Question

If the line (3x – y + 5) + K(2x - 3y - 4) = 0 is parallel to y-axis, then K= -1 (A (D) O ( (B) i Tim (C) 1: C 5​

Answer 1

Step-by-step explanation:

We have,

$(3x - y + 5) + K(2x - 3y - 4) = 0$

$\implies \: 3x - y + 5 + 2 K x - 3 K y - 4 K = 0$

$\implies \:( 3 + 2K) x - (1 + 3K) y + (5 - 4 K) = 0$

Slope of this line is $0$, because the line is parallel to x-axis,

$\frac{3 + 2K }{1 + 3 K } = 0$

$\implies \: K = - \frac{3}{2}$

so, the required line is,

$\implies \: \bigg \{3 - 2 \bigg(\frac{3}{2} \bigg) \bigg \} x - \bigg \{1 - 3 \bigg( \frac{3}{2} \bigg) \bigg \} y + \bigg \{5 + 4 \bigg( \frac{3}{2} \bigg)\bigg \} = 0$

$\implies \: \{3 - 3 \} x - \bigg \{1 - \frac{9}{2} \bigg \} y + \{5 + 6 \} = 0$

$\implies \: - \bigg \{ - \frac{7}{2} \bigg \} y + 11 = 0$

$\implies \: 7 y + 22 = 0$