Gọi pt AB có dạng:
\(a\left(x-1\right)+b\left(y-4\right)=0\Leftrightarrow ax+by-a-4b=0\)
BC vuông góc AB nên pt BC có dạng:
\(b\left(x+5\right)-ay=0\Leftrightarrow bx-ay+5b=0\)
\(BC=d\left(P;AB\right)=\frac{\left|2a+2b-a-4b\right|}{\sqrt{a^2+b^2}}=\frac{\left|a-2b\right|}{\sqrt{a^2+b^2}}\)
\(AB=d\left(Q;BC\right)=\frac{\left|b-8a+5b\right|}{\sqrt{a^2+b^2}}=\frac{\left|8a-6b\right|}{\sqrt{a^2+b^2}}\)
\(BC.AB=5\)
\(\Leftrightarrow\frac{\left|\left(a-2b\right)\left(8a-6b\right)\right|}{a^2+b^2}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}8a^2+12b^2-22ab=5a^2+5b^2\\8a^2+12b^2-22ab=-5a^2-5b^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3a^2+7b^2-22ab=0\\13a^2+17b^2-22ab=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=7b\\3a=b\end{matrix}\right.\) \(\Rightarrow\) chọn \(\left(a;b\right)=\left(7;1\right);\left(1;3\right)\)
Có 2 trường hợp thỏa mãn...
