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

we know thatA 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[tex]A=\sqrt{p(p-a)(p-b)(p-c)}[/tex]   wherea,b.c are the length sides of the trianglep is half the perimeter of the triangleLet[tex]A(-4,1)\\B(-7,5)\\C(0,1)[/tex]Step 1Find the distance ABwe know thatthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]we have[tex]A(-4,1)\\B(-7,5)[/tex]substitute the values in the formula[tex]d=\sqrt{(5-1)^{2}+(-7+4)^{2}}[/tex][tex]d=\sqrt{(4)^{2}+(-3)^{2}}[/tex][tex]d=\sqrt{25}[/tex][tex]dAB=5\ units[/tex]Step 2Find the distance BCwe know thatthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]we have[tex]B(-7,5)\\C(0,1)[/tex]substitute the values in the formula[tex]d=\sqrt{(1-5)^{2}+(0+7)^{2}}[/tex][tex]d=\sqrt{(-4)^{2}+(7)^{2}}[/tex][tex]d=\sqrt{65}[/tex][tex]dBC=8.06\ units[/tex]Step 3Find the distance ACwe know thatthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]we have[tex]A(-4,1)\\C(0,1)[/tex]substitute the values in the formula[tex]d=\sqrt{(1-1)^{2}+(0+4)^{2}}[/tex][tex]d=\sqrt{(0)^{2}+(4)^{2}}[/tex][tex]d=\sqrt{16}[/tex][tex]dAC=4\ units[/tex]Step 4Find half the perimeter[tex]p=\frac{1}{2}(AB+BC+AC)[/tex]substitute the values[tex]p=\frac{1}{2}(5+8.06+4)[/tex][tex]p=8.53\ units[/tex]Step 5Find the areaApplying the Heron's Formula[tex]A=\sqrt{8.53(8.53-5)(8.53-8.06)(8.53-4)}[/tex]   [tex]A=\sqrt{8.53(3.53)(0.47)(4.53)}[/tex]     [tex]A=\sqrt{64.11}[/tex]  [tex]A=8\ units^{2}[/tex]  thereforethe answer isthe area of the triangle is [tex]8\ units^{2}[/tex]