Homology: completing a perspectively transformed panel
A flat metal panel carries a logo that is drawn in perspective on a vertical picture plane. To check the logo, the draughtsman sets up a homology on the drawing sheet: the axis e is the horizontal line y = 0, the centre O is at (0, 4), and a point A(2, 1) of the true shape has its image A′(4, −2) on the same ray from O. A second point B(3, 3) must be transferred. A colleague, used to homotheties, proposes to place B′ by scaling B from O with the ratio OA′/OA = 2, which would give (6, 2). The checker suspects this is wrong because the image of B must lie on the image of line AB and because the behaviour of points near the axis differs from the behaviour of points far from it. The drawing must also locate the vanishing line of the homology and state whether B′ is finite.
1. Construct B′ using the standard homology construction. Give the auxiliary lines and the resulting coordinates.
Read the worked answer
Join A and B. A(2, 1) and B(3, 3): slope = (3 − 1)/(3 − 2) = 2, so the line is y = 2x − 3. It meets the axis y = 0 at R(1.5, 0).
The axis is pointwise fixed in a homology, so R is its own image. Join A′ to R: A′(4, −2) to R(1.5, 0). Its slope is (−2 − 0)/(4 − 1.5) = −0.8, so the image line is y = −0.8x + 1.2.
The centre O, B and B′ must be collinear. Line OB through O(0, 4) and B(3, 3) has slope −1/3: y = 4 − x/3.
Intersect the two lines:
4 − x/3 = −0.8x + 1.2
2.8 = −0.4667x
x = −6, y = 4 − (−6)/3 = 6.
Therefore B′ = (−6, 6). Check: O, B and B′ are collinear because O→B = (3, −1) and O→B′ = (−6, 2) = −2(3, −1). B′ also lies on the image line A′R.
2. Locate the vanishing line and determine whether B′ is finite. Use a test point on y = 2 to justify the result.
Read the worked answer
Use C(1, 2). Its ray from O is y = 4 − 2x, meeting the axis at Q(2, 0). Line AC through A(2, 1) and C(1, 2) has equation y = −x + 3, meeting the axis at R(3, 0). The image line A′R through A′(4, −2) and R(3, 0) has slope −2, so y = −2x + 6.
Intersect this image line with the ray OC:
4 − 2x = −2x + 6, which gives 4 = 6, an impossibility. The lines are parallel, so the image of C is at infinity. Hence the vanishing line is the horizontal line through C, y = 2. Points on y = 2 map to infinity; points on one side of it map to finite images on one side, and points on the other side map to finite images on the opposite side.
B(3, 3) lies above y = 2, so B′ is finite, as the construction confirmed: B′ = (−6, 6). The colleague’s point (6, 2) lies exactly on the vanishing line and is therefore not a finite image of B; it is a different point entirely.
3. Explain why the homothety ratio OA′/OA = 2 is not a valid shortcut. What is invariant in the homology, and how does the error show up numerically?
Read the worked answer
A homology is not a homothety. In a homothety every point is scaled from the same centre O by the same signed ratio, so B would map to O + 2(B − O) = (6, 2). But the homology construction places B′ at the intersection of the image line A′R with OB, giving (−6, 6).
The numerical check is decisive: the point (6, 2) does not lie on the image line A′R, since at x = 6 that line gives y = −0.8(6) + 1.2 = −3.6, not 2. Also, A maps with factor +2 along ray OA, while B maps with factor −2 along ray OB; the signed scale from O is not constant. What is preserved is the projective correspondence defined by the axis, the centre and one pair of corresponding points: corresponding points are collinear with O, corresponding lines meet on the axis, and the cross ratio with the axis is preserved.
The correct workflow is therefore: join the point to a known corresponding point, find the axis intersection, draw the image line through the corresponding image and the axis intersection, then intersect with the ray from O. Never replace this by a single scale factor.