Properties of Triangles 3 Question 1
1. 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?
(2019 Main, 10 Jan II)
(a) Fourth
(b) Third
(c) Second
(d) First
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Answer:
Correct Answer: 1. (c)
Solution:
- Let $A B C$ be a given triangle with vertices $B(0,2), C(4,3)$ and let third vertex be $A(a, b)$
$(0,2)$
Also, let $D, E$ and $F$ are the foot of perpendiculars drawn from $A, B$ and $C$ respectively.
Then, $A D \perp B C \Rightarrow \frac{b-0}{a-0} \times \frac{3-2}{4-0}=-1$
[if two lines having slopes $m _1$ and $m _2$, are
$$ \begin{array}{lc} \Rightarrow & b+4 a=0 \\ \text { and } & C F \perp A B \\ \Rightarrow & \frac{b-2}{a-0} \times \frac{3-0}{4-0}=-1 \\ \Rightarrow & 3 b-6=-4 a \\ \Rightarrow & 4 a+3 b=6 \end{array} $$
From Eqs. (i) and (ii), we get
$$ -b+3 b=6 \Rightarrow 2 b=6 $$
$$ \begin{array}{lll} \Rightarrow & b & =3 \\ \text { and } & & a=-\frac{3}{4} \end{array} $$
So, the third vertex
$(a, b) \equiv-\frac{3}{4}, 3$, which lies in II quadrant.
