What is an elegant way of calculating this result (programmatically) in mathematica?

Let $X$ be a random variable with PDF

$$f_X(x) =
begincases
x/4, & textif $1<xleq3$ \
0, & textotherwise
endcases$$

Let $A$ be the event $Xgeq2$

Find $f_A(x)$

Here is the answer done manually for convenience:

$$f_A(x) =
begincases
fracf_X(x)P(A), & textif $2leq xleq3$ \
0, & textotherwise
endcases$$

$$P(A)=P(Xgeq2)=int_2^3f_X(x) dx=5/8$$

$$implies f_A(x) =
begincases
frac2x5, & textif $2leq xleq3$ \
0, & textotherwise
endcases$$

I am looking for a programmatic solution that utilizes [Conditioned] or something else as high-level.

TransformedDistribution and ProbabilityDistribution look like promising avenues, but I am unable to state my problem in terms of them.

I know I could write out the problem explicitly in low level code. But I am looking for an implicit, declarative implementation. Surely mathematica with all its high level abstraction power can do this?

here is a failed attempt

A = X >= 2;
conditionalDist = TransformedDistribution[X [Conditioned] A, X [Distributed]ProbabilityDistribution[x/4, x, 1, 3] ] ;
PDF[conditionalDist, y]

You can use combination of ProbabilityDistribution and TruncatedDistribution as follows:

f[x_] := x/4
dist = ProbabilityDistribution[f[x], x, 1, 3];
td = TruncatedDistribution[2, Infinity, dist];

PDF[td, x]

$begincases
frac2 x5 & 2<x<3 \
0 & textTrue
endcases$

Probability[Conditioned[x <= t, x > 2], Distributed[x, dist]] // TeXForm

$begincases
1 & tgeq 3 \
frac15 left(t^2-4right) & 2<t<3
endcases$

Here's another method that uses the CDF to directly define the desired distribution and which might be a bit more intuitive/self-evident:

f[x_] := x/4
dist = ProbabilityDistribution[f[x], x, 1, 3];

ProbabilityDistribution[
 
 "CDF", 
 Probability[x < y [Conditioned] x > 2, x [Distributed] dist]
 ,
 y, -[Infinity], [Infinity]
];

ProbabilityDistribution[Piecewise[(2*[FormalX])/5, 2 < [FormalX] < 3, 0], [FormalX], -Infinity, Infinity]

In other words: we just compute the CDF with Probability and then plonk it into ProbabilityDistribution while telling it that it's a CDF rather than a PDF. This is a useful trick to keep in mind, since the CDF is sometimes easier to work with since it's a proper probability.

I learned from @kglr that Probability can give you a conditional CDF.
One can take the derivative of this to compute the desired conditional PDF.

A = X >= 2;
conditionalCDF = Probability[X <= y [Conditioned] A, X [Distributed] ProbabilityDistribution[x/4, x, 1, 3]];
D[conditionalCDF, y]

Sometimes "compact" can be considered elegant:

(x/4)/Integrate[x/4, x, 2, 3]
(* (2 x)/5 *)

If you need it in terms of Mathematica's ProbabilityDistribution:

d = ProbabilityDistribution[(x/4)/Integrate[x/4, x, 2, 3], x, 2, 3]
PDF[d, x]

CDF[d, x]