1. quadrilateral abcd has vertices a(-1, 1), b(2, 3), c(6, 0) and d(3, -2). determine using coordinate geometry whether or not the diagonals* of abcd a. bisect each other. b. are congruent. c. are perpendicular.1. quadrilateral abcd has vertices a(-1, 1), b(2, 3), c(6, 0) and d(3, -2). determine using coordinate geometry whether or not the diagonals* of abcd a. bisect each other. b. are congruent. c. are perpendicular.

The Conclusion is

Diagonals AC and BD,

a. Bisect each other

b. Not Congruent

c. Not Perpendicular

Step-by-step explanation:

Given:

[]ABCD is Quadrilateral having Vertices as

A(-1, 1),

B(2, 3),

C(6, 0) and

D(3, -2).

So the Diagonal are AC and BD

To Check

The diagonals AC and BD

a. Bisect each other. B. Are congruent. C. Are perpendicular.

Solution:

For a. Bisect each other

We will use Mid Point Formula,

If The mid point of diagonals AC and BD are Same Then

Diagonal, Bisect each other,

For mid point of AC

Substituting the coordinates of A and C we get

Similarly, For mid point of BD

Substituting the coordinates of B and D we get

Therefore The Mid point of diagonals AC and BD are Same

Hence Diagonals,

a. Bisect each other

B. Are congruent

For Diagonals to be Congruent We use Distance Formula

For Diagonal AC

Substituting A and C we get

Similarly ,For Diagonal BD

Substituting Band D we get

Therefore Diagonals Not Congruent

For C. Are perpendicular.

For Diagonals to be perpendicular we need to have the Product of slopes must be - 1

For Slope we have

Substituting A and C we get

Similarly, for BD we have

The Product of slope is not -1

Hence Diagonals are Not Perpendicular.

- 1.613 = -1.6 or -1.61

-1.522 = -1.5 or -1.52

0.587 = 0.6 or 0.59

3.079 = 3.1 or 3.08

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: