Let us consider that the original point is (x, y) and new transformed point is (x′, y′). Further, Shx and Shy are shearing factors in x and y directions. If we perform the y-direction shear relative to x = xref then the transformed point is given by _______.
x′ = x
y′ = Shy (x – xref) + y
A shear slants an object. A y-direction shear changes only the y coordinate, by an amount proportional to the point's distance from a reference line; the x coordinate is left unchanged.
Step 1 — For a y-direction shear relative to the vertical line \(x = x_{ref}\), points on that line must stay fixed, so the displacement is proportional to \(x - x_{ref}\).
Step 2 — The transformation is \(x' = x\) and \(y' = y + Sh_y\,(x - x_{ref})\), which in matrix form is \(\begin{bmatrix}1 & 0 & 0 \\ Sh_y & 1 & -Sh_y\,x_{ref} \\ 0 & 0 & 1\end{bmatrix}\).
Step 3 — Check: a point with \(x = x_{ref}\) gives \(y' = y\), so the reference line is unchanged, as required.
The pair that changes x by \(Sh_x (y - y_{ref})\) and keeps y is an x-direction shear about \(y = y_{ref}\). Multiplying y by \(Sh_x\) is a scaling, not a shear, and the pair that multiplies y by \((x - x_{ref})\) is not a linear shear at all.
x′ = x, y′ = Shy (x – xref) + y option is correct.
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