Question

Find the length of the perpendicular from A to BC given that A(-4,2), B(8,-2) and C(2,6).

Answer 1

The distance from A to C is
sqrt(horizontal^2+vertical^2)

this is sqrt(4^2+6^2)=2sqrt(13)

In vector form, the lines A^C and C^B are
2x+3y
and
3x+4y
respectively

the angle between these two are given by use of the scalar product;

arccos(1/sqrt(13*25))

and so mapping the perpendicular distance from A to the line B^C is given by multiplying the distance A to C with the cosine of the smallest angle left in the sub triangle we've created. (i.e. 180 - 90 - theta (above))

which gives a final answer of 2sqrt(13)*cos(180-90-~87)=7.2

Answer 2

excuse me, the lines A^C and C^B are
3x + 2y
and
3x - 4y
respectively

Answer 3

Thanks!