# Revision history [back]

Your approach will not work since you try to play with plots. Instead, you have to play with vectors, not their plots.

To my understanding of the picture, D is the orthogonal projection of A to the line (B,C).

If v1 denotes the vector BA and v2 denotes the vector BC, the length j is the inner product of v1 by v2 divided by the norm of v2. Here is how to do it in Sage:

sage: A = (9, 5); B = (2, 4); C = (16, -2); sage: v1 = vector(A)-vector(B) sage: v2 = vector(C)-vector(B) sage: v1.inner_product(v2)/v2.norm() 23/29*sqrt(58)

If you want a floating approximation, you can do:

sage: RDF(v1.inner_product(v2)/v2.norm()) 6.040095911547238

Your approach will not work since you try to play with plots. Instead, you have to play with vectors, not their plots.

To my understanding of the picture, D is the orthogonal projection of A to the line (B,C).

If v1 denotes the vector BA and v2 denotes the vector BC, the length j is the inner product of v1 by v2 divided by the norm of v2. Here is how to do it in Sage:

sage: A = (9, 5); B = (2, 4); C = (16, -2);
sage: v1 = vector(A)-vector(B)
sage: v2 = vector(C)-vector(B)
sage: v1.inner_product(v2)/v2.norm()
23/29*sqrt(58)23/29*sqrt(58)


If you want a floating approximation, you can do:

sage: RDF(v1.inner_product(v2)/v2.norm())
6.0400959115472386.040095911547238


Your approach will not work since you try to play with plots. Instead, you have to play with vectors, not their plots.

To my understanding of the picture, D is the orthogonal projection of A to the line (B,C).

If v1 denotes the vector BA and v2 denotes the vector BC, the length j is the inner product of v1 by v2 divided by the norm of v2. Here is how to do it in Sage:

sage: A = (9, 5); B = (2, 4); C = (16, -2);
sage: v1 = vector(A)-vector(B)
sage: v2 = vector(C)-vector(B)
sage: v1.inner_product(v2)/v2.norm()
23/29*sqrt(58)


If you want a floating approximation, you can do:

sage: RDF(v1.inner_product(v2)/v2.norm())
6.040095911547238


*EDIT : * When you write

sage: plot(x,(x,2,5),color='red')


You ask Sage to plot the function x (i.e. the identity function), for x varying from 2 to 5.

To compute the coordinates of D, you can do:

sage: BA = -AB
sage: D = vector(B) + BC*BA.inner_product(BC)/BC.norm()/BC.norm()
sage: D
(219/29, 47/29)


To add the point D on the picture, you can do:

sage: P += point(D)


If you want the x-axis and the y-axis to have the same scale (so that the right angles are visible), you can do:

sage: P.set_aspect_ratio(1)


Your approach will not work since you try to play with plots. Instead, you have to play with vectors, not their plots.

To my understanding of the picture, D is the orthogonal projection of A to the line (B,C).

If v1 denotes the vector BA and v2 denotes the vector BC, the length j is the inner product of v1 by v2 divided by the norm of v2. Here is how to do it in Sage:

sage: A = (9, 5); B = (2, 4); C = (16, -2);
sage: v1 = vector(A)-vector(B)
sage: v2 = vector(C)-vector(B)
sage: v1.inner_product(v2)/v2.norm()
23/29*sqrt(58)


If you want a floating approximation, you can do:

sage: RDF(v1.inner_product(v2)/v2.norm())
6.040095911547238


*EDIT : *EDIT : When you write

sage: plot(x,(x,2,5),color='red')


You ask Sage to plot the function x (i.e. the identity function), for x varying from 2 to 5.

To compute the coordinates of D, you can do:

sage: BA = -AB
sage: D = vector(B) + BC*BA.inner_product(BC)/BC.norm()/BC.norm()
sage: D
(219/29, 47/29)


To add the point D on the picture, you can do:

sage: P += point(D)


If you want the x-axis and the y-axis to have the same scale (so that the right angles are visible), you can do:

sage: P.set_aspect_ratio(1)