Question

# Find the equations of the medians of a triangle, the equations of whose sides are :$$3 x + 2 y + 6 = 0,2 x - 5 y + 4 = 0$$  and  $$x - 3 y - 6 = 0.$$

Solution

## The equations of the sides of a triangle are given by the following equations$$3x+2y+6=0\quad \longrightarrow \left( 1 \right)$$$$2x-5y+4=0\quad \longrightarrow \left( 2 \right)$$$$x-3y-6=0\quad \longrightarrow \left( 3 \right)$$Solving the equations taking two at a time we can find the vertices of the triangle whose sides are given by $$(1)$$, $$(2)$$ and $$(3)$$Solving $$(1)$$ \$ $$(2)$$$$3x+2y+6=0$$ and $$2x-5y+4=0$$we get, $$x=-2$$ and $$y=0$$Solving $$(2)$$ & $$(3)$$$$x-3y-6=0$$$$2x-5y+4=0$$    } $$\Rightarrow$$ $$x=-42$$ and $$y=-16$$Solving $$(3)$$ & $$(1)$$$$x-3y-6=0$$$$3x+2y+6=0$$    } $$\Rightarrow$$ $$x=\dfrac { -6 }{ 11 }$$ and $$y=\dfrac { -24 }{ 11 }$$So, the vertices are$$A(-2,0)$$  $$B=(-42,-16)$$  $$C=\left( \dfrac { -6 }{ 11 } ,\dfrac { -24 }{ 11 } \right)$$let the midpoint of the side $$AB$$ be $$P$$, which has coordinates $$(-22,-8)$$let the midpoint of the side $$BA$$ be $$Q$$, which has coordinates $$\left( \dfrac { -468 }{ 11 } ,\dfrac { -200 }{ 11 } \right)$$let the midpoint of the side $$CA$$ be $$R$$, which has coordinates $$\left( \dfrac { -28 }{ 11 } ,\dfrac { -24 }{ 11 } \right)$$Now the equation of a line in the two-point form is given by $$y-{ y }_{ 1 }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \left( x-{ x }_{ 1 } \right)$$So the equation of the median $$PC$$ is given by : $$y+\dfrac { 24 }{ 11 } =\dfrac { -8+\dfrac { 24 }{ 11 } }{ -22+\dfrac { 6 }{ 11 } } \left( x+\dfrac { 6 }{ 11 } \right)$$$$\Rightarrow 11y+24=\dfrac { 64 }{ 236 } \left( 11x+6 \right)$$$$\Rightarrow 11y+24=\dfrac { 16 }{ 59 } \left( 11x+6 \right)$$$$\Rightarrow \boxed { 176x-649y-1032=0 }$$the equation of the median $$RB$$ is given by $$y+\dfrac { 24 }{ 11 } =\dfrac { -16+24/11 }{ -42+28/11 } \left( x+\dfrac { 28 }{ 11 } \right) \Rightarrow 2387y+5208=836x+2128$$                                                                 $$\Rightarrow \boxed { 836x-2387y-3080=0 }$$the equation of the median $$AQ$$ is given by $$y-0=\dfrac { -200/11 }{ \dfrac { -468 }{ 11 } +2 } \left( x+2 \right)$$$$\Rightarrow \boxed { 100x-223y+200=0 }$$Maths

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