Question

find the equation of the line drawn perpendicular to X/6 - y/4 =1 through the point where it meets the X axis.​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ equation \: of \: straight \: line \\ \\ so \: a \: certain \: line \: is \: perpendicular \\ to \: line \: \: \: \frac{x}{6} - \frac{y}{4} = 1 \\ \\ \frac{x}{6} - \frac{y}{4} = 1 \: \: can \: be \: written \: as \\ 4x - 6y = 24 \\ \\ 2x - 3y = 12 \\ on \: comparing \: the \: equation \\ with \: \: ax + by + c = 0 \\ we \: get \\ a = 2 \\ b = - 3 \\ c = - 12 \\ \\ so \: we \: know \: that \\ m = \frac{ - a}{b} \\ \\ = \frac{ - 2}{ - 3} \\ \\ m1= \frac{2}{3}$

$since \: both \: lines \: are \: perpendicular \\ m1.m2 = - 1 \\ \\ \frac{2}{3} .m2 = - 1 \\ \\ m2 = \frac{ - 3}{2}$

$similarly \: as \: the \: required \: line \\ meets \: \: the \: x - axis \\ (x1,y1) = (p,0) \\ \\ we \: know \: that \\ x - intercept = \frac{ - c}{a} \\ \\ = \frac{ - ( - 12)}{2} \\ \\ = \frac{12}{2} \\ \\ p= 6 \\ \\ (x1,y1) = (6,0)$

$so \: equation \: of \: required \: line \\ can \: be \: expressed \: in \\ slope - intercept \: \: form \: ie \: in \\ y - y1 = m(x - x1) \\ \\ so \: then \\ y - 0 = - \frac{3}{2} (x - 6) \\ \\ 2y - 3(x - 6) \\ \\ 2y = - 3x + 18 \\ \\ 3x + 2y = 18 \\ \\ or \\ \\ 3x + 2y - 18 = 0$