If three non-zero real numbers a, b, c are in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $a, b, c$ are in harmonic progression,we have $\frac{1}{b} = \frac{1}{2} (\frac{1}{a} + \frac{1}{c})$,which implies $\frac{2}{b} = \fra

Vedclass · Accepted Answer

$(1, -2)$

Vedclass · Answer

(A) Given that $a, b, c$ are in harmonic progression,we have $\frac{1}{b} = \frac{1}{2} (\frac{1}{a} + \frac{1}{c})$,which implies $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.Rearranging this,we get $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$.The given equation of the line is $\frac{x}{a} + \frac{y}{b} - \frac{2}{c} = 0$.Comparing this with the condition $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$,we can rewrite the line equation as $\frac{1}{a}(x) + \frac{1}{b}(y) + \frac{1}{c}(-2) = 0$.For this to be true for all $a, b, c$ satisfying the harmonic progression condition,the coefficients must satisfy the same linear relation.By inspection,if we set $x = 1$ and $y = -2$,the equation becomes $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$,which is exactly the condition for harmonic progression.Thus,the lines are concurrent at the point $(1, -2)$.

If three non-zero real numbers $a, b, c$ are in harmonic progression,then the straight lines $\frac{x}{a} + \frac{y}{b} - \frac{2}{c} = 0$ are concurrent at the point

Similar Questions

The equation of the straight line perpendicular to the straight line $3x + 2y = 0$ and passing through the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$ is

The equation of the line passing through the intersection of the lines $x - y = 4$ and $3x + y = 7$ and parallel to the line $x + 2y = 1$ is:

The value of $k$ such that the lines $2x - 3y + k = 0$,$3x - 4y - 13 = 0$,and $8x - 11y - 33 = 0$ are concurrent,is

The lines $15x - 18y + 1 = 0$,$12x + 10y - 3 = 0$ and $6x + 66y - 11 = 0$ are

The lines $x-2y+1=0$,$2x-3y-1=0$,and $3x-y+k=0$ are concurrent. The angle between the lines $3x-y+k=0$ and $mx-3y+6=0$ is $45^{\circ}$. If $m$ is an integer,then $m-k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module