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  • The vertices of a triangle are <img alt="\left[a t_1 t_2, a\left(t_l+t_2\right)\right],\left[a t_2 t_3, a\left(t_2+t_3\right)\right],\left[a t_3 t_l, a\left(t_3+t_I\right)\right] \text {.

The vertices of a triangle are

\left[a t_1 t_2, a\left(t_l+t_2\right)\right],\left[a t_2 t_3, a\left(t_2+t_3\right)\right],\left[a t_3 t_l, a\left(t_3+t_I\right)\right] \text {. }

Find the abscissa of orthocentre of the triangle.

Option: 1

\mathrm{a t_1 t_2 t_3}


Option: 2

\mathrm{a\left(t_1+t_2+t_3\right)}


Option: 3

\mathrm{-a}


Option: 4

\mathrm{a}


Answers (1)

best_answer

Solution
Let A, B, C be the vertices

\begin{aligned} \mathrm{A} & \mathrm{\equiv\left[a t_1 t_2, a\left(t_1+t_2\right)\right]} \\ \mathrm{B} &\mathrm{ \equiv\left[a t_2 t_3, a\left(t_2+t_3\right)\right]} \\ \mathrm{C} &\mathrm{ \equiv\left[a t_3 t_1, a\left(t_3+t_l\right)\right]} \end{aligned}

Slope of \mathrm{B C=\frac{a\left(t_3+t_1\right)-a\left(t_2+t_3\right)}{a t_3 t_l-a t_2 t_3}=\frac{I}{t_3}}

Slope of altitude \mathrm{A D=-t_3}

Equation of AD is

\mathrm{y-a\left(t_1+t_2\right)=-t_3\left(x-a t_1 t_2\right) \quad \quad \quad \dots(1)}

Similarly equation of altitude B E is

\mathrm{y-a\left(t_2+t_3\right)=-t_1\left(x-a t_2 t_3\right) \quad \quad \quad \dots(2)}

The coordinates of orthocentre O are obtained by simultaneously solving equation (1) and equation (2).

Subtracting equation (2) from (1)

\mathrm{a\left(t_z-t_I\right)=-x\left(t_z-t_I\right) \Rightarrow x=-a }

Posted by

manish painkra

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