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Chapter 14: Problem 1263

The angle between the lines \(\mathrm{x}=3\) and \(\sqrt{3} \mathrm{x}-\mathrm{y}+5=0\) is \(\ldots \ldots\) (a) \((\pi / 6)\) (b) \((\pi / 3)\) (c) \((\pi / 4)\) (d) \((\pi / 2)\)

Short Answer

Expert verified
The angle between the lines $\mathrm{x}=3$ and $\sqrt{3}\mathrm{x}-\mathrm{y}+5=0$ is $\boxed{\frac{\pi}{4}}$ (option c).

Step by step solution

01

Slope of the vertical line

The given equation for the first line is \(\mathrm{x}=3\). This is a vertical line with an undefined slope. For our calculations, we will consider the slope of a vertical line to be infinite.
02

Slope of the second line

The equation of the second line is given as \(\sqrt{3}\mathrm{x}-\mathrm{y}+5=0\). To find the slope, rearrange the equation in the form of \(\mathrm{y=mx+b}\), where \(\mathrm{m}\) is the slope. First, move the y term to the left side: \(\mathrm{y}=\sqrt{3}x-5\) The slope of the second line is \(\sqrt{3}\).
03

Find the angle between the lines using the tangent formula

The formula for finding the angle \(\theta\) between two lines with slopes \(\mathrm{m_1}\) and \(\mathrm{m_2}\) is given by: \[\tan(\theta) = \frac{m_1 - m_2}{1 + m_1m_2}\] For the given lines, \(\mathrm{m_1}\) is infinite and \(\mathrm{m_2}\) is \(\sqrt{3}\). Plug in the slopes into the formula: \[\tan(\theta) = \frac{\infty - \sqrt{3}}{1 + \infty \cdot \sqrt{3}}\] Since the denominator will also be infinite, this simplifies to: \[\tan(\theta) = \frac{\infty - \sqrt{3}}{\infty} \approx 1\] Now, calculate the angle by taking the arctangent of both sides: \[\theta = \arctan(1)\] \[\theta = \frac{\pi }{4}\]
04

Compare the result with the given options

Our calculated angle between the lines is \(\frac{\pi }{4}\). Compare this angle to the given options: (a) \((\pi / 6) \ne \frac{\pi }{4}\) (b) \((\pi / 3) \ne \frac{\pi }{4}\) (c) \((\pi / 4) = \frac{\pi }{4}\) (d) \((\pi / 2) \ne \frac{\pi }{4}\) Our result matches the option (c), which is the correct answer.

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Most popular questions from this chapter

The equation of the lines with slope \(-2\) and intersecting \(\mathrm{x}\) -axis at points distance 3 unit from the origin is...... (a) \(2 \mathrm{x}+\mathrm{y} \pm 6=0\) (b) \(x+2 y \pm 6=0\) (c) \(2 \mathrm{x}+\mathrm{y} \pm 3=0\) (d) \(x+2 y \pm 3=0\)

If \(\mathrm{P}(-1,0), \mathrm{Q}(0,0)\) and \(\mathrm{R}(3,3 \sqrt{3})\), then the equation of bisector of \(\angle P Q R\) is \(\ldots \ldots .\) (a) \((\sqrt{3} / 2) \mathrm{x}+\mathrm{y}=0\) (b) \(x+(\sqrt{3} / 2) y=0\) (c) \(\sqrt{3 x+y}=0\) (d) \(x+\sqrt{3} y=0\)

The equation of a line passing through \((4,3)\) and the sum of whose intercepts is \(-1\) is........ (a) \((\mathrm{x} / 2)+(\mathrm{y} / 3)=1,(\mathrm{x} / 2)+(\mathrm{y} / 1)]=1\) (b) \((\mathrm{x} / 2)+(\mathrm{y} / 3)=-1,[\mathrm{x} /(-2)]+(\mathrm{y} / 1)=1\) (c) \((\mathrm{x} / 2)+(\mathrm{y} / 3)=-1,\\{\mathrm{x} /(-2)\\}+(\mathrm{y} / 1)=-1\) (d) \((\mathrm{x} / 2)-(\mathrm{y} / 3)=1,\\{\mathrm{x} /(-2)\\}+(\mathrm{y} / 1)=1\)

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) \(\pi\)

Shifting origin at which point the transformed form of \(\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-8 \mathrm{y}-85=0\) would be \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{k}\) ? (a) \((2,4)\) (b) \((-2,-4)\) (c) \((2,-4)\) (d) \((-2,4)\)
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