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Chapter 14: Problem 1281

The line \(3 \mathrm{x}-4 \mathrm{y}+7=0\) is rotated through an angle \((\pi / 4)\) in the clockwise direction about the point \((-1,1)\). The equation of the line in its new position is (a) \(7 \mathrm{y}+\mathrm{x}-6=0\) (b) \(7 \mathrm{y}-\mathrm{x}-6=0\) (c) \(7 \mathrm{y}+\mathrm{x}+6=0\) (d) \(7 \mathrm{y}-\mathrm{x}+6=0\)

Short Answer

Expert verified
The correct answer is \(7y - x - 6 = 0\).

Step by step solution

01

Determine the slope of the given equation

First, rewrite the given equation as \(y = mx + c\): \[3x - 4y + 7 = 0\] \[-4y = -3x + 7\] \[y = \frac{3}{4}x - \frac{7}{4}\] The slope of the given line is \(m = \frac{3}{4}\).
02

Rotate the slope by \(\pi/4\) in the clockwise direction

When a line is rotated clockwise by an angle \(\alpha\), its slope, \(m\), changes according to this formula: \[m_\text{rotated} = \frac{m \cos(\alpha) + \sin(\alpha)}{-m \sin(\alpha) + \cos(\alpha)}\] Since we are rotating by an angle of \(\pi/4\), substitute \(\alpha = \pi/4\) into the formula: \[m_\text{rotated} = \frac{(\frac{3}{4}) \cos(\pi/4) + \sin(\pi/4)}{-(\frac{3}{4}) \sin(\pi/4) + \cos(\pi/4)}\] Calculate the rotated slope using these values: \[m_\text{rotated} = \frac{(\frac{3}{4}) (\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}}{-(\frac{3}{4}) (\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}}\] \[m_\text{rotated} = \frac{6 - \sqrt{2}}{4}\]
03

Find the equation of the new line using the rotated slope and the given point \((-1, 1)\)

Now, we need to find the equation of the new line using the rotated slope and the point \((-1, 1)\). Substitute these values into the equation \(y = m_\text{rotated} x + c\): \[1 = (\frac{6 - \sqrt{2}}{4})(-1) + c\] Solve for \(c\): \[c = 1 + \frac{6 - \sqrt{2}}{4} = \frac{10 - \sqrt{2}}{4}\] So, the equation of the new line is: \[y = \frac{6 - \sqrt{2}}{4}x + \frac{10 - \sqrt{2}}{4}\]
04

Verify which of the answer choices matches the new line equation

Finally, compare this equation to the given answer choices and find the one that matches (after clearing fractions and simplifying): (a) \[7y + x - 6 = 0\] (b) \[7y - x - 6 = 0\] (c) \[7y + x + 6 = 0\] (d) \[7y - x + 6 = 0\] Choice (b) resembles our result: \[7y - x - 6 = 0\] \[-x + 7y - 6 = 0\] \[x = 7y - 6\] Therefore, the correct answer is (b), \(7y - x - 6 = 0\).

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Most popular questions from this chapter

Consider a square OPQR having the length of side a, where \(\mathrm{O}(0,0)\). The sides \(\underline{\mathrm{OP}}\) and \(\underline{\mathrm{OR}}\) are along the positive \(\mathrm{X}\) -axis and Y-axis respectively. If \(\mathrm{A}\) and \(\mathrm{B}\) are the mid points of \(\underline{\mathrm{PQ}}\) and \(Q \underline{R}\) respectively, then the angle between \(\underline{\mathrm{OA}}\) and \(\underline{\mathrm{OB}}\) would be \(\ldots \ldots \ldots\) (a) \(\cos ^{-1}(3 / 5)\) (b) \(\tan ^{-1}(4 / 3)\) (c) \(\cos ^{-1}(3 / 4)\) (d) \(\sin ^{-1}(3 / 5)\)

The angle between the lines \(\\{(\mathrm{x}, 0) /(\mathrm{x} \in \mathrm{R})\\}\) and \(\\{(0, y) /(y \in R)\\}\) is \(\ldots \ldots\) (a) \((\pi / 2)\) (b) \(-(\pi / 2)\) (c) 0 (d) \(\pi\)

If the points \([k, 2-2 k],[1-k, 2 k]\) and \([-k-4,6-2 k]\) are collinear, the possible value of \(\mathrm{k}\) are (a) \(-(1 / 2), 1\) (b) \((1 / 2),-1\) (c) \(1.2\) (d) \(1.3\)

The image of origin in the line \(\mathrm{x}+4 \mathrm{y}=1 \mathrm{is}\) (a) \([(2 / 17),-(8 / 17)]\) (b) \([-(2 / 17),-(8 / 17)]\) (c) \([-(2 / 17),(8 / 17)]\) (d) \([(2 / 17),(8 / 17)]\)

The \(\mathrm{p}-\alpha\) form of the line \(\mathrm{x}+\sqrt{(3) \mathrm{y}-4}=0 \mathrm{is}\) (a) \(x \cos (\pi / 6)+\operatorname{ysin}(\pi / 6)=2\) (b) \(x \cos (\pi / 3)+y \sin (\pi / 3)=2\) (c) \(x \cos [-(\pi / 3)]+y \sin [-(\pi / 3)]=2\) (d) \(x \cos [-(\pi / 6)]+y \sin [-(\pi / 6)]=2\)
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