If (a, b) be the orthocentre of the triangle whose vertices are (1,2),(2,3) and (3,1), and…

Question

If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x, \mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$, then $36 \frac{\mathrm{I}_1}{\mathrm{I}_2}$ is equal to :

Accepted Answer

Equation of CE

$$\begin{aligned}
& y-1=-(x-3) \
& x+y=4
\end{aligned}$$

orthocentre lies on the line $x+y=4$

so, $a+b=4$

$$I_1=\int_\limits a^b x \sin (x(4-x)) d x\quad$$ ..... (i)

Using king rule

$$I_1=\int_\limits a^b(4-x) \sin (x(4-x)) d x\quad$$ .... (ii)

$$\begin{aligned}
& \text { (i) }+ \text { (ii) } \
& 2 \mathrm{I}_1=\int_\limits{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \
& 2 \mathrm{I}_1=4 \mathrm{I}_2 \
& \mathrm{I}_1=2 \mathrm{I}_2 \
& \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \
& \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72
\end{aligned}$$

Solutions · 1