Question

draw the graph of the equations 5x-y=5 and 3x-y=3. determinethe area of the triangle formed by these line and the y-axis.

Answer 1

find area of triangle by your self

Answer 2

Thus, Area of $\triangle ABC$ is $0.9978$ square units

Step-by-step explanation:

Given,

Two equation $5x-y=5$ and $3x-y=3$

As shown in graph point $A,B$ and $C$ makes a triangle.

Point $A$ is $(1,0)$

Point $B$ is $(0,-3)$

And point $C$ is $(0,-5)$

∴ $AB=\sqrt{(0-1)^{2}+(-3-0)^{2} }$

⇒$AB=\sqrt{1+9} =\sqrt{10}$ units

And $BC=\sqrt{(0-0)^{2} +(-5+3)^{2} }$

   ⇒$BC=\sqrt{4}=2$ units

Also, $AC=\sqrt{(0-1)^{2}+(-5-0)^{2} }$

⇒$AC=\sqrt{26}$ units

From Heron's Formula,

Area of $\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Where, $s=\frac{a+b+c}{2}$

In $\triangle ABC$,

$a=AB,b=BC$ and $c=AC$

∴ $s=\frac{\sqrt{10}+2+\sqrt{26} }{2}$

Area of $\triangle ABC=\sqrt{\frac{\sqrt{10}+2+\sqrt{26} }{2}(\frac{\sqrt{10}+2+\sqrt{26} }{2}-\sqrt{10} )(\frac{\sqrt{10}+2+\sqrt{26} }{2}-2)(\frac{\sqrt{10}+2+\sqrt{26} }{2}-\sqrt{26} )}$

           $=\sqrt{\frac{15.930}{16} }$

          $=0.9978$ square units

∴ Area of $\triangle ABC$ is $0.9978$ square units