Question

If the straight line 3x + 4y + 5 - k(x+y+3) = 0 is parallel to y-axis

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ value \: of \: k \\ \\ so \: we \: know \: that \\ slope \: of \: y - axis \: is \: undefined \\ ie \: \: \frac{1}{0} \\ \\ let \: then \: here \\ m1 = \frac{1}{0} \\ \\ moreover \: given \: straight \: line \: is \\ 3x + 4y + 5 - k( x + y + 3) = 0 \\ \\ 3x + 4y + 5 - kx - ky - 3k = 0 \\ \\ x(3 - k) + y(4 - k) + 5 - 3k = 0 \\ \\ comparing \: with \: now \\ \: ax + by + c = 0 \\ we \: get \\ a = 3 - k \\ b = 4 - k \\ \\ we \: have \: \\ \\ slope \: of \: this \: line = \frac{ - a}{b} \\ \\ m2 = \frac{ - (3 - k)}{4 - k} \\ \\ but \: we \: know \: that \\ whenever \: two \: lines \\ are \: parallel \\ their \: corresponding \: slopes \: \\ are \: equal \\ \\ thus \: then \: \\ m1 = m2 \\ \\ \frac{1}{0} = \frac{ - (3 - k)}{4 - k} \\ \\ 4 - k = 0( - 3 + k) \\ \\ 4 - k = 0 \\ \\ k = 4$