Question

find the equations of the diagonals of a rectangle whose sides are x=-1,x=4,y=-1,y=2​

Answer 1

Answer:

ANSWER

Given,

The equations of sides of a rectangle are

x

1

=−1,x

2

=2,y

1

=2,y

2

=6.

These lines form a rectangle when they intersect at A, B, C, D respectively

Now,

The co-ordinates of A, B, C and D will be (−1,−2),(2,−2),(2,6),(−1,6) respectively.

And, AC and BD are its diagonals.

Slope of the diagonal AC

=

x

2

−x

1

y

2

−y

1

=

(2+1)

(6+2)

=

3

8

=m

So, the equation of AC will be

y−y

1

=m(x−x

1

)

y+2=

3

8

(x+1)

3y+6=8x+8

⇒8x−3y+2=0

Slope of the diagonal BD

=

x

2

−x

1

y

2

−y

1

=

(−1−2)

(6+2)

=

3

−8

=m

So, the equation of BD will be

y−y

1

=m(x−x

1

)

y−6=

3

−8

(x+1)

3y−18=−8x−8

⇒8x−3y−10=0