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3 years ago

PLEASE HELPPP!!!!!! Write an equation of the line passing through point P that is perpendicular to the given line.

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:(d): y=ax+b (a≠0); (d1)y=-4x+13

because of (d)⊥(d1) --> a.(-4)=-1 --> a= $\frac{1}{4}$

(d): y= $\frac{1}{4}x$ +b. (d) pass through P(-1;-2) so we have

-2=(1/4).(-1)+b ---> b= $\frac{-7}{4}$

(d) y= $\frac{1}{4}x+ \frac{-7}{4}$

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<h2>Answer:</h2>

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$\text{Perimeter\ of\ square}=4s$

Now, it is given that:

The perimeters of square region S and rectangular region R are equal.

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$4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}$

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$\text{Area\ of\ square}=s^2$

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$\text{Area\ of\ Rectangle}=L\times B$

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$\text{Area\ of\ Rectangle}=6x^2$

Hence,

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$\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)$

Angle θ is given by
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$\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\ \\ -4(x^2+81)+9(x^2-10x+41)=0 \\ \\ -4x^2-324+9x^2-90x+369=0 \\ \\ 5x^2-90x+45=0 \\ \\ x^2-18x+9=0 \\ \\ x=9\pm6 \sqrt{2}$

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Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.

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