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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

A square of side 'a' lies above the \(\mathrm{x}\) -axis and has one vertex at the origin. The side passing through the origin makes an angle o. \([0<\alpha<(\pi / 4)]\) with the positive direction of \(\mathrm{x}\) -axis. The equation of the diagonal not passing through the origin is (a) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a\) (b) \(y(\cos \alpha+\sin \alpha)+x(\sin \alpha-\cos \alpha)=a\) (c) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha+\cos \alpha)=a\) (d) \(y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a\)

In a triangle \(\mathrm{ABC}\), coordinates of \(\mathrm{A}\) are \((1,2)\) and the equations of the medians through \(\mathrm{B}\) and \(\mathrm{C}\) are \(\mathrm{x}+\mathrm{y}=5\) and \(\mathrm{x}=4\) respectively. Then coordinates of \(\mathrm{B}\) and \(\mathrm{C}\) will be (a) \((-2,7),(4,3)\) (b) \((7,-2),(4,3)\) (c) \((2,7),(-4,3)\) (d) \((2,-7),(3,-4)\)

Find the slope of the line passing through the point \((1,2)\) and the point of intersection of this line with the line \(\mathrm{x}+\mathrm{y}+3=0\) is at a distance \(3 \sqrt{2}\) units from \((1,2)\). (a) \((1 / \sqrt{3})\) (b) 1 (c) \(\sqrt{3}\) (d) \([(\sqrt{3}-1) / 2]\)

If the \(\mathrm{y}\) -intercept of the perpendicular bisector of the segment obtained by joining \(\mathrm{P}(1,4)\) and \(\mathrm{Q}(\mathrm{k}, 3)\) is \(-4\) then \(\mathrm{k}=\ldots \ldots\) (a) 1 (b) 2 (c) \(-2\) (d) \(-4\)

For any values of \(\mathrm{p}\) and \(\mathrm{q}\), the line \((p+2 q) x+(p-3 q) y-p+q\) passes through which fixed point? (a) \([(3 / 2),(5 / 2)]\) (b) \([(2 / 5),(2 / 5)]\) (c) \([(3 / 5) /(3 / 5)]\) (d) \([(2 / 5),(3 / 5)]\)
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