Question

Show from the gradients that the points (4,1),(1,2),(2,5) and (5,4) are the vertices of a square.
(This question is from the chapter Co-ordinate geometry)
Please help me to solve this and no spams please or else the answer will be reported ​

Answer 1

Question :-

Show from the gradients that the points (4,1),(1,2),(2,5) and (5,4) are the vertices of a square.

Solution :-

Let the vertices are,

R(4 , 1) , A(1 , 2) , J(2 , 5) , S(5 , 4)

We know that,

Distance formula, XY = √(x2 - x1)² + (y2 - y1)²

Case (I),

•R(4 , 1) , A(1 , 2)

↪ RA = √(1 - 4)² + (2 - 1)²

↪ RA = √(3)² + (-1)²

↪ RA = √9 + 1

↪ RA = √10 _________(1)

Case (II),

•A(1 , 2) , J(2 , 5)

↪ AJ = √(2 - 1)² + (5 - 2)²

↪ AJ = √(-1)² + (3)²

↪ AJ = √1 + 9

↪ AJ = √10_________(2)

Case (III),

•J(2 , 5) , S(5 , 4)

↪ JS = √(5 - 2)² + (4 - 5)²

↪ JS = √(3)² + (-1)²

↪ JS = √9 + 1

↪ JS = √10________(3)

Case (IV),

•R(4 , 1) , S(5 , 4)

↪ RS = √(5 - 4)² + (4 - 1)²

↪ RS = √(1)² + (3)²

↪ RS = √1 + 9

↪ RS = √10__________(4)

From eqn. (1) , (2) , (3) and (4) , We get ;

↪ RA = AJ = JS = RS

[ .°. All sides are equal ]

Therefore,

□ RAJS is a square.

Answer 2

```solution ```

let the vertices are

R ( 4,1 ) , A ( 1,2) , J (2,5), S (5,4)

we know that

$distance \: formula \: xy = \sqrt{( {x}^{2} } - {x}^{1} ) + ( {y}^{2} - {y}^{1} {)}^{2}$

case ( 1 )

$ra = \sqrt{(1 - {4)}^{2} } + (2 - {1)}^{2} \\ ra = \sqrt{ {(3)}^{2} } + { (- 1)}^{2} \\ ra = \sqrt{9 + 1 } \\ ra = \sqrt{10} (1)$

case (2)

$aj = \sqrt{(2 - {1)}^{2} } + (5 - {2)}^{2} \\ aj \sqrt{ ( - {1)}^{2} + {(3)}^{2} } \\ aj = \sqrt{1 + 9 } \\ aj = \sqrt{10} (2)$

case (3)

$js = \sqrt{ {(5 - 2)}^{2} + {(4 - 5)}^{2} } \\ js = \sqrt{ {(3)}^{2} + {( - 1)}^{2} } \\ js = \sqrt{9 + 1} \\ js = \sqrt{10} (3)$

case (4)

$rs = \sqrt{ {(5 - 4)}^{2} + {(4 - 1)}^{2} } \\ rs = \sqrt{ {(1)}^{2} + ( {3)}^{2} } \\ rs = \sqrt{1 + 9} \\ rs = \sqrt{10} (4)$

from equation (1) ,(2 ),(3)and ( 4 )

we get

RA = AJ = JS = RS

( all sides are equal )

therefore

.......RAJS is a square......