First slide

Equation of line in Straight lines

Question

$The set of real values of k for which the lines x + 3 y + 1 = 0, k x + 2 y - 2 = 0 and 2 x - y + 3 = 0 form a triangle is$

Moderate

A

$R - \{- 4, \frac{2}{3}\}$
B

$R - \{- 4, \frac{- 6}{5}, \frac{2}{3}\}$
C

$R - \{\frac{- 2}{3}, 4\}$
D

$R$

Solution

$Lines form triangle. Therefore, x + 3 y + 1 = 0 is not parallel to k x + 2 y - 2 = 0 \Rightarrow \frac{- k}{2} x - y + 1 = 0$
$\begin{array}{l} ∴ \frac{- k}{2} \neq \frac{- 1}{3} \Rightarrow k \neq \frac{2}{3} \\ Also line 2 x - y + 3 = 0 is not parallel to k x + 2 y - 2 = 0 \end{array}$
$\frac{2 x}{3} - \frac{y}{3} + 1 = 0 \frac{- k x}{2} - y + 2 = 0 \Rightarrow \frac{- 3 k}{4} \neq 3 \Rightarrow k \neq - 4$
$\begin{aligned} Further lines must not be concurrent. \\ ∴ |\begin{array}{rrr} 1 & 3 & 1 \\ k & 2 & - 2 \\ 2 & - 1 & 3 \end{array}| \neq 0 \Rightarrow k \neq \frac{- 6}{5} \end{aligned}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App