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2 years ago

Find the equation of a line passing through (-4,-3) and perpendicular to 3 x + 2 y = 14.

1 answer:

3 0

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$3x+2y=14\implies 2y=-3x+14\implies y=\cfrac{-3x+14}{2}\implies y = \cfrac{-3x}{2}+\cfrac{14}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so therefore

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}$

so we're really looking for the equation of a line whose slope is 2/3 and passes through (-4 , -3)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-4)}) \implies y+3=\cfrac{2}{3}(x+4) \\\\\\ y+3=\cfrac{2}{3}x+\cfrac{8}{3}\implies y=\cfrac{2}{3}x+\cfrac{8}{3}-3\implies y=\cfrac{2}{3}x-\cfrac{1}{3}$

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(b) Now we have to see when it will land. At land or ground level height h will be equal to 0. So simply plug 0 in h place in equation made in part (a)

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Now we have to solve this quadratic. We will use quadratic formula method to solve this equation.

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Plugging these values in quadratic formula we get

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For t =2 seconds, $h = -16(2)^2+150(2)+100 =336 feet$

For t =4 seconds, $h = -16(4)^2+150(4)+100 = 444 feet$

For t =6 seconds, $h = -16(6)^2+150(6)+100 = 424 feet$

For t =8 seconds, $h = -16(8)^2+150(8)+100 = 276 feet$

For t =10 seconds, $h = -16(10)^2+150(10)+100 = 0 feet$

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$x = \frac{-b}{2a}$

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t = 4.6875 is axis of symmetry line

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$x = \frac{-b}{2a}$

so t = 4.6875

then to find y coordinate we will plug this value of t as 4.6875 in equation made in part (a)

For t =4.6875, $h = -16(4.6875)^2+150(4.6875)+100 = 451.563$

so vertex is at (4.6875, 451.563)

(f) As the firework is launched so in starting time is t=0, we cannot have time before t=0 (negative values) practically. Also we cannnot have firework going down into the ground so we cannot have h value negative physically.

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