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Chapter 14: Problem 1359

The lines \(a x+2 y+1=0, b x+3 y+1=0\) and \(c x+4 y+1=0\) are concurrent then (a) \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in A.P. (b) \(a, b, c\) are in G.P. (c) \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in H.P. (d) None of these

Short Answer

Expert verified
(a) a, b, c are in Arithmetic Progression

Step by step solution

01

Calculate the intersection point of each line pair

First, we need to find the common point where these lines meet. To do this, we will calculate the intersection point for each pair of lines and check if they are the same. Line 1: \(a x+2 y+1 = 0\) Line 2: \(b x+3 y+1 = 0\) Line 3: \(c x+4 y+1 = 0\) Find intersection point of Line 1 and Line 2: \(ax + 2y + 1 = 0\) \(bx + 3y + 1 = 0\) At the intersection point, the value of x will be the same in both equations. Therefore, we can write the value of x in terms of y from the first equation and substitute it into the second equation. \(x = \frac{-2y - 1}{a}\) Substitute the value of x in the second equation: \(b\left(\frac{-2y-1}{a}\right) + 3y + 1 = 0\)
02

Solve for y

Now, solve for y: \(- 2by - b + 3ay + a^2 = 0\) Rearrange the terms: \(2by - 3ay = a^2 - b\) Divide both sides by y: \(2b - 3a = \frac{a^2 - b}{y}\) Now we will find the intersection point of Line 2 and Line 3 in the same way.
03

Calculate intersection point of Line 2 and Line 3

Line 2: \(b x+3 y+1 = 0\) Line 3: \(c x+4 y+1 = 0\) Find intersection point of Line 2 and Line 3: \(bx + 3y + 1 = 0\) \(cx + 4y + 1 = 0\) At the intersection point, the value of x will be the same in both equations. Therefore, we can write the value of x in terms of y from the second equation and substitute it into the third equation. \(x = \frac{-3y - 1}{b}\) Substitute the value of x in the third equation: \(c\left(\frac{-3y-1}{b}\right) + 4y + 1 = 0\)
04

Solve for y

Now, solve for y: \(- 3cy - c + 4by + b^2 = 0\) Rearrange the terms: \(3cy - 4by = b^2 - c\) Divide both sides by y: \(3c - 4b = \frac{b^2 - c}{y}\)
05

Compare both expressions with y

For the lines to be concurrent, both expressions with y should be equal: \(2b - 3a = 3c - 4b\)
06

Check if it satisfies the given options

Rearrange the equation: \(6b = 3a + 3c\) Divide both sides by 3: \(2b = a + c\) Now, the above equation states a relationship between a, b, and c. This actually represents an Arithmetic Progression where the common difference between each term is 'b'. Therefore, the given answer is: (a) a, b, c are in Arithmetic Progression

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Most popular questions from this chapter

For any values of \(\mathrm{p}\) and \(\mathrm{q}\), the line \((p+2 q) x+(p-3 q) y-p+q\) passes through which fixed point? (a) \([(3 / 2),(5 / 2)]\) (b) \([(2 / 5),(2 / 5)]\) (c) \([(3 / 5) /(3 / 5)]\) (d) \([(2 / 5),(3 / 5)]\)

The equation of line containing the angle bisector of the lines \(3 x-4 y-2=0\) and \(5 x-12 y+2=0\) is ....... (a) \(7 x+4 y-18=0\) (b) \(4 x-7 y-1=0\) (c) \(4 \mathrm{x}-7 \mathrm{y}+1=0\) (d) None of these

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) \(24 x+7 y-38=0\)

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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) \((1,-2)\)
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