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Mathematics, 12.07.2019 19:40 happysage12

What is the area of a triangle with vertices at (−4, 1) , (−7, 5) , and (0, 1) ? enter your answer in the box.

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Answers: 2

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Answer from: fooligan3448

The area of a shape is the amount of space it occupies.

The area of the triangle is 8 unit square.

The vertices are given as:

$\mathbf{A = (-4,1)}$

$\mathbf{B = (-7,5)}$

$\mathbf{C = (0,1)}$

The area of the triangle is:

$\mathbf{A= \frac{1}{2} \times |A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|}$

So, we have:

$\mathbf{A= \frac{1}{2} \times |-4(5 - 1) -7(1 - 1) + 0(1 - 5)|}$

$\mathbf{A= \frac{1}{2} \times |-4(4) -7(0) + 0|}$

$\mathbf{A= \frac{1}{2} \times |-4(4) -0 + 0|}$

$\mathbf{A= \frac{1}{2} \times |-16|}$

Remove absolute brackets

$\mathbf{A= \frac{1}{2} \times 16}$

$\mathbf{A= 8}$

Hence, the area of the triangle is 8 unit square.

Read more about areas at:

link

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Answer from: Aced

we know that

A method for calculating the area of a triangle when you know the lengths of all three sides is the Heron's Formula.

The Heron's Formula states that

$A=\sqrt{p(p-a)(p-b)(p-c)}$

where

a,b.c are the length sides of the triangle

p is half the perimeter of the triangle

Let

$A(-4,1)\\B(-7,5)\\C(0,1)$

Step 1

Find the distance AB

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$A(-4,1)\\B(-7,5)$

substitute the values in the formula

$d=\sqrt{(5-1)^{2}+(-7+4)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$dAB=5\ units$

Step 2

Find the distance BC

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$B(-7,5)\\C(0,1)$

substitute the values in the formula

$d=\sqrt{(1-5)^{2}+(0+7)^{2}}$

$d=\sqrt{(-4)^{2}+(7)^{2}}$

$d=\sqrt{65}$

$dBC=8.06\ units$

Step 3

Find the distance AC

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$A(-4,1)\\C(0,1)$

substitute the values in the formula

$d=\sqrt{(1-1)^{2}+(0+4)^{2}}$

$d=\sqrt{(0)^{2}+(4)^{2}}$

$d=\sqrt{16}$

$dAC=4\ units$

Step 4

Find half the perimeter

$p=\frac{1}{2}(AB+BC+AC)$

substitute the values

$p=\frac{1}{2}(5+8.06+4)$

$p=8.53\ units$

Step 5

Find the area

Applying the Heron's Formula

$A=\sqrt{8.53(8.53-5)(8.53-8.06)(8.53-4)}$

$A=\sqrt{8.53(3.53)(0.47)(4.53)}$

$A=\sqrt{64.11}$

$A=8\ units^{2}$

therefore

the answer is

the area of the triangle is $8\ units^{2}$

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Answer from: Quest

answer: kljdcdhjbdlchjad oac apoich a

step-by-step explanation:

Which set of data could be used for the box-and-whisker plot shown below? 23, 21, 20, 22, 24, 17, 1

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Answer from: Quest

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = $\frac{-8+3}{4-7}$ = - $\frac{5}{3}$

y = - $\frac{5}{3}$ x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - $\frac{20}{3}$ + c ⇒ c = - $\frac{4}{3}$

y = - $\frac{5}{3}$ x - $\frac{4}{3}$ ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form

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