Question

if a vertex of a parallelogram is (2, 3) and the diagonals cut at (3, -2) find the opposite vertex
​

Answer 1

Answer:

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points (x

1

,y

1

) and (x

2

,y

2

) in the ratio m:n is

(x,y)=(

m+n

mx

2

+nx

1

,

m+n

my

2

+ny

1

)

O is the midpoint of AC and BD

∴(2,−5)= mid point of AC

(2,−5)=(

2

x

1

+3

,

2

y

1

+2

)

2

x

1

+3

=2

⇒x

1

+3=4

∴x

1

=1

Also,

2

y

1

+2

=−5

y

1

+2=−10

y

1

=−12

∴c=(1,−12)

(2,−5)= Mid point of BD

(2,−5)=(

2

−1+x

2

,

2

0+y

2

)

⇒

2

−1+x

2

=2

⇒−1+x

2

=4

∴x

2

=5

⇒

2

0+y

2

=−5

∴y

2

=−10

∴D=(5,−10)

Answer 2

Answer:

If a vertex of a parallelogram is (2, 3) and the diagonals cut at (3, -2), D=(5,−10) at the opposite vertex.

Explanation:

The coordinates of the points that internally split the line segment connecting the points $(x^1,y^1)$ and $(x^2,y^2)$ in the ratio m:n are known as the section formula points:

$(x,y) = (m+nmx^2+nx^1,m+nmy^2+ny^1)$

O is the midpoint of AC and BD

∴(2,−5) = midpoint of AC

$(2,−5)=(2x^1+3,2y^1+2)$

$2x^1+3=2$

⇒$x^1+3=4$

∴$x^1=1$

Also,$2y^1+2= -5$

$y^1+2= -10y^1= -12$

∴c=(1,−12)

(2,−5)= Midpoint of BD

$(2,-5)=(2-1+x^2,20+y^2)*2=1+x^2=2$

$-1+x^2=4$

∴$x^2=5$

⇒$20+y^2=−5$

∴$y^2=−10$

∴D=(5,−10)

If a vertex of a parallelogram is (2, 3) and the diagonals are cut at (3, -2), D=(5,−10) at the opposite vertex.

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