Question

the three vertices of a parallelogram p(3,-4) q(-1,-3) and. r(-6,2). find the coordinates of the fourth vertex S. also find the area of parallelogram PQRS.

Answer 1

Answer:

Step-by-step explanation:

(-2,1)

Answer 2

Answer:

Forth vertex:S(-2,1)

Area:15 sq-unit

Step-by-step explanation:

To find the coordinates of the fourth vertex S, if the three vertices of a parallelogram P(3,-4),Q(-1,-3) and R(-6,2).

let the vertex S(x,y)

we all know that, Diagonal of parallelogram bisect each other

So, Apply mid-point formula to find midpoint of PR,let the Mid-point is N

$N (x)= \frac{3 - 6}{2} = \frac{ - 3}{2} \\\\ N(y) = \frac{ - 4 + 2}{2} = - 1 \\ \\ N = ( \frac{ - 3}{2}, - 1)$

It will be the Mid-point of QS

$\frac{ - 3}{2} = \frac{ - 1 + x}{2} \\ \\ = > - 3 = - 1 + x \\ \\ x = - 2 \\ \\ - 1 = \frac{ - 3 + y}{2} \\ \\ - 2 = - 3 + y \\ \\ y = 1$

Point S(-2,1)

2) Area of Parallelogram:Base×height

Since ,height is not given.

So,apply the formula of area of triangle

when three vertex are given.

(ar∆PQR)

$= \frac{1}{2} |3( - 3 - 2) - 1(2 + 4) - 6( - 4 + 3)| \\ \\ = \frac{1}{2} | - 15 - 6 + 6| \\ \\ = \frac{15}{2}$

ar(∆PRS)

$= \frac{1}{2} |3(2 - 1) - 6(1 + 4) - 2( - 4 - 2)| \\ \\ = \frac{1}{2} |3 - 30 + 12| \\ \\ = \frac{15}{2}$

Area of Parallelogram PQRS=(ar∆PQR)+ar(∆PRS)

=

$\frac{15}{2} + \frac{15}{2} \\ \\ = 15 \: unit ^{2}$

Hope it helps you.