Question

What is the area of a triangle whose vertices are J(−2, 1) , K(4, 3) , and L(−2, −5) ?

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Answer 1

we know that

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

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

where

a,b,c -----> are the lengths of the sides of a triangle

p ----> is half the perimeter

we have

$J(-2,1)\ K(4,3)\ L(-2,-5)$

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

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

Step 1

Find the distance JK

$J(-2,1)\ K(4,3)$

Substitute the values in the formula of distance

$d=\sqrt{(3-1)^{2}+(4+2)^{2}}$

$d=\sqrt{(2)^{2}+(6)^{2}}$

$dJK=\sqrt{40}\ units$

Step 2

Find the distance KL

$K(4,3)\ L(-2,-5)$

Substitute the values in the formula of distance

$d=\sqrt{(-5-3)^{2}+(-2-4)^{2}}$

$d=\sqrt{(-8)^{2}+(-6)^{2}}$

$dKL=10\ units$

Step 3

Find the distance JL

$J(-2,1)\ L(-2,-5)$

Substitute the values in the formula of distance

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

$d=\sqrt{(-6)^{2}+(0)^{2}}$

$dJL=6\ units$

Step 4

Find the perimeter of the triangle

we know that

the perimeter of a triangle is the sum of the length sides of the triangle

so

$P=dJK+dKL+dJL$

substitute the values

$P=\sqrt{40}\ units+10\ units+6\ units=22.32\ units$

Find half the perimeter

$p=22.32/2=11.16\ units$

Step 5

Find the area of the triangle

Applying the Heron's Formula

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

we have

$p=11.16\ units$

$a=dJK=\sqrt{40}\ units=6.32\ units$

$b=dKL=10\ units$

$c=dJL=6\ units$

substitute the values

$Area=\sqrt{11.16(11.16-6.32)(11.16-10)(11.16-6)}$

$Area=\sqrt{11.16(4.84)(1.16)(5.16)}$

$Area=17.98\ units^{2}$

$Area=18\ units^{2}$

therefore

the answer is

The area of the triangle is $18\ units^{2}$

Answer 2

Answer:

here is a screenshot of the answer this is the real asnwer

Step-by-step explanation: