Question

SHOW THAT THE POINTS (-1,0) (0,3) (1,3) (0,0) are vertices of a paralleogram

Answer 1

SOLUTION

Refer to the attachment

hope it helps ☺️

Answer 2

let the points are A(-1,0),B(0,3),C(1,3) and D(0,0)

now,,,,

AB=

$\sqrt{( - 1 - 0) {}^{2} + (0 - 3) {}^{2} } = \sqrt{10}$

BC=

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

CD=

$\sqrt{(1 - 0) {}^{2} + (3 - 0) {}^{2} } = \sqrt{10}$

DA=

$\sqrt{(0 + 1) {}^{2} + (0 - 0) {}^{2} } = 1$

therefore. ......

AB=CD;BC=DA ( the opposite sides are equal)...(I)

now ......

Any diagonal .....AC=

$\sqrt{( - 1 - 1) {}^{2} + (0 - 3) {}^{2} } = \sqrt{13}$

AB^2+BC^2=

$( \sqrt{10) } {}^{2} + (1) {}^{2} \\ = 11$

AC^2=13

therefore....AB^2+BC^2 not equal to AC^2

so angle <B is not equal to 90°....(ii)

so from (I) and (I) we can say that A,B,C,D are the vertices of a parallelogram