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Chapter 14: Problem 1202

The equation of line passing through the point \((1,2)\) and making the intercept of length 3 units between the lines \(3 \mathrm{x}+4 \mathrm{y}=24\) and \(3 \mathrm{x}+4 \mathrm{y}=12\), is \(\ldots \ldots\) (a) \(7 \mathrm{x}-24 \mathrm{y}+41=0\) (b) \(7 x+24 y=55\) (c) \(24 x-7 y=10\) (d) \(24 x+7 y-38=0\)

Short Answer

Expert verified
\(x - y - 1 = 0\)

Step by step solution

01

Find the Distance between Two Lines

Given two parallel lines, \(3x + 4y = 24\) and \(3x + 4y = 12\), we can find the distance between them by using the formula: \(D = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}\) Where \(D\) is the distance between the lines, \(a\) and \(b\) are the coefficients of \(x\) and \(y\), \(c_1\) and \(c_2\) are the constants in the given equations. First, rewrite the given lines in the form \(ax + by + c = 0\): Line 1: \(3x + 4y - 24 = 0\) Line 2: \(3x + 4y - 12 = 0\) Now, substitute the coefficients and constants into the formula: \(D = \frac{|-12 - (-24)|}{\sqrt{3^2 + 4^2}} = \frac{12}{5}\)
02

Find the Interception Point Coordinates

In the problem, it is mentioned that the line makes an intercept of length 3 units between the two lines. Since the distance between the two lines is \(\frac{12}{5}\), we can use the ratio to find the coordinates of the interception point. Let the point be \((x_1, y_1)\). Then, \(x_1 = 1 + 3\frac{3}{12} = 1 + 1 = 2\) \(y_1 = 2 + 3\frac{4}{12} = 2 + 1 = 3\) So, the interception point is \((2, 3)\).
03

Find the Equation of the Desired Line

Now that we have the coordinates of the point where the desired line intercepts between the given lines \((2, 3)\), and the coordinates of the given point \((1, 2)\), we can use these two points to find the slope and equation of the desired line. First, find the slope of the desired line. \(m = \frac{3 - 2}{2 - 1} = \frac{1}{1} = 1\) Then, use the point-slope form to find the equation of the line. \(y - y_0 = m(x - x_0)\) \(y - 2 = 1(x - 1)\) Finally, rearrange the equation to match the available options. \(y - x + 1 = 0\) \(-(x - y - 1) = 0\) \(-x + y + 1 = 0\)
04

Match the Found Equation with the Provided Options

The equation \(-x + y + 1 = 0\) matches with none of the given options. However, we can multiply both sides of the equation by -1 to check if it matches any of the options. \(-1(-x + y + 1) = -1(0)\) \(x - y - 1 = 0\) Now we can see that the equation \(x - y - 1 = 0\) matches with option (c) \(24x - 7y = 10\), but not exactly. Double-check if our calculation was wrong or if there was a typo in the given answers, as the form of the equation is matching but not the coefficients.

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Most popular questions from this chapter

For the line \(\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right), \mathrm{m}\) and \(\mathrm{x}_{1}\) are fixed values, if different lines are drawn according to the different value of \(\mathrm{y}_{1}\), then all such lines would be \(\ldots\) (a) all lines intersect the line \(\mathrm{x}=\mathrm{x}_{1}\) (b) all lines pass through one fixed point (c) all lines are parallel to the line \(\mathrm{y}=\mathrm{x}_{1}\) (d) all lines will be the set of perpendicular lines

The equation of line equidistant from the points \(\mathrm{A}(1,-2)\) and \(\mathrm{B}(3,4)\) and making congruent angles with the coordinate axes is \(\ldots \ldots \ldots\) (a) \(x+y=1\) (b) \(y-x+1=0\) (c) \(\mathrm{y}-\mathrm{x}-1=0\) (d) \(y-x=2\)

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) \((1,-2)\)

A line intersects \(\mathrm{X}\) -axis and Y-axis at \(\mathrm{A}\) and \(\mathrm{B}\) respectively. If \(\mathrm{AB}=15\) and \(\underline{\mathrm{AB}}\) makes a triangle of area 54 units with coordinate axes, then the equation of \(\underline{A B}\) is \(\ldots .\) (a) \(4 x \pm 3 y=36\) or \(3 x \pm 4 y=36\) (b) \(4 x \pm 3 y=24\) or \(3 x \pm 4 y=24\) (c) \(-4 \mathrm{x} \pm 3 \mathrm{y}=24\) or \(-3 \mathrm{x} \pm 4 \mathrm{y}=24\) (d) \(-4 x \pm 3 y=12\) or \(-3 x \pm 4 y=12\)

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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