Question

show that the points A(3,0), B(4,5), C(-1,4) and D(-2,-1) taken in order are the vertices of a rhombus

Answer 1

Answer:

Step-by-step explanation:

Answer 2

★Solution:-⠀⠀

⠀⠀⠀⠀⠀⠀•Finding distance between AB by distance formula :-

$\sf \longrightarrow AB= \sqrt{(4-3)^2+(5-0)^2}= \sqrt{1+25} = \sqrt{26}$•

Finding distance between BC

$\sf \longrightarrow BC= \sqrt{(-1-4)^2+(4-5)^2}= \sqrt{25+1}= \sqrt{26}$•

Finding distance between CD

$\sf \longrightarrow CD= \sqrt{(-2+1)^2+(-1-4)^2} = \sqrt{1+25} = \sqrt{26}$•

Finding distance between DA

$\sf \longrightarrow DA= \sqrt{(3+2)^2+(0+1)^2}= \sqrt{25+1}= \sqrt{26}$

This above results shows us that ⇛AB=BC=CD=DA

↬All the sides are equal ⠀

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⠀Thus , the given points taken is order are the vertices of a rhombus ⠀

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⠀$\sf\longrightarrow AC= \sqrt{(-1-3)^2+(4-0)^2}= \sqrt{16+16} = \sqrt{32} = 4 \sqrt{2}$⠀⠀

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$\sf \longrightarrow BD= \sqrt{(-2-4)^2+(-1-5)^2}= \sqrt{36+36} = \sqrt{72} = 6 \sqrt{2}$

Therefore the given points forms a rhombus

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More to know;

↬The distance of the point P(0,0) from the origin is √(x²+y²)

↬The distance between the points P(x1,y1) and Q(x2,y2) =PQ=√[(x2-x1)²+(y2-y1)²]

↬This above formula is also known as distance formula which is used for finding distance between two points in coordinate.