JEE MainMathematicsToughStraight Lines

Let equation of two sides of a triangle are 4x + 5y = 20 and 3x
- 2y + 6 = 0. If orthocentre of triangle is (1, 1) then the equation of
third side of triangle is

Show answer and solution

4x + 5y = 20..........(i)
3x - 2y + 6 = 0..........(ii)
Orthocenter (1, 1)
Line

\bot

to 4x + 5y = 20 and passes (1, 1) is 5x - 4y = 1..........(iii)
and line

\bot

to 3x - 2y + 6 = 0 and passes (1, 1) is 2x + 3y = 5..........(iv)
on solving (i) & (iv) p

\left( \frac{35}{2}, - 10 \right)

on solving (ii) & (iii) pt

\left( - 13, - \frac{33}{2} \right)

side BC is y + 10 =

\frac{- \frac{33}{2} + 10}{- 13 - \frac{35}{2}}\left( x + \frac{35}{2} \right)

\Rightarrow y - \frac{23}{2} = \frac{33 - 92}{28 + 55}\left( x + \frac{55}{8} \right)

y + 10 = \frac{- 13}{61}\left( x - \frac{35}{2} \right)

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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