(A) Let the point be $(x, y)$ at time $t$ seconds.Given the initial position $(1, 2)$ and velocities along the axes,the coordinates at time $t$ are:$x

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(A) Let the point be $(x, y)$ at time $t$ seconds.Given the initial position $(1, 2)$ and velocities along the axes,the coordinates at time $t$ are:$x = 1 + 3t$$y = 2 + 2t$From the first equation,$t = \frac{x - 1}{3}$.Substituting this into the second equation:$y = 2 + 2 \left( \frac{x - 1}{3} \right)$$3y = 6 + 2x - 2$$3y = 2x + 4$$2x - 3y + 4 = 0$Thus,the locus is $2x - 3y + 4 = 0$.

Class 11 Mathematics Straight Line Locus of Point

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$A$ point starts moving from $(1, 2)$ and its projections on $x$ and $y$-axes are moving with velocities of $3 \ m/s$ and $2 \ m/s$ respectively. Its locus is

A
$2x - 3y + 4 = 0$
B
$3x - 2y + 1 = 0$
C
$3y - 2x + 4 = 0$
D
$2y - 3x + 1 = 0$

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Straight Line

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Locus of Point

If the sum of the distances of a point from two perpendicular lines in a plane is $1$ unit, then its locus is

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If $A = (a, 0)$ and $B = (-a, 0)$,then the locus of a point $P = (x, y)$ such that $PA^2 - PB^2 = a^2$ is.

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Let $A (2, 3)$ and $B (-4, 5)$ be two fixed points. $A$ point $P$ moves in such a way that the area of $\Delta PAB = 12 \, \text{sq. units}$. Find its locus.

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The locus of the mid-point of the segment intercepted between the axes by the variable line $x \cos \alpha + y \sin \alpha = p$,where $p$ is a constant,is

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$A$ line passing through the point $(1, 2)$ meets the axes at $P$ and $Q$ such that it forms a triangle $OPQ$,where $O$ is the origin. If the area of triangle $OPQ$ is minimum,then the slope of the line $PQ$ is:

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$A$ point starts moving from $(1, 2)$ and its projections on $x$ and $y$-axes are moving with velocities of $3 \ m/s$ and $2 \ m/s$ respectively. Its locus is

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If the sum of the distances of a point from two perpendicular lines in a plane is $1$ unit, then its locus is

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