JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 13)
Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant :
third
fourth
second
first
Explanation
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)
_10th_January_Evening_Slot_en_13_1.png)
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.
$$ \Rightarrow $$ $$\left( {{{3 - 2} \over {4 - 0}}} \right) \times \left( {{{b - 0} \over {a - 0}}} \right) = - 1$$
$$ \Rightarrow $$ 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.
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