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Chapter 14: Problem 1198

The equation of line passing through the point \((-5,4)\) and making the intercept of length \((2 / \sqrt{5})\) between the lines \(x+2 y-1=0\) and \(x+2 y+1=0\) is \(\ldots \ldots\) (a) \(2 \mathrm{x}-\mathrm{y}+4=0\) (b) \(2 x-y-14=0\) (c) \(2 \mathrm{x}-\mathrm{y}+14=0\) (d) None of these

Short Answer

Expert verified
The equation of the line passing through the point \((-5,4)\) and making the intercept of length \((2 / \sqrt{5})\) between the lines \(x+2 y-1=0\) and \(x+2 y+1=0\) is (d) None of these.

Step by step solution

01

Find the slope of the given lines

: We are given two lines: Line 1: \(x + 2y - 1 = 0\) Line 2: \(x + 2y + 1 = 0\) The slope-intercept form of a line is: \(y = mx + b\), where m is the slope and b is the y-intercept. Rewriting the given lines in slope-intercept form, we get: Line 1: \(y = (-1/2)x + (1/2)\) Line 2: \(y = (-1/2)x - (1/2)\) The slope of both given lines is \(-1/2\). Since both lines have the same slope, they are parallel to each other.
02

Determine the slope of the required line

: The required line should have the same slope as the given lines since intercept of length \((2/\sqrt{5})\) is between the parallel lines. So, Required line slope: \(m = -1/2\)
03

Write the equation of the required line in point-slope form

: We have the slope of the required line, and we are given a point it passes through, which is \((-5, 4)\). The point-slope form of a line is: \(y - y_1 = m(x - x_1)\) Plugging in the given values and the calculated slope: \(y - 4 = -\frac{1}{2}(x - (-5))\) Now, let's simplify the equation: \(y - 4 = -\frac{1}{2}(x + 5)\)
04

Choose the correct answer from the given options

: Now, convert the simplified equation into the standard form and compare it with the given options. \(y - 4 = -\frac{1}{2}x -\frac{5}{2}\) \(2y - 8 = -x -5\) \(2x + 2y - 3= 0\) Comparing this with the given options, we see that the correct answer is not among the given options, so the answer is: (d) None of these

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Most popular questions from this chapter

For a \(+b+c=0\), the line \(3 \mathrm{ax}+4 \mathrm{by}+\mathrm{c}=0\) passes through the fixed point \(\ldots \ldots .\) (a) \([(1 / 3),-(1 / 4)]\) (b) \([-(1 / 3),(1 / 4)]\) (c) \([(1 / 3),(1 / 4)]\) (d) \([-(1 / 3),-(1 / 4)]\)

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) \(3 \mathrm{x}+\mathrm{y}-21=0, \mathrm{x}-3 \mathrm{y}-13=0\)

The locus of mid points of the segment intercepted between the axes by the line xseca \(+\) ytana \(=p\) is \(\ldots \ldots\) (a) \(\left[\mathrm{p}^{2} /\left(4 \mathrm{x}^{2}\right)\right]=1+\left[\mathrm{p}^{2} /\left(4 \mathrm{y}^{2}\right)\right]\) (b) \(\left(\mathrm{x}^{2} / \mathrm{p}^{2}\right)+\left(\mathrm{y}^{2} / \mathrm{p}^{2}\right)=4\) (c) \(\left(\mathrm{p}^{2} / \mathrm{x}^{2}\right)=1+\left(\mathrm{p}^{2} / \mathrm{y}^{2}\right)\) (d) \(\left[\mathrm{p}^{2} /\left(4 \mathrm{x}^{2}\right)\right]+\left[\mathrm{p}^{2} /\left(4 \mathrm{y}^{2}\right)\right]=1\)

If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(\mathrm{x}+2 \mathrm{y}=1\) and \(2 \mathrm{x}+4 \mathrm{y}=15\), then the range of \(\mathrm{t}\) is \(\ldots \ldots\) (a) \(0<\mathrm{t}<[5 /(6 \sqrt{2})]\) (b) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<0\) (c) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<[(5 \sqrt{2}) / 6]\) (d) None of these

The locus of the variable point whose distance from \((-2,0)\) is \((2 / 3)\) times its distance from the line \(\mathrm{x}=-(9 / 2)\) is (a) ellipse (b) parabola (c) circle (d) hyperbola
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