Geo2.37

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Show that the feet of the perpendicular from (-2,8)\, to the lines x-2y+6=0,x-7y+6=0,x+3y-4=0\, are collinear.

Foot of the perpendicular to x-2y+6=0\, be (h,k)\,

{\frac  {h+2}{1}}={\frac  {k+8}{-2}}={\frac  {-(-2+16+6)}{1+4}}\,

(h,k)=(-6,0)\,

Foot of the perpendicular to x-7y+6=0\, be (a,b)\,

{\frac  {a+2}{1}}={\frac  {b+8}{-7}}={\frac  {-(-2+56+6)}{1+49}}\,

(a,b)=(-{\frac  {16}{5}},{\frac  {2}{5}})\,

Foot of the perpendicular to x-2y+6=0\, be (x_{1},y_{1})\,

{\frac  {x_{1}+2}{1}}={\frac  {y_{1}+8}{3}}=-{\frac  {-2-24-4}{1+9}}\,

(x_{1},y_{1})=(1,1)\,

Now {\begin{vmatrix}-6&-{\frac  {16}{5}}&1&-6\\0&{\frac  {2}{5}}&1&0\end{vmatrix}}=-{\frac  {12}{5}}-{\frac  {16}{5}}-{\frac  {2}{5}}+6=-6+6=0\,

Therefore, the three points are collinear.

Main Page:Geometry:Straight Lines-I