Problem
GEO-B2-M09-P014 Projection onto a Side
#14
★★★☆☆ Level 3 of 5
In triangle \(A(0,0)\), \(B(6,0)\), \(C(2,5)\), point \(P\) is the foot of the perpendicular from \(A\) to \(BC\). Find the ratio \(BP:PC\).
Write \(P=B+t(C-B)\) and use the condition \(AP\perp BC\).
Let \(P=B+t(C-B)=(6,0)+t(-4,5)\). Then \(\overrightarrow{AP}=P\), and \(\overrightarrow{BC}=(-4,5)\). The perpendicularity condition gives \(((6,0)+t(-4,5))\cdot(-4,5)=0\). Thus \(-24+41t=0\), so \(t=\frac{24}{41}\). Hence \(BP:PC=t:(1-t)=24:17\).
A side parameter often saves computation.