If is the orthocenter of the triangle ABC with vertices A ( 3 , - 7 ) , B ( - 1 , 2 ) and C ( 4 , 5 ) , then 9 - 6+ 60 is equal to [2023]

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Solution

Q.
If $(α, β)$ is the orthocenter of the triangle ABC with vertices $A (3, - 7)$ , $B (- 1, 2)$ and $C (4, 5)$ , then $9 α - 6 β + 60$ is equal to [2023]

1 30

2 35

3 25

4 40

Ans.
(3)

Draw altitudes BM, CN and AP, then their intersection point is the orthocentre.

Now, let us find the equation of line BM, CN and AP.

Slope of $A C = \frac{12}{1}$

So, slope of $B M = \frac{- 1}{12}$

Equation of BM: $(y - 2) = \frac{- 1}{12} (x + 1)$

$\Rightarrow 12 y - 24 = - x - 1 \Rightarrow 12 y + x = 23$ ...(i)

Now, slope of $A B = \frac{9}{- 4}$

So, slope of $C N = \frac{4}{9}$

Equation of CN: $(y - 5) = \frac{4}{9} (x - 4) \Rightarrow 9 y - 45 = 4 x - 16$

$\Rightarrow 4 x - 9 y - 16 + 45 = 0 \Rightarrow 4 x - 9 y + 29 = 0 ...(ii)$

$Similarly, equation of A P : 5 x + 3 y = - 6 ...(iii)$

$Now, solving (i) and (ii), we get y = \frac{121}{57}, x = \frac{- 141}{57}$

$Also,$ $(\frac{- 141}{57}, \frac{121}{57})$ $satisfies (iii)$

$So, orthocentre, (α, β) = (\frac{- 141}{57}, \frac{121}{57})$

and $9 α - 6 β + 60 = 9 \times (\frac{- 141}{57}) - \frac{6 \times 121}{57} + 60$

$= \frac{- 423}{19} - \frac{242}{19} + \frac{1140}{19} = \frac{475}{19} = 25$

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