p, x_1, x_2, ldots, x_n and q, y_1, y_2, ldot | vedclass

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

Question

(B) Given that $p, x_1, x_2, \ldots, x_n$ is an $A$.$P$. with common difference $a$,we have $x_k = p + ka$. Thus,$x_1 = p+a$ and $x_n = p+na$. The ari

Vedclass · Accepted Answer

$b(x-p)=a(y-q)$

Vedclass · Answer

(B) Given that $p, x_1, x_2, \ldots, x_n$ is an $A$.$P$. with common difference $a$,we have $x_k = p + ka$. Thus,$x_1 = p+a$ and $x_n = p+na$. The arithmetic mean $\alpha$ of $x_1, \ldots, x_n$ is given by $\alpha = \frac{x_1 + x_n}{2} = \frac{p+a + p+na}{2} = \frac{2p + a(n+1)}{2} \quad (i)$.Similarly,for the $A$.$P$. $q, y_1, \ldots, y_n$ with common difference $b$,we have $y_1 = q+b$ and $y_n = q+nb$. The arithmetic mean $\beta$ is $\beta = \frac{y_1 + y_n}{2} = \frac{q+b + q+nb}{2} = \frac{2q + b(n+1)}{2} \quad (ii)$.From $(i)$,$\frac{2\alpha - 2p}{a} = n+1$. From $(ii)$,$\frac{2\beta - 2q}{b} = n+1$.Equating the two expressions for $n+1$,we get $\frac{2(\alpha - p)}{a} = \frac{2(\beta - q)}{b}$,which simplifies to $b(\alpha - p) = a(\beta - q)$.Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $b(x-p) = a(y-q)$.

$p, x_1, x_2, \ldots, x_n$ and $q, y_1, y_2, \ldots, y_n$ are two arithmetic progressions with common differences $a$ and $b$ respectively. If $\alpha$ and $\beta$ are the arithmetic means of $x_1, x_2, \ldots, x_n$ and $y_1, y_2, \ldots, y_n$ respectively,then the locus of $P(\alpha, \beta)$ is

Similar Questions

If a point $(x, y) = (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

The locus of the point $P(x, y)$ such that the area of the $\triangle PAB$ is $7$,where $A(4, 5)$ and $B(-2, 3)$ are given points,is

The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. units. The locus of the point is:

If the point $(a, a^2)$ lies inside the angle formed by the lines $y = \frac{x}{2}$ $(x > 0)$ and $y = 3x$ $(x > 0)$,then $a$ belongs to:

$A$ line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module