD = DA = √ 34

and diagonal AC = diagonal BD = √ 68

Hence, ABCD is a square and given points are vertices of a square.