Straight LineHardBloom L3
Question
If $P \equiv \left( \frac{1}{x_{p}},p \right);Q = \left( \frac{1}{x_{q}},q \right);R = \left( \frac{1}{x_{r}},r \right)$
where $x_{k} \neq 0$, denotes the $k^{\text{th}\text{~}}$ terms of a H.P. for $k \in N$, then:
Options
A.ar. $( \bigtriangleup PQR) = \frac{p^{2}q^{2}r^{2}}{2}\sqrt{(p - q)^{2} + (q - r)^{2} + (r - p)^{2}}$
B.$\bigtriangleup PQR$ is a right angled triangle
C.the points $P,Q,R$ are collinear
D.None of these
Solution
Sol. [Key Concept: Points $(\frac{1}{x_{k}},k)$ from a HP are collinear (the area determinant vanishes).]
$\frac{1}{x_{p}} = a + (p - 1)d$;
$${\frac{1}{x_{q}} = a + (q - 1)d;\ \frac{1}{x_{r}} = a + (r - 1)d }{\Rightarrow \left| \begin{matrix} a + (p - 1)d & p & 1 \\ a + (q - 1)d & q & 1 \\ a + (r - 1)d & r & 1 \end{matrix} \right| = 0}$$
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