Question

(1,2) (3,6)are two opposite vertices of a rectangle and if the other two vertices lie on the line 2y=x+6, then other two vertices are​

Answer 1

Given:

(1,2) and (3,6) are two opposite vertices of a rectangle

The other two vertices lie on the line 2y = x+6

To find:

The other two vertices of the rectangle

Calculation:

As the other vertices lie on the line 2y = x+6, Let P(2h,(h+3) be one of the vertices of the rectangle

The sides of a rectangle will be perpendicular to each other. So the product of their slopes will be equal to -1

Consider two sides having vertices (1,2),(2h,(h+3) and (3,6),(2h,(h+3)

       $\frac{(h+3)-2 }{2h-1} \times \frac{(h+3)-6}{2h-3} =-1\\ \\\frac{h+1}{2h-1} \times \frac{h-3}{2h-3} =-1 \\ \\(h+1)(h-3)=-(2h-1)(2h-3) \\ \\h^2-2h-3+4h^2-6h-2h+3=0 \\ \\5h^2-10h=0\\ \\5h(h-2)=0\\ \\h=0,h=2$

By substituting the values in the point P, the other two vertices of the rectangle are (0,3) and (4,5)

Final answer:

The other two vertices of the triangle are (0,3) and (4,5)

Answer 2

Given : (1,2) (3,6)are two opposite vertices of a rectangle . other two vertices lie on the line 2y=x+6

To find : other two vertices  

Solution:

Diagonals of rectangle bisect each other  and equal in length

Hence mid point of

( 1 ,2 )  &  ( 3 , 6)  

= (1 + 3)/2 , (2 + 6)/2

= 2 , 4

Distance of  ( 2, 4)  from ( 1, 2)

= √(2 -1 )² + (4 - 2)²

= √5

Let say point on line 2y  = x  + 6

is   (2h ,  h + 3  ) vertex of rectangle

Distance from mid point

√(2h -2 )² + (h+3 - 4)²  = √5

=> 4h² - 8h + 4 + h² - 2h + 1 = 5

=> 5h²  - 10h = 0

= 5h(h - 2) =0

=> h = 0    , h  = 2

=> 2h =  0  ,   4

 h + 3 = 3   ,   5

( 0 , 3)   &  ( 4 , 5 )

other two vertices are​  ( 0 , 3)   &  ( 4 , 5 )  

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