1. name the type of triangle pqr formed by the points p(√2,√2) , q(-√2,-√2) and r(-√6,√6)

2. abcd is a parallelogram with vertices a(x1,y1) b(

x2,y2) and c(x3,y3). find the coordinated of the fourth vertex d in terms of x1,x2,x3,y1,y2,y3

we were assigned 30 sums as homework.. i dint know how to solve these two out of them. please help me out....

Let Δ PQR be the given triangle.

Applying the distance formulae, we get

Since, PQ = QR = RP = 4

∴ ∆PQR is an equilateral triangle.

(2)

Given, ABCD is a parallelogram.

Let the coordinates of the vertex D be (*x*_{4}, *y*_{4}).

We know that, diagonals of a parallelogram bisect each other.

∴ Coordinate of mid point of diagonal BD = Coordinate of mid point of diagonal AC

Coordinate of mid point of diagonal

Coordinate of mid point of diagonal

Thus, coordinates of the fourth vertex D are (*x*_{1} + *x*_{3} – *x*_{2}, *y*_{1} + *y*_{3} – *y*_{2})

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