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Chapter 14: Problem 1206

A line passing through \(0(0,0)\) intersect the parallel lines \(4 x+2 y=0\) and \(2 x+y+6=0\) at \(P\) and \(Q\) respectively, then in what ratio does 0 divide \(\underline{\text { PQ from }} \mathrm{P}\) ? (a) \(1: 2\) (b) \(3: 4\) (c) \(2: 1\) (d) \(4: 3\)

Short Answer

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Please recheck the given exercise and ensure all the given information is accurate.

Step by step solution

01

Finding Point P

To find the point P, we need to solve the equations of the given line and the equation of the first parallel line. Let the given line be L1: y = m1x Given parallel line: \[4x + 2y = 0\] Substituting L1 into the parallel line's equation, we get: \[4x + 2(m1x) = 0\] \[(4 + 2m1)x = 0\] Since this is not the trivial case, we can say that x = 0 Now, substituting x = 0 into the equation for L1, we get y = 0. So point P is (0,0).
02

Finding Point Q

To find the point Q, we need to solve the equations of the given line and the equation of the second parallel line. Given line L1: y = m1x Given parallel line: \[2x + y + 6 = 0\] Substituting L1 into the second parallel line's equation, we get: \[2x + m1x + 6 = 0\] \[(2 + m1)x = -6\] Since this is not the trivial case, we can say that x ≠ 0 and m1 ≠ -2. Solving this, we get x = -6 / (2 + m1) and y = m1 * (-6 / (2 + m1)) So point Q is (-6 / (2 + m1), m1 * (-6 / (2 + m1)))
03

Applying the Section Formula

Now, applying the section formula to find the ratio in which O divides PQ. Let's define the ratio as k:1 x-coordinate: \[0 = \frac{k(-6 / (2 + m1)) + 1 * 0}{k + 1}\] Since -6 / (2 + m1) ≠ 0, we can say that k = 0. y-coordinate: \[0 = \frac{k(m1 * (-6 / (2 + m1))) + 1 * 0}{k + 1}\] Since k = 0, we only need to solve y-coordinate equation: \[0 = m1 * (-6 / (2 + m1))\] Since m1 ≠ 0, we can say that -6 / (2 + m1) = 0 and m1 = -2 However, our previous assumption about m1 ≠ -2 was wrong. The problem statement seems to have some inconsistency. Please recheck the given exercise and ensure all the given information is accurate.

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Most popular questions from this chapter

The angle between the lines \(\\{(\mathrm{x}, 0) /(\mathrm{x} \in \mathrm{R})\\}\) and \(\\{(0, y) /(y \in R)\\}\) is \(\ldots \ldots\) (a) \((\pi / 2)\) (b) \(-(\pi / 2)\) (c) 0 (d) \(\pi\)

The line \(3 \mathrm{x}-4 \mathrm{y}+7=0\) is rotated through an angle \((\pi / 4)\) in the clockwise direction about the point \((-1,1)\). The equation of the line in its new position is (a) \(7 \mathrm{y}+\mathrm{x}-6=0\) (b) \(7 \mathrm{y}-\mathrm{x}-6=0\) (c) \(7 \mathrm{y}+\mathrm{x}+6=0\) (d) \(7 \mathrm{y}-\mathrm{x}+6=0\)

\(\sqrt{(3) x+y}=2\) is the equation of line containing one of the sides of an equilateral triangle and if \((0,-1)\) is one of the vertices, then the length of the side of the triangle is \(\ldots \ldots\) (a) \(\sqrt{3}\) (b) \(2 \sqrt{3}\) (c) \((\sqrt{3} / 2)\) (d) \((2 / \sqrt{3})\)

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) \((1,-2)\)

The equation of line passing through the point of intersection of the lines \(3 \mathrm{x}-2 \mathrm{y}=0\) and \(5 \mathrm{x}+\mathrm{y}-2=0\) and making the angle of measure \(\tan ^{-1}(-5)\) with the positive direction of \(\mathrm{x}\) -axis is \(\ldots \ldots\) (a) \(3 x-2 y=0\) (b) \(5 x+y-2=0\) (c) \(5 \mathrm{x}+\mathrm{y}=0\) (d) \(3 \mathrm{x}+2 \mathrm{y}+1=0\)
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