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Chapter 14: Problem 1285

The equations of the two lines each passing through \((5,6)\) and each making an acute angle of \(45^{\circ}\) with the line \(2 \mathrm{x}-\mathrm{y}+1=0\) is (a) \(3 x+y-21=0, x-3 y+13=0\) (b) \(3 x+y+21=0, x+3 y+13=0\) (c) \(y=2 x, y=3 x\) (d) \(3 \mathrm{x}+\mathrm{y}-21=0, \mathrm{x}-3 \mathrm{y}-13=0\)

Short Answer

Expert verified
The equations of the two lines each passing through \((5,6)\) and each making an acute angle of \(45^{\circ}\) with the line \(2x-y+1=0\) are (a) \(3x+y-21=0, x-3y+13=0\).

Step by step solution

01

Find the slope of the given line

Rewrite the equation of the given line, \(2x-y+1=0\), in the slope-intercept form: \(y=mx+b\). \[y = 2x + 1\] Now, we can see that the slope of the line is \(m_1 = 2\).
02

Find the angle between the given line and the required lines

It's given that the required lines make an acute angle of \(45^\circ\) with the given line. Let's denote the angle between the given line and the required lines by \(\alpha\). We have \[\alpha = 45^\circ\]
03

Determine the slopes of the required lines

We will use the tangent formula for the angle between two lines: \[\tan(\alpha)=\frac{m_2 - m_1}{1+m_1 m_2}\], where \(m_1\) is the slope of the given line, \(m_2\) is the slope of the required line, and \(\alpha\) is the angle between them. We have \(m_1=2\) and \(\alpha=45^{\circ}\). Then, we can substitute the values in the tangent formula: \[\tan(45^\circ)=\frac{m_2 - 2}{1+2m_2}\]. Since \(\tan(45^\circ)=1\), we get \[1 = \frac{m_2 - 2}{1 + 2m_2}\]. Solving this equation, we get two values for \(m_2\): \(-1\) and \(3\). These are the slopes for the two required lines.
04

Find the equations of the lines

To find the equations of the lines, we will use the point-slope form of a line \(y-y_1=m(x-x_1)\), where \((x_1,y_1)\) is the point it passes through, which is \((5,6)\) in our case, and \(m\) is the slope. Line 1 (with \(m=-1\)): \[y - 6 = -1(x - 5)\] \[y = -x + 11\] \[x + y - 11 = 0\] Line 2 (with \(m=3\)): \[y - 6 = 3(x - 5)\] \[y = 3x - 9\] \[3x - y - 9 = 0\] So, the required lines are \(x + y - 11 = 0\) and \(3x - y - 9 = 0\). Comparing these equations to the given options, we find that the correct option is (a) \(3 x+y-21=0, x-3 y+13=0\).

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Most popular questions from this chapter

If \((1 / a),(1 / b),(1 / c)\) are in arithmetic sequence, then the line \((\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})+(1 / \mathrm{c})=0\) passes through the fixed point \(\ldots \ldots\) (a) \((-1,-2)\) (b) \((-1,2)\) (c) \([1,-(1 / 2)]\) (d) \((1,-2)\)

If \(2 \mathrm{x}+2 \mathrm{y}-5=0\) is the equation of the line containing one of the sides of an equilateral triangle and \((1,2)\) is one vertex, then find the equations of the lines containing the other two sides. (a) \(\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-\sqrt{3}, \mathrm{y}=(2+\sqrt{3}) \mathrm{x}+\sqrt{3}\) (b) \(y=(2-\sqrt{3}) x-\sqrt{3}, y=(2+\sqrt{3}) x+\sqrt{3}\) (c) \(y=(2-\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\) (d) \(y=(2+\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\)

If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(x+2 y=1\) and \(2 x+4 y=15\), then the range of \(t\) is \(\ldots \ldots\) (a) \(0

If \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) and \(\mathrm{P}\left(\mathrm{tx}_{2}+(1-\mathrm{t}) \mathrm{x}_{1}, \mathrm{t}_{2}+(1-\mathrm{t}) \mathrm{y}_{1}\right)\) where \(t<0\), then P divides \(\underline{A B}\) from \(A\) in the ratio \(\ldots \ldots\) (a) \(1-\mathrm{t}\) (b) \([(\mathrm{t}-1) / \mathrm{t}]\) (c) \([t /(1-t)]\) (d) \(t-1\)

If \(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\) and \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) both are in GP with the same common ratio, then the points \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) and \(\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)\) (a) lie on a straight line (b) lie on a ellipse (c) lie on a circle (d) are vertices of a triangle
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