given the three points A = (-2,6) , B = (4,-6) and C = (1,2), consider the triangle ABC and let M be the point on AC so that BM is the altitude through B perpen

Vertices, A = (-2,6), B = (4,-6), C = (1,2)
Transform into Positional vector,
A = -2i + 6j
B = 4i - 6j
C = i + 2j

Point M is on Line AC, such that Line BM perpendicular to AC.
Therefore using scalar vector projection give us length of AM or CM
[tex]|\overline{AM}|=\frac{\overline{AB}.\overline{AC}}{|\overline{AC}|}[/tex]
AB = i(4+2) + j(-6 - 6) = 6i - 12j
AC = i(1+2) + j(2 - 6) = 3i - 4j
|AC| = 5
[tex]|\overline{AM}|=\frac{18+48}{5}[/tex]
[tex]|\overline{AM}|=\frac{66}{5}[/tex]
Distance between point A and M are 66/5
[tex]\frac{\overline{AC}}{|\overline{AC}|}= \frac{3}{5}i-\frac{4}{5}j[/tex]
Therefore vector AM,
[tex]\overline{AM}= \frac{198}{25}i-\frac{264}{25}j[/tex]
Point M,
[tex]A+\overline{AM}= \frac{-50+198}{25}i+\frac{150-264}{25}j[/tex]
[tex]M= \frac{148}{25}i-\frac{114}{25}j[/tex]
Which is not on Line AC.

By projecting line AC to AB,
AB = 6i - 12j
AC = 3i - 4j
|AB| = √(144+36) = 6√5
[tex]|\overline{AN}|=\frac{\overline{AB}.\overline{AC}}{|\overline{AB}|}[/tex]
[tex]|\overline{AN}|=\frac{66}{6\sqrt{5}}= \frac{11\sqrt{5}}{5} [/tex]

[tex]\frac{\overline{AB}}{|\overline{AB}|}= \frac{\sqrt{5}}{5}i-\frac{2\sqrt{5}}{5}j[/tex]
Vector AN,
[tex]\frac{11\sqrt{5}}{5} \frac{\overline{AB}}{|\overline{AB}|} = \frac{55}{25}i - \frac{110}{25}j [/tex]
[tex] \overline{AN} = \frac{55}{25}i - \frac{110}{25}j [/tex]
Point N,
[tex] A + \overline{AN} = \frac{-50+55}{25}i + \frac{150-110}{25}j [/tex]
[tex] N = \frac{1}{5}i + \frac{40}{25}j [/tex]