Question

Find the coordinates of C in a parallelogram ABCD in which the coordinates of the vertices A( – 1, 0), B(1, 3) and D(3, 5).

Answer 1

Answer:

(5, 8)

Step-by-step explanation:

Define the coordinate:

Let the coordinate of C be (x, y)

It is a parallelogram

⇒ Slope of AB = Slope of CD

⇒ Slope of AD = Slope of BC

Find the Slope of AB:

Slope = (Y2 - Y1)/(X2 - X1)

Slope of AB = (3 - 0)/(1 - (-1) ) = 3/2

Find the slope of CD:

Slope = (Y2 - Y1)/(X2 - X1)

Slope of CD = (y - 5)/( x - 3)

Slope of AB = Slope of CD

(y - 5)/( x - 3)  = 3/2

2(y - 5) = 3(x - 3)

2y - 10 = 3x - 9

2y = 3x + 1

Find the slope of AD:

Slope = (Y2 - Y1)/(X2 - X1)

Slope of AD = (5 - 0)/(3 - (-1) ) = 5/4

Find the slope of BC:

Slope = (Y2 - Y1)/(X2 - X1)

Slope of BC = (y - 3)(x - 1)

Slope of AD = Slope of BC

(y - 3)(x - 1) = 5/4

4(y - 3) = 5(x - 1)

4y - 12 = 5x - 5

4y = 5x + 7

Solve for x:

2y = 3x + 1   ----------- [ 1 ]

4y = 5x + 7 ----------- [ 2 ]

[ 1 ] x 2 :

4y = 6x + 2 ----------- [ 1a ]

[ 1a ] - [ 2 ]:

x - 5 = 0

x = 5

Solve for y:

2y = 3x + 1

2y = 3(5) + 1

2y = 16

y = 8

Fid the coordinate of c:

x = 5

y = 8

Coordinate of C = (5, 8)

Answer: (5, 8)

Answer 2

Answer:

(5,8)

Step-by-step explanation:

Concept:

The diagonals of a parallelogram bisect each other.

Midpoint formula:

The midpoint of the line joining $(x_1,y_1) and (x_2,y_2) \:is\:(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Given:

A(-1,0), B(1,3), D(3,5)

Let the third vertex C be (x,y)

since diagonals bisect each other,

we have

Midpoint of diagonal AC = Midpoint of of diagonal BD

$(\frac{-1+x}{2},\frac{0+y}{2})=(\frac{1+3}{2},\frac{3+5}{2})\\\\(\frac{-1+x}{2},\frac{y}{2})=(\frac{4}{2},\frac{8}{2})\\\\(\frac{-1+x}{2},\frac{y}{2})=(2,\:4)$

Equaating corresponding coordinates

we get

$\frac{-1+x}{2}=2, \frac{y}{2}=8$

-1+x = 4, y=8

x=4+1,y=8

x=5, y=8

Therefore the coordinates of C is (5,8)