prove that P(2, -4), Q(-1,2), R(4,8) and S(7,2) are the vertices of a rectangle.
Please answer it and the correct one will be marked as brainliest. Wrong answers will be reported.
Answers
Answered by
1
Answer:
PQ=√{(2--1)^2+(-4-2)^2}=√(9+36)=√45
RS=√{7-4)^2+(2-8)^2}=√(9+36)=√45
PR=√{(4-2)^2+(8+4)^2}=√(4+144=√148
SP=√{(2-7)^2+(-4-2)^2}=√25+36)=√61
QR=√{(4+1)^2+(8-2)^2}=√(25+36)=√61
QS=√{(7+1)^2+(2-2)^2}=√64
You can see that opposite sides are equal. But diagonals are not equal. one diagonal is √148 and the other is√64. In a rectangle diagonals should be equal.
Answered by
2
Step-by-step explanation:
hit A(−4,−1), B(−2,4), C(4,0) and D(2,3) be the given points (circles)
Now, AD=
(2+4)
2
+(3+1)
2
=
52
BC=
(4+2)
2
+(0+4)
2
=
52
AB=
(−2+4)
2
+(−4+1)
2
=
4+9
=
13
CD=
(2−4)
2
+(3−0)
2
=
13
⇒ AD=BC and AB=CD
Hence ABCD is a parallelogram
Now, AC=
(4+4)
2
+(0+1)
2
=
65
BD=
(2+2)
2
+(3+4)
2
=
65
Clearly AC
2
=65=AB
2
+BC
2
=(
52
)
2
+(
13
)
2
=65
and (BD)
2
=BC
2
+CD
2
65=52+13=65
Hence, ABCD is a rectangle
⇒ The points (−4,−1, )(4,0) and (2,3) are which of rectangle.
Hope its help..
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