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

Accepted Solution

we know thatThe 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[tex]Area=\sqrt{p(p-a)(p-b)(p-c)}[/tex]where a,b,c -----> are the lengths of the sides of a trianglep ----> is half the perimeterwe have[tex]J(-2,1)\ K(4,3)\ L(-2,-5)[/tex]The formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]Step 1Find the distance JK[tex]J(-2,1)\ K(4,3)[/tex]Substitute the values in the formula of distance[tex]d=\sqrt{(3-1)^{2}+(4+2)^{2}}[/tex][tex]d=\sqrt{(2)^{2}+(6)^{2}}[/tex][tex]dJK=\sqrt{40}\ units[/tex]Step 2Find the distance KL[tex]K(4,3)\ L(-2,-5)[/tex]Substitute the values in the formula of distance[tex]d=\sqrt{(-5-3)^{2}+(-2-4)^{2}}[/tex][tex]d=\sqrt{(-8)^{2}+(-6)^{2}}[/tex][tex]dKL=10\ units[/tex]Step 3Find the distance JL[tex]J(-2,1)\ L(-2,-5)[/tex]Substitute the values in the formula of distance[tex]d=\sqrt{(-5-1)^{2}+(-2+2)^{2}}[/tex][tex]d=\sqrt{(-6)^{2}+(0)^{2}}[/tex][tex]dJL=6\ units[/tex]Step 4Find the perimeter of the trianglewe know thatthe perimeter of a triangle is the sum of the length sides of the triangleso[tex]P=dJK+dKL+dJL[/tex]substitute the values[tex]P=\sqrt{40}\ units+10\ units+6\ units=22.32\ units[/tex]Find half the perimeter[tex]p=22.32/2=11.16\ units[/tex]Step 5Find the area of the triangleApplying the Heron's Formula[tex]Area=\sqrt{p(p-a)(p-b)(p-c)}[/tex]we have[tex]p=11.16\ units[/tex][tex]a=dJK=\sqrt{40}\ units=6.32\ units[/tex][tex]b=dKL=10\ units[/tex][tex]c=dJL=6\ units[/tex]substitute the values[tex]Area=\sqrt{11.16(11.16-6.32)(11.16-10)(11.16-6)}[/tex][tex]Area=\sqrt{11.16(4.84)(1.16)(5.16)}[/tex][tex]Area=17.98\ units^{2}[/tex][tex]Area=18\ units^{2}[/tex]thereforethe answer isThe area of the triangle is [tex]18\ units^{2}[/tex]