Question

find the equation of the line passing through the point (5,4) and perpendicular to the line 3y+2x=1

Answer 1

Answer:

y= 1/2x +3

Step-by-step explanation:

3y + 2x =1

3y = -2x +1 (-2 is the gradient of this equation)

when a line is perpendicular we negative reciprocate the gradient

gradient = 1/2

y=mx+c

5=1/2 (4) +c

5-2= c

c=3

y=mx+c

y=1/2x +3

Answer 2

Answer:

hey here is your solution

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Step-by-step explanation:

$so \: here \: let \\ (5,4) = (x1,y1) \\ \\ so \: let \: the \: slope \: of \: the \: \: line \\ \: whose \: equation \: is \: to \: be \: formed \: \\ be \: \: m1 \\ \\ so \: here \: another \: line \: is \: \\ 2x + 3y = 1 \\ \\ comparing \: it \: with \\ \: ax + by = c \\ we \: have \\ \\ a = 2 \\ b = 3 \\ c = 1 \\ \\ so \: we \: know \: that \\ \\ slope = \frac{ - a}{b} \\ \\ m2 = \frac{ - 2}{3} \\ \\ so \: as \: \: 2x + 3y = 1 \: is \\ perpendicular \: to \: \: line \: with \\ slope \: \: m1\\ \\ m1 \times m2 = - 1 \\ \\ m1 \times ( \frac{ - 2}{3} ) = - 1 \\ \\ m1 = \frac{3}{2}$

$so \: we \: know \: that \\ slope - point \: form \: equation \: of \\ straight \: line \: is \: \\ given \: by \\ \\ y - y1 = m(x - x1) \\ \\ y - 4 = \frac{3}{2} (x - 5) \\ \\ 2(y - 4) = 3(x - 5) \\ \\ 2y - 8 = 3x - 15 \\ \\ 3x - 2y = 7 \\ \\ or \\ \\ 3x - 2y - 7 = 0$

$hence \: the \: required \: equation \: of \\ straight \: line \: with \: points \: \: (5,4) \\ and \: slope \: \frac{3}{2} \: \: is \\ \\ 3x - 2y = 7$