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3 years ago

Line E passes through the points (2, 3) and (4, -3). What is the slope of a line perpendicular to line E?

1 answer:

7 0

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 2 &,& 3~) 
% (c,d)
&&(~ 4 &,& -3~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-3-3}{4-2}\implies \cfrac{-6}{2}\implies \cfrac{-3}{1}$

now, a line perpendicular to that one, will have a "negative reciprocal" slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for}\quad \cfrac{-3}{1}\\\\
negative\implies +\cfrac{3}{ 1}\qquad reciprocal\implies +\cfrac{ 1}{3}\implies \cfrac{1}{3}$

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PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

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The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$

Thus, both sides of the equation are equal.

Hence, $x=\frac{5}{2}$ is a true solution.

Thus, solutions are not extraneous.

Hence, Option C is the correct answer.

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