If (a, b) is the orthocentre of the triangle | vedclass

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(A) Let the vertices be $A(1, 2), B(2, 3)$,and $C(3, 1)$.First,find the orthocentre $(a, b)$.The slope of $BC = \frac{1 - 3}{3 - 2} = -2$. The altitud

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(A) Let the vertices be $A(1, 2), B(2, 3)$,and $C(3, 1)$.First,find the orthocentre $(a, b)$.The slope of $BC = \frac{1 - 3}{3 - 2} = -2$. The altitude from $A$ to $BC$ has slope $\frac{1}{2}$.Equation of altitude from $A$: $y - 2 = \frac{1}{2}(x - 1) \implies x - 2y + 3 = 0$.The slope of $AC = \frac{1 - 2}{3 - 1} = -\frac{1}{2}$. The altitude from $B$ to $AC$ has slope $2$.Equation of altitude from $B$: $y - 3 = 2(x - 2) \implies 2x - y - 1 = 0$.Solving $x - 2y = -3$ and $2x - y = 1$,we get $x = \frac{5}{3}, y = \frac{7}{3}$. So $(a, b) = (\frac{5}{3}, \frac{7}{3})$.Note that $a + b = \frac{5}{3} + \frac{7}{3} = 4$.Now,$I_1 = \int_{a}^{b} x \sin(4x - x^2) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$:$I_1 = \int_{a}^{b} (a + b - x) \sin(4(a + b - x) - (a + b - x)^2) dx$Since $a + b = 4$,$I_1 = \int_{a}^{b} (4 - x) \sin(4(4 - x) - (4 - x)^2) dx = \int_{a}^{b} (4 - x) \sin(16 - 4x - (16 - 8x + x^2)) dx = \int_{a}^{b} (4 - x) \sin(4x - x^2) dx$.Thus,$I_1 = 4 \int_{a}^{b} \sin(4x - x^2) dx - I_1 \implies 2I_1 = 4I_2 \implies \frac{I_1}{I_2} = 2$.Therefore,$36 \frac{I_1}{I_2} = 36 	imes 2 = 72$.

If $(a, b)$ is the orthocentre of the triangle whose vertices are $(1, 2), (2, 3)$ and $(3, 1)$,and $I_1 = \int_{a}^{b} x \sin(4x - x^2) dx$,$I_2 = \int_{a}^{b} \sin(4x - x^2) dx$,then $36 \frac{I_1}{I_2}$ is equal to:

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