Question

Three consecutive vertices of a parallelogram ABCD are A(1,2), B(1,0)and C(4,0). Find the fourth vertex ‘D’.

Answer 1

D(4,2)

GivEn:-

• ABCD is a parallelogram.

• Vertices of A = (1,2)

• Vertices of B = (1,0)

• Vertices of C = (4,0)

To find:-

• Vertices of D = ?

Solution:-

Let vertices of D is (x,y) and 0 is the mid - point of AC and BD

Since, the diagonals of a parallelogram bisect each other.

Therefore,

Cordinates of mid - point of AC = Cordinates of mid - point of BD

Now,

$\sf{ \dfrac{1+4}{2} , \dfrac{0+2}{2} = \dfrac{1+x}{2} , \dfrac{0+y}{2}}$

$\implies \sf{ \dfrac{5}{2} , \dfrac{2}{2} = \dfrac{1+x}{2} , \dfrac{y}{2}}$

Therefore,

$\sf{ \dfrac{5}{2} = \dfrac{1+x}{2}}$

$\implies \sf{10 = 2 + 2x}$ ....[cross multiplication]

$\implies \sf{8 = 2x}$

$\implies \sf{x = 4}$

Now,

$\sf{ \dfrac{y}{2} = 1}$

$\sf{y = 2}$

Therefore, (x,y) = (4,2)

★ Hence, Vertices of D is (4,2).

$\rule{200}{2}$

Answer 2

Answer:-

The coordinates of vertex D is (4,2) ,

Given:

Three consecutive vertices of a parallelogram

ABCD are:

• => A(1,2)

• => B(1,0)

• => C(4,0)

To find:

• Coordinates of vertex 'D'.

Solution:

Opposite side of parallelogram are parallel.

$\sf{\therefore}$ Slope of the opposite sides will also be equal.

$\boxed{\sf{Slope \ of \ line=\frac{y2-y1}{x2-x1}}}$

Let the coordinates of vertex D be (x,y)

Slope of AB = Slope of DC

$\sf{\therefore{\frac{0-2}{1-1}=\frac{0-y}{4-x}}}$

$\sf{\therefore{\frac{0-y}{4-x}=\frac{-2}{0}}}$

$\sf{\therefore}$ -2(4-x)=0

$\sf{\therefore}$ 2x=8

$\sf{\therefore{x=\frac{8}{2}}}$

$\sf{\therefore}$ x=4

Slope of AD = Slope of BC

$\sf{\therefore{\frac{y-2}{x-1}=\frac{0-0}{4-1}}}$

$\sf{\therefore{\frac{y-2}{x-1}=0}}$

$\sf{\therefore}$ y-2=0

$\sf{\therefore}$ y=2

The coordinates of vertex D is (4,2) .