Show that A(-4,-7) B(-1,2) C(8,5) P(5,-4) are the vertices of a Parallelogram.
Soln :
Answers
Step-by-step explanation:
Given vertices are A(-4,-7), B(-1,2),C(8,5) and D(5,-4).
We know that Distance : \sqrt{(x2 - x1)^2 + (y2 - y1)^2}Distance:
(x2−x1)
2
+(y2−y1)
2
Now,
AB = \sqrt{(-1 + 4)^2 + (2 + 7)^2}AB=
(−1+4)
2
+(2+7)
2
= \ \textgreater \ \sqrt{3^2 + 9^2}= \textgreater
3
2
+9
2
= \ \textgreater \ \sqrt{9 + 81}= \textgreater
9+81
= \ \textgreater \ \sqrt{90}= \textgreater
90
= \ \textgreater \ 3 \sqrt{10}= \textgreater 3
10
BC = \sqrt{(8 + 1)^2 + (5 - 2)^2 }BC=
(8+1)
2
+(5−2)
2
= \ \textgreater \ \sqrt{(9)^2 + (3)^2}= \textgreater
(9)
2
+(3)
2
= \ \textgreater \ \sqrt{81 + 9}= \textgreater
81+9
= \ \textgreater \ \sqrt{90}= \textgreater
90
= \ \textgreater \ 3 \sqrt{10}= \textgreater 3
10
CD = \sqrt{(5 - 8)^2 + (-4 - 5)^2}CD=
(5−8)
2
+(−4−5)
2
= \ \textgreater \ \sqrt{(-3)^2 + (-9)^2}= \textgreater
(−3)
2
+(−9)
2
= \ \textgreater \ \sqrt{9 + 81}= \textgreater
9+81
= \ \textgreater \ \sqrt{90}= \textgreater
90
= \ \textgreater \ 3 \sqrt{10}= \textgreater 3
10
AD = \sqrt{(5 + 4)^2 + (-4 + 7)^2}AD=
(5+4)
2
+(−4+7)
2
= \ \textgreater \ \sqrt{9^2 + 3^2}= \textgreater
9
2
+3
2
= \ \textgreater \ \sqrt{81 + 9}= \textgreater
81+9
= \ \textgreater \ \sqrt{90}= \textgreater
90
= \ \textgreater \ 3\sqrt{10}= \textgreater 3
10
Now,
Diagonals AC = \sqrt{(8 + 4)^2 + (5 + 7)^2}DiagonalsAC=
(8+4)
2
+(5+7)
2
= \ \textgreater \ \sqrt{(12)^2 + (12)^2}= \textgreater
(12)
2
+(12)
2
= \ \textgreater \ \sqrt{144 + 144}= \textgreater
144+144
= \ \textgreater \ \sqrt{288}= \textgreater
288
= \ \textgreater \ 12 \sqrt{2}= \textgreater 12
2
Diagonals BD = \sqrt{(5 + 1)^2 + (-4 - 2)^2}DiagonalsBD=
(5+1)
2
+(−4−2)
2
= \ \textgreater \ \sqrt{(6)^2 + (-6)^2}= \textgreater
(6)
2
+(−6)
2
= \ \textgreater \ \sqrt{36 + 36}= \textgreater
36+36
= \ \textgreater \ \sqrt{72}= \textgreater
72
= \ \textgreater \ 6 \sqrt{2}= \textgreater 6
2
Now,
Given that sides are equal but diagonals are not equal. Thus ABCD is a rhombus.
Hope this helps!