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Chapter 14: Problem 1321

If non-zero numbers \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{HP}\), then the straight line \((\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})+(1 / \mathrm{c})=0\) always passes through a fixed point that point is (a) \([1,-(1 / 2)]\) (b) \((1,-2)\) (c) \((-1,-2)\) (d) \((-1,2)\)

Short Answer

Expert verified
The fixed point through which the straight line (x / a) + (y / b) + (1 / c) = 0 always passes is (b) (1, -2).

Step by step solution

01

Denote the reciprocals and the arithmetic progression of 1/a, 1/b, and 1/c

: Let p = 1/a, q = 1/b, and r = 1/c. Since a, b, and c are in HP, the reciprocals p, q, and r are in an arithmetic progression with a common difference. That implies: q - p = r - q
02

Express r in terms of p and q and substitute in the given equation

: From step 1, we know that r - q = q - p, which can be written as: r = 2q - p Now replace p, q and r with their respective reciprocals in the given equation of the line: (x / a) + (y / b) + (1 / c) = 0 => (x * p) + (y * q) + (1 * r) = 0 Substitute r = 2q - p in the equation: (x * p) + (y * q) + (1 * (2q - p)) = 0
03

Rewrite equation in terms of x and y

: Now rearrange and simplify the equation from step 2: xp + yq + 2q - p = 0 => p(x - 1) + q(y + 2) = 0 Since a, b, and c are non-zero, their reciprocals p, q, and r, are also non-zero. So we have a linear equation in x and y with non-zero coefficients.
04

Find the fixed point

: Now we have the equation p(x - 1) + q(y + 2) = 0. To find the fixed point, let's set values of p and q that satisfy the equation: p = 1 q = -1/2 Now, substituting these values into the equation: (1)(x - 1) + (-1/2)(y + 2) = 0 x - 1 - (y + 2) / 2 = 0 2x - 2 - y - 2 = 0 Simplifying the equation and solving for x and y: 2x - y - 4 = 0 => y = 2x - 4 Now, we can see that the coordinates of the fixed point are (x, y) = (1, -2). Therefore, the correct option is: (b) (1, -2)

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Most popular questions from this chapter

The angle between the lines \(\mathrm{x}=3\) and \(\sqrt{3} \mathrm{x}-\mathrm{y}+5=0\) is \(\ldots \ldots\) (a) \((\pi / 6)\) (b) \((\pi / 3)\) (c) \((\pi / 4)\) (d) \((\pi / 2)\)

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) \((1,-2)\)

In triangle \(\mathrm{ABC}\), equation of right bisectors of the sides \(\underline{A B}\) and \(\underline{A C}\) are \(x+y=0\) and \(y-x=0\) respectively. If \(\mathrm{A}=(5,7)\) then equation of side \(\mathrm{BC}\) is (a) \(7 \mathrm{y}=5 \mathrm{x}\) (b) \(5 x=y\) (c) \(5 \mathrm{y}=7 \mathrm{x}\) (d) \(5 \mathrm{y}=\mathrm{x}\)

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