Question

help me with this problem please right away​

Answer 1

Answer:

$6\frac{1}{2} = A$

Step-by-step explanation:

The first step is to use the Distance Formula to figure out the lengths of the two legs:

$\sqrt{[-x_1 + x_2]^{2} + [-y_1 + y_2]^{2}} = D$

R(1, 1) and P(−1, −2) ↷

$\sqrt{[1 + 1]^{2} + [2 + 1]^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}\\ \\ \sqrt{13} = RP$

Q(2, −4) and P(−1, −2) ↷

$\sqrt{[1 + 2]^{2} + [2 - 4]^{2}} = \sqrt{[-2]^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13} \\ \\ \sqrt{13} = QP$

* So, by my calculations, we have an isosceles right triangle, and according to the 45°-45°-90° triangle theorem, we automatically know that the hypotenuse, RQ, is $\sqrt{26}$:

30°-60°-90° triangle theorem

2x, x, x√3

↑ ↑ ↑

h leg leg

y

p

o

t

e

n

u

s

e

45°-45°-90° triangle theorem

x√2, x, x

↑ ↑ ↑

h legs

y

p

o

t

e

n

u

s

e

So now, we have to find the area of the triangle by taking half of either height or base, then multiplying that by the height or base, but since this is an isosceles right triangle, it does not matter:

$[\frac{h}{2}][b] = A \: \:OR\: \: [\frac{b}{2}][h] = A \: \:OR \: \: \frac{hb}{2} = A$

$[\frac{\sqrt{13}}{2}][\sqrt{13}] = \frac{13}{2} = 6\frac{1}{2} \\ \\ 6\frac{1}{2} = A$

I am joyous to assist you anytime.