Two vertices of a triangle are (0,2) and (4,3).

Question

Two vertices of a triangle are $(0,2)$ and $(4,3)$. If its orthocentre is at the origin, then its third vertex lies in which quadrant?

Accepted Answer

mBD $$ \times $$ mAD = - 1

$$ \Rightarrow $$  $$\left( {{{3 - 2} \over {4 - 0}}} \right) \times \left( {{{b - 0} \over {a - 0}}} \right) =  - 1$$

$$ \Rightarrow $$  b + 4a = 0      . . . . (i)

mAB $$ \times $$ mCF = - 1

$$ \Rightarrow $$  $$\left( {{{\left( {b - 2} \right)} \over {a - 0}}} \right) \times \left( {{3 \over 4}} \right) =  - 1$$

$$ \Rightarrow $$  3b - 6 = - 4a

$$ \Rightarrow $$  4a + 3b = 6      . . . . .(ii)

From (i) and (ii)

a = $${{ - 3} \over 4}$$,  b = 3

$$ \therefore $$  IInd quadrant.

Solutions · 1