Question

The vertices ofa parellogram are given in quadinate A B C D the coordinate are A(-2,3) B(3,-1) C(p,q)
D(-1,9)​

Answer 1

Given:-

A = (-2,3) => (x1, y1) , B = (3, -1) => (x2, y2)

C = (p, q) => (x3, y3) , D = (-1, 9) => (x4, y4)

To find:-

coordinates of point C.

Solution:-

• By slope formula

slope of line AB = (y2 - y1) / (x2 - x1)

= (-1 - 3) / (3 - (-2))

= (-4) / ( 3 + 2)

= -4/5,, ....(1)

Here the slope of line AB is -4/5.

• According to one of the property of parellelogram.

AB = CD

•°• slope of line AB = slope of line DC = -4/5 ..(2)

• By slope formula

Slope of line DC = -4/5

(y4 - y3) /(x4 - x3) = -4/5

(9 - q) / (-1 - p) = -4/5

5(9 - q) = -4(-1 - p)

45 - 5q = 4 + 4p

45 - 4 = 4p + 5q

4p + 5q = 41 ....(3)

now,

AD = BC ...{sides of parallelogram}

• By slope formula

Slope of line AD = (y4 - y1) / (x4 - x1)

= (9 - 3) / (-1 - (-2))

= 6/( -1 + 2) = 6/1

slope oglf line = 6,, ....(4)

•°• slope of line BC = 6

(y3 - y2) / (x3 - x2) = 6

(q - (-1)) / (p - 3) = 6

(q + 1) / (p - 3) = 6/1

6(p - 3) = 1(q + 1)

6p - 18 = q + 1

6p - q = 1 + 18

6p - q = 19 ......(5)

• Solving equation (3) and (5)

4p + 5q = 41 ....(3)

6p - q = 19 ......(5)

• Multiplying the eq (5) by 5

•°• 30p - 5q = 95 .... (6)

• Adding the equation (3) and (6)

•°• 30p - 5q = 95

+ 4p + 5q = 41

=> 26p = 54

p = 54/26 = 27/13

p = 27/13 = 2.07

• Substituting in eq (3)

•°• (4 × 27/ 13) - 5q = 41

108/ 13 - 41 = 5q

(108 - 533)/13 = 5q

425/13 = 5q

425/13 × 5 = q

85/13 = 6.53 = q

Answer the coordinate of C is (27/13, 85/13) i.e (2.07, 6.53)