A line moves such that the portion of it inte | vedclass

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

Question

(C) Let $A = (p, 0)$ and $B = (0, q)$ be the points where the line intersects the $x$-axis and $y$-axis respectively.Let $P(h, k)$ be the midpoint of

Vedclass · Accepted Answer

$x^2+y^2=\frac{a^2}{4}$

Vedclass · Answer

(C) Let $A = (p, 0)$ and $B = (0, q)$ be the points where the line intersects the $x$-axis and $y$-axis respectively.Let $P(h, k)$ be the midpoint of the line segment $\overline{AB}$.Given that the length of the segment $\overline{AB} = a$.Since $P(h, k)$ is the midpoint of $\overline{AB}$,we have:$h = \frac{p+0}{2} \implies p = 2h$$k = \frac{0+q}{2} \implies q = 2k$The length of the segment $\overline{AB}$ is given by the distance formula:$\sqrt{(p-0)^2 + (0-q)^2} = a$$\sqrt{p^2 + q^2} = a$Substituting $p = 2h$ and $q = 2k$ into the equation:$\sqrt{(2h)^2 + (2k)^2} = a$$\sqrt{4h^2 + 4k^2} = a$Squaring both sides:$4h^2 + 4k^2 = a^2$$h^2 + k^2 = \frac{a^2}{4}$Replacing $(h, k)$ with $(x, y)$,the locus of the midpoint is:$x^2 + y^2 = \frac{a^2}{4}$Thus,the correct option is $C$.

$A$ line moves such that the portion of it intercepted between the coordinate axes is of constant length $a$. Then,the locus of the midpoint of that line segment is

Similar Questions

$A$ variable line passes through the fixed point $(\alpha, \beta)$. The locus of the foot of the perpendicular from the origin on the line is

$p, x_1, x_2, \ldots, x_n$ 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

If a point $P$ on the line $3x + 5y = 15$ is equidistant from the coordinate axes,then $P$ lies

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

$A$ variable line passing through a fixed point $(\alpha, \beta)$ intersects the coordinate axes at $A$ and $B$. If $O$ is the origin,then the locus of the centroid of the $\triangle OAB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module