Straight Line And Pair Of Straight Lines Ques 63
- The vertices of a triangle are
$\left[a t _1 t _2, a\left(t _1+t _2\right)\right],\left[a t _2 t _3, a\left(t _2+t _3\right)\right]$, $\left[a t _3 t _1, a\left(t _3+t _1\right)\right]$.
Find the orthocentre of the triangle.
(1983, 3M)
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Answer:
Correct Answer: 63.$\left[\left(-a, a\left(t _1+t _2+t _3+t _1 t _2 t _3\right)\right]\right.$
Solution:
Formula:
Equation Of Lines Parallel And Perpendicular To A Given Line :
- Let $A B C$ be a triangle whose vertices are $A\left[a t _1 t _2, a\left(t _1+t _2\right)\right], \quad B\left[a t _2 t _3, a\left(t _2+t _3\right)\right] \quad$ and $C\left[a t _1 t _3, a\left(t _1+t _3\right)\right]$.
Then, Slope of $B C=\frac{a\left(t _2+t _3\right)-a\left(t _1+t _3\right)}{a t _2 t _3-a t _1 t _3}=\frac{1}{t _3}$
Slope of $A C=\frac{a\left(t _1+t _3\right)-a\left(t _1+t _2\right)}{a t _1 t _3-a t _1 t _2}=\frac{1}{t _1}$
So, the equation of a line through $A$ perpendicular to $B C$ is $\quad y-a\left(t _1+t _2\right)=-t _3\left(x-a t _1 t _2\right)$ and the equation of a line through $B$ perpendicular to $A C$ is
$$ y-a\left(t _2+t _3\right)=-t _1\left(x-a t _2 t _3\right) $$
The point of intersection of Eqs. (i) and (ii), is the orthocentre.
On subtracting Eq. (ii) from Eq. (i), we get $x=-a$.
On putting $x=-a$ in Eq. (i), we get
$$ y=a\left(t _1+t _2+t _3+t _1 t _2 t _3\right) $$
Hence, the coordinates of the orthocentre are $\left[-a, a\left(t _1+t _2+t _3+t _1 t _2 t _3\right)\right]$.
BETA
