Question

when we plot the points A(1,4) B (-1,0) C (1,-4)
and D (3,0) the figure .
we had obtained is
a​

Answer 1

look.at the picture.it may help you

Answer 2

$\Large{\underline{\underline{\bold{✯Given:}}}}$

• Point A = (1,4) $→\;{\sf{(X_1,Y_1)}}$
• Point B = (-1,0) $→\;{\sf{(X_2,Y_2)}}$
• Point C = (1,-4) $→\;{\sf{(X_3,Y_3)}}$
• Point D = (3,0) $→\;{\sf{(X_4,Y_4)}}$

$\Large{\underline{\underline{\red{\bold{➸To\:Find:}}}}}$

• Figure Formed by joining the points.

$\Large{\underline{\underline{\blue{\bold{✄Solution:}}}}}$

$\boxed{\green{\bf{Distance\; between\;two\;points = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}\;units}}}$

AB:-

$\sqrt{ {( - 1 - 1)}^{2} + {(0 - 4)}^{2} } \\ \sqrt{ { (- 2)}^{2} + {( - 4)}^{2} } \\ \sqrt{4 + 16} \\ \sqrt{20} \\ 2 \sqrt{5}$

BC:-

$\sqrt{ {(1 - ( - 1 ))}^{2} + {( - 4 - 0)}^{2} } \\ \sqrt{( {2})^{2} + {( - 4)}^{2} } \\ \sqrt{4 + 16} \\ \sqrt{20} \\ 2 \sqrt{5}$

CD:-

$\sqrt{ {(3 - 1)}^{2} + {( - 0 - 4)}^{2} } \\ \sqrt{ {(2)}^{2} + {( - 4)}^{2} } \\ \sqrt{4 + 16} \\ \sqrt{20} \\ 2 \sqrt{5}$

DA:-

$\sqrt{ {(1 - 3)}^{2} + {(4 - 0)}^{2} } \\ \sqrt{ {( - 2)}^{2} + {(4)}^{2} } \\ \sqrt{4 + 16} \\ \sqrt{20} \\ 2 \sqrt{5}$

☞ AB = BC = CD = DA → $\sf{2\sqrt{5}}\;units$

• Since All Sides are equal it might be a Square or Rhombus.

AC:-

$\sqrt{ {(1 - 1)}^{2} + {( - 4 - 4)}^{2} } \\ \sqrt{0 + {( - 8)}^{2} } \\ \sqrt{64} \\ \sqrt{ {8}^{2} } \\ 8$

BD:-

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

AC ≠ BD

• Since, The diagonals are not equal it is a Rhombus

ABCD is a ☞ $\Large\sf{\color{salmon}янοмϐυѕ}$

☆ $\pink{\bf{Refer\;The\; Attachment}}$

$\Large{\tt{@MrSovereign}}$

Hope This Helps!!