The point with the co-ordinates $(2 a, 3 a), (3 b, 2 b)$ & $(c, c)$ are collinear _____________ .

for no value of

a, b, c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

for all values of

a, b, c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a, \frac{c}{5}, b

are in

H . P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a, \frac{2}{5} c, b

are in

H . P

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D if $a, \frac{2}{5} c, b$ are in $H . P$ .
$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 a & 3 a & 1 3 b & 2 b & 1 c & c & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow (2 a - c) (2 b - c) - (3 a - c) (3 b - c) = 0$

$\Rightarrow 4 a b - 2 a c - 2 b c + c^{2} = (9 a b - 3 a c - 3 b c + c^{2}) = 0$

$\Rightarrow a c + b c - 5 a b = 0$

$\frac{1}{a} + \frac{1}{b} = \frac{5}{c} \Rightarrow \frac{1}{a} + \frac{1}{b} = 2 (\frac{5}{2 c})$

$∵$ $\frac{1}{a}, \frac{5}{2 c} a n d \frac{1}{b}$ are in $A . P .$

$∴ a, \frac{2 c}{5}, b$ are in H.P.

Suggest Corrections

Similar questions

(2 a, 3 a)

(3 b, 2 b)

(c, c)

If a, b, c are three distinvt positive real numbers which are in H.P., then $\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}$ is

a, b, c

\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}

If a, b, c are three distinct positive real numbers which are in H.P., then $\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}$ is

\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}

Join BYJU'S Learning Program