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Chapter 14: Problem 1303

The nearest point on the line \(3 \mathrm{x}+4 \mathrm{y}=1\) from origin is (a) \([(7 / 25),(4 / 25)]\) (b) \([(7 / 25),(2 / 25)]\) (c) \([(3 / 25),(4 / 25)]\) (d) \([(1 / 25),(3 / 25)]\)

Short Answer

Expert verified
The nearest point on the line \(3x + 4y = 1\) from the origin is (c) \(\left(\dfrac{3}{25}, \dfrac{4}{25}\right)\).

Step by step solution

01

Find the slope of the given line

To find the slope, we rewrite the given equation (\(3x + 4y = 1\)) into the slope-intercept form. The slope-intercept form is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. 1. Solve for \(y\): \(4y = -3x + 1\) \(y = -\dfrac{3}{4}x + \dfrac{1}{4}\) Now that we have the equation in slope-intercept form, we see that the slope, \(m\) is \(-\dfrac{3}{4}\).
02

Find the slope of the perpendicular line and its equation

Since the perpendicular line has a slope that is a negative reciprocal of the given line's slope, we find the negative reciprocal of \(-\dfrac{3}{4}\): Perpendicular line's slope = \(\dfrac{4}{3}\) Since the line passes through the origin (0,0), the equation of the perpendicular line is \(y = \dfrac{4}{3}x\).
03

Solve the system of equations

Now, we will solve for the point of intersection between the two lines: \( \begin{cases} 3x + 4y = 1\\ y = \dfrac{4}{3}x \end{cases} \) 1. Substitute \(\frac{4}{3}x\) for \(y\) in the first equation: \(3x + 4\left(\dfrac{4}{3}x\right) = 1\) 2. Solve for x: \(3x + \dfrac{16}{3}x = 1\) \(\Rightarrow (3+\dfrac{16}{3})x = 1\) \(\Rightarrow (\dfrac{25}{3})x = 1\) \(\Rightarrow x = \dfrac{3}{25}\) 3. Substitute \(x = \dfrac{3}{25}\) into the equation of the perpendicular line to find y: \(y = \dfrac{4}{3} \left(\dfrac{3}{25}\right)\) \(\Rightarrow y = \dfrac{4}{25}\) The point of intersection is \(\left(\dfrac{3}{25}, \dfrac{4}{25}\right)\).
04

Verify the result with given options

Now we check which of the given options matches our found point: (a) \(\left(\dfrac{7}{25}, \dfrac{4}{25}\right)\) (b) \(\left(\dfrac{7}{25}, \dfrac{2}{25}\right)\) (c) \(\left(\dfrac{3}{25}, \dfrac{4}{25}\right)\) (d) \(\left(\dfrac{1}{25}, \dfrac{3}{25}\right)\) The correct answer is (c) \(\left(\dfrac{3}{25}, \dfrac{4}{25}\right)\) as it matches the point we found.

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Most popular questions from this chapter

If \(2 \mathrm{x}+3 \mathrm{y}=8\) is perpendicular to the line \((\mathrm{x}+\mathrm{y}+1)+\lambda(2 \mathrm{x}-\mathrm{y}-1)=0\), then \(\lambda=?\) (a) \(-5\) (b) \((3 / 2)\) (c) 5 (d) 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) \((\mathrm{x} / 2)-(\mathrm{y} / 3)=1,\\{\mathrm{x} /(-2)\\}+(\mathrm{y} / 1)=1\)

Consider a square OPQR having the length of side a, where \(\mathrm{O}(0,0)\). The sides \(\underline{\mathrm{OP}}\) and \(\underline{\mathrm{OR}}\) are along the positive \(\mathrm{X}\) -axis and Y-axis respectively. If \(\mathrm{A}\) and \(\mathrm{B}\) are the mid points of \(\underline{\mathrm{PQ}}\) and \(Q \underline{R}\) respectively, then the angle between \(\underline{\mathrm{OA}}\) and \(\underline{\mathrm{OB}}\) would be \(\ldots \ldots \ldots\) (a) \(\cos ^{-1}(3 / 5)\) (b) \(\tan ^{-1}(4 / 3)\) (c) \(\cos ^{-1}(3 / 4)\) (d) \(\sin ^{-1}(3 / 5)\)

The sum of squares of intercepts on the axes cut off by the tangents to the curve \(\mathrm{x}^{(2 / 3)}+\mathrm{y}(2 / 3)=\mathrm{a}^{(2 / 3)}(\mathrm{a}>0)\) at \([(\mathrm{a} / 8),(\mathrm{a} / 8)]\) is \(2 .\) Thus a has the value. (a) 1 (b) 2 (c) 4 (d) 8

The circumcentre of the triangle formed by the lines \(\mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0\) and \(\mathrm{x}-7=0\) is \(\ldots \ldots\) (a) \((7,0)\) (b) \((3.5,0)\) (c) \((0,7)\) (d) \((3.5,3.5)\)
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