Question

The line x=c cuts the trTheiangle with vertices (0,0),(1,1) and (9,1) into two regions. For the area of two regions to be the same, c must be equal to​

Answer 1

Answer:

c=3

Step-by-step explanation:

Let us assume that the three vertices of the triangle be O(0,0), A(9,1) and B(1,1).

So, equation of OA is given by $y=\frac{1-0}{9-0}x+C$

⇒$y=\frac{x}{9}$ { As the line passes through origin, so, C=0}

Similarly, the equation of OB is given by, $y=x$

And the equation of AB is given by, $y=1$

Let us assume again that the line x=c cuts ΔOAB between A and B.

Now, the area of the right hand region will be

=$\int\limits^9_c {(1-\frac{x}{9} )} \, dx$

=$(9-\frac{9}{2})-c+\frac{c^{2} }{18}$

=$\frac{c^{2} }{18}-c+\frac{9}{2}$

Again, the area of left hand region will be

=$\int\limits^1_0 {(x-\frac{x}{9} )} \, dx+\int\limits^c_1 {(1-\frac{x}{9} )} \, dx$

=$\frac{8*1^{2} }{9*2}+c-\frac{c^{2}}{18}-1+\frac{1}{18}$

=$\frac{4}{9}+c-\frac{c^{2} }{18}-1+\frac{1}{18}$

Given that the area of those two region will be same.

So, $\frac{c^{2} }{18}-c+\frac{9}{2}$ = $\frac{4}{9}+c-\frac{c^{2} }{18}-1+\frac{1}{18}$

⇒$\frac{c^{2} }{9}-2c+5=0$

⇒$c^{2}-18c+45=0$

⇒$(c-15)(c-3)=0$

⇒c=3 {Since c can not be greater than 9}

(Answer)