Geo2.17

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A(10,4),B(-4,9),C(-2,-1)\, are the vertices of a triangle.Find the equations of i)AB. ii)Median through A. iii)Altitude through B. iv)Perpendicular bisector of side AB.

Equation of AB is

y-4={\frac  {9-4}{-4-10}}(x-10)\,

-14y+56=5x-50\,

5x+14y-106=0\,

Median of the traingle is

\left({\frac  {10-4-2}{3}},{\frac  {4+9-1}{3}}\right)\,

{\frac  {4}{3}},4\,

Now the equation of the line passing through A and median is

y-4={\frac  {4-4}{{\frac  {4}{3}}-10}}(x-10)\,

y-4=0\,

The slope of AC is

{\frac  {-1-4}{-2-10}}={\frac  {5}{12}}\,

Now the slope of the altitude is -12/5

Now the equation of altitude through B is

y-9={\frac  {-12}{5}}(x+4)\,

5y-45=-12x-48\,

12x+5y+3=0\,

Perpendicular bisector of side AB

The mid point of AB is {\frac  {10-4}{2}},{\frac  {4+9}{2}}\,

The slope of AB is -5/14,the slope of line perpendicular to this is 14/5.

hence the equation of perpendicular bisector is

y-{\frac  {13}{2}}={\frac  {14}{5}}(x-3)\,

10y-65=28x-84\,

28x-10y-19=0\,

Main page:Geometry:Straight Lines-I