Question

a triangle on coordinate plame has vertices (-8,3) (3,9) and (9,3) what is the area in square units of the triangle

Answer 1

Given :-

Vertices of triangle are (-8 , 3) , (3 , 9)  , (9 ,3)

To find :-

Area of triangle

Diagram :-

$\setlength{\unitlength}{2.5mm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(0,0)(0,0)(8,17)\qbezier(0,0)(0,0)(18,0)\qbezier(18,0)(18,0)(8,17)\put(8,17.8){\sf A=(-8,3)}\put( - 1, - 1){\sf B=(3,9)}\put(18,-1){\sf C=(9,3)}\put(8, - 1.5){\sf }\put(15,8.1){\sf }\put( - 0,8.1){\sf }\end{picture}$

Solution :-

We have formula to find the Area of triangle that is

$\dfrac{1}{2} \bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$

$x_1 = -8\\x_2 = 3\\x_3 = 9 \\y_1 = 3\\y_2 = 9 \\y_3 = 3$

Substituting the values ,

$= \dfrac{1}{2} \bigg|-8(9-3)+3(3-3)+9(3-9)\bigg|$

$= \dfrac{1}{2} \bigg|-8(6)+3(0)+9(-6)\bigg|$

$= \dfrac{1}{2} \bigg|-48+0-54\bigg|$

$= \dfrac{1}{2} \bigg|-102\bigg|$

$= \dfrac{1}{2} (102)$

$= 51 sq.units$

$\purple{\boxed{Area\:of\:triangle\:formed\:by\:those\:points\:are \:51 sq.units}}$

Method -2 :-

This method is about the using the concept of determinants We have formula that is

$\dfrac{1}{2} \left|\begin{array}{cc}x_1&y_1\\x_2&y_2\\\end{array}\right| +\left|\begin{array}{cc}x_2&y_2\\x_3&y_3\\\end{array}\right|+\left|\begin{array}{cc}x_3&y_3\\x_1&y_1\\\end{array}\right|$

Substituting the values ,

$\dfrac{1}{2} \left|\begin{array}{cc}-8&3\\3&9\\\end{array}\right| +\left|\begin{array}{cc}3&9\\9&3\\\end{array}\right|+\left|\begin{array}{cc}9&3\\-8&3\\\end{array}\right|$

Solving by formula ad-bc

$=\dfrac{1}{2}\bigg| [(-8)(9)-(3)(3)] +3(3)-9(9) +9(3)-3(-8)\bigg|$

$=\dfrac{1}{2}\bigg| (-72-9)+(9-81) +(27+24)\bigg|$

$=\dfrac{1}{2}\bigg| -81-72+51\bigg|$

$=\dfrac{1}{2}\bigg| -153+51\bigg|$

$=\dfrac{1}{2} \bigg|-102\bigg|$

$=\dfrac{1}{2} (102)$

$=51 sq.units$

$\red{\boxed{Area\:of\:triangle\:formed\:by\:those\:points\:are \:51 sq.units}}$