Question

The three vertices of a parallelogram ABCD are A(3, -4), B(-1, -3), and C(-6, 2).
Find the coordinates of vertex d and find the area of ABCD

Answer 1

=1/2{(-9-2+24)-(6+18+4)}
=1/2(13-28)
=1/2(-15)
=Area will be 7.5 square unit as area can not be negative

Answer 2

Answer:

Coordinate of 4th vertex is (-2,1)

Step-by-step explanation:

Given three of the vertices of a square are the points whose coordinates are A(3, -4), B(-1, -3), and C(-6, 2)

we have to find the coordinates of the fourth vertex.

By mid-point formula, if (a,b) and (c,d)  are the vertices of line segment then the coordinates of mid-point are

$(\frac{a+b}{2},\frac{c+d}{2})$

As the diagonals of parallelogram bisect each other therefore the mid point of both the diagonal are same.

∴ mid-point of AC=mid-point of BD

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

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

Comparing

-1+x=-3 and -3+y=-2

x=-2 and y=1

Hence, coordinate of 4th vertex is (-2,1)