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  • Using method of integration find the area of the triangle ABC, co-ordinates of whose vertices are A(1, –2), B(3, 5) and C(5, 2).       <div style="float:left; w

Using method of integration find the area of the triangle ABC, co-ordinates of whose vertices are A(1, –2), B(3, 5) and C(5, 2).

 

 

 

 
 
 
 
 

Answers (1)

Equation of AC= y-2= \frac{2-\left ( -2 \right )}{5-1}\left ( x-8 \right )
               \Rightarrow y= x-3
Equation of  AB= y-5= \frac{5-\left ( -2 \right )}{3-1}\left ( x-3 \right )\Rightarrow y-5= \frac{7}{2}\left ( x-3 \right )
                      \Rightarrow y= \frac{7x-11}{2}
Equation of BC= y-5= \frac{5-2}{3-5}\left ( x-3 \right )
              \Rightarrow y= \frac{19-3x}{2}


Required area
\Rightarrow \int_{1}^{3}\frac{7x}{2}-\frac{11}{2}-\int_{1}^{3}\left ( x-3 \right )dx+\int_{3}^{5}(\frac{19}{2}-\frac{3x}{2})dx-\int _{3}^{5}\left ( x-3 \right )dx
\Rightarrow \int_{1}^{3}\left ( \frac{5x}{2}-\frac{5}{2} \right )dx+\int_{3}^{5}\left ( \frac{25}{2}-\frac{5x}{2} \right )dx
\Rightarrow [\frac{5}{2}\times \frac{x^{2}}{2}-\frac{5}{2}x]^{3}_{1}+[\frac{25}{2}x\, -\frac{5x^{2}}{4}]^{5}_{3}
\Rightarrow \frac{5}{2}\times9-\frac{15}{2}-\left ( \frac{5}{4} -\frac{5}{2}\right )+\frac{125}{2}-\frac{125}{4}-\left ( \frac{75}{2}-\frac{45}{4} \right )
\Rightarrow \frac{45}{4}-\frac{15}{2}-\frac{5}{4}+\frac{5}{2}+\frac{125}{2}-\frac{125}{4}-\frac{75}{2}+\frac{45}{4}
\Rightarrow \frac{45-30-5+10+250-125-150+45}{4}=\frac{40}{4}= 10

Posted by

Ravindra Pindel

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