Problem
GEO-B2-M10-P010 Projection by Coordinates
#10
★★★☆☆ Level 3 of 5
In triangle \(A(0,0)\), \(B(6,0)\), \(C(0,8)\), point \(P\) is the foot of the perpendicular from \(A\) to \(BC\). Find \(BP:PC\).
Write \(P=B+t(C-B)\) and use \(\overrightarrow{AP}\cdot\overrightarrow{BC}=0\).
We have \(P=(6,0)+t(-6,8)=(6-6t,8t)\). The perpendicularity condition is \((6-6t,8t)\cdot(-6,8)=0\). Thus \(-36+36t+64t=0\), so \(t=\frac{9}{25}\). Hence \(BP:PC=t:(1-t)=9:16\).
Coordinates are shorter here than trying to build several similar triangles.