Question

Equations of perpendicular bisectors of side AB and AC of a triangle ABC are \[x-y+5=0\] and \[x+2y=0\]. If the vertex A (1, -2), then find the equation of side BC.

Answer 1

Hint: To solve this question we will first compute mid – points of side AB & AC of \[\Delta ABC\] and that can be calculated as line, \[x-y+5=0\] and \[x+2y=0\] are perpendicular bisectors. So, the point in mid – point of AB & AC. After computing D & E mid – points of AB & AC we will compute co – ordinates of B and C using slope of line \[x+y+c=0\] as \[\dfrac{dy}{dx}=\dfrac{-y}{x}\]. Also we will use the product by two slopes of perpendicular line is -1.

Complete step by step answer:
Let us first construct \[\Delta ABC\],

The co – ordinates of A are (1, -2).
And let D and E be the points on line AB & AC on which perpendicular bisectors are drawn.
The equation of line at D is \[x-y+5=0\] and the equation of line at E is \[x+2y=0\].
Let the co – ordinates of B be \[\left( {{x}_{1}},{{y}_{1}} \right)\] and the co – ordinates of C be \[\left( {{x}_{2}},{{y}_{2}} \right)\].
   

Because the perpendicular bisector of side AB is at D then D is mid – point of side AB.
\[\Rightarrow \] Co – ordinates of D = \[\left( \dfrac{1+{{x}_{1}}}{2},\dfrac{{{y}_{1}}-2}{2} \right)\].
Similarly E is also mid – point of side AC,
\[\Rightarrow \] Co – ordinates of D = \[\left( \dfrac{{{x}_{2}}+1}{2},\dfrac{{{y}_{2}}-2}{2} \right)\]
As D lies on the line \[x-y+5=0\].
So its co – ordinates satisfy the equation \[x-y+5=0\].
Substituting \[x=\dfrac{1+{{x}_{1}}}{2}\] and \[y=\dfrac{{{y}_{1}}-2}{2}\] we get,
\[\begin{align}
  & \left( \dfrac{1+{{x}_{1}}}{2} \right)-\left( \dfrac{{{y}_{1}}-2}{2} \right)+5=0 \\
 & \Rightarrow 1+{{x}_{1}}-{{y}_{1}}+2+10=0 \\
\end{align}\]
\[\Rightarrow {{x}_{1}}-{{y}_{1}}+13=0\] - (1)
Similarly as E lies on the line \[x+2y=0\].
\[\Rightarrow \] Co – ordinates of E satisfy \[x+2y=0\].
Substituting \[x=\dfrac{1+{{x}_{2}}}{2}\] and \[y=\dfrac{{{y}_{2}}-2}{2}\] we get,
\[\begin{align}
  & \left( \dfrac{{{x}_{2}}+1}{2} \right)+2\left( \dfrac{{{y}_{2}}-2}{2} \right)=0 \\
 & \Rightarrow \dfrac{{{x}_{2}}+1}{2}+{{y}_{2}}-2=0 \\
 & \Rightarrow {{x}_{2}}+1+2{{y}_{2}}-4=0 \\
\end{align}\]
\[\Rightarrow {{x}_{2}}+2{{y}_{2}}-3=0\] - (2)
Now the slope of an equation \[ax+by+c=0\] is given by \[\dfrac{dy}{dx}\].
Then slope of line \[x-y+5=0\] is obtained by differentiating,
\[x-y+5=0\]
Differentiating with respect to x we have,
\[1-\dfrac{dy}{dx}+0=0\]
\[\Rightarrow \dfrac{dy}{dx}=1\] = slope of line \[x-y+5=0\].
Similarly slope of line \[x+2y=0\].
\[1+2\dfrac{dy}{dx}=0\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{2}\] = slope of line \[x+2y=0\].
Now we will find the slope of line AB and of AC.
When co – ordinates of two points are \[\left( {{x}_{1}},{{y}_{1}} \right)\] & \[\left( {{x}_{2}},{{y}_{2}} \right)\].
Then slope is \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].
Applying this we have slope of line AB is, \[\dfrac{{{y}_{2}}+2}{{{x}_{1}}-1}\].
Now because two lines which are perpendicular have the product of slopes as ‘-1’.
Then as AB and line \[x-y+5=0\] are perpendicular,
\[\Rightarrow \] (slope of AB) \[\times \] (slope of \[x-y+5=0\]) = -1
\[\Rightarrow \left( \dfrac{{{y}_{1}}+2}{{{x}_{1}}-1} \right)\left( 1 \right)=-1\]
\[\Rightarrow {{y}_{1}}-2=-{{x}_{1}}+1\Rightarrow {{x}_{1}}+{{y}_{1}}-1=0\] - (3)
Similarly slope of line AC is \[\dfrac{{{y}_{2}}+2}{{{x}_{2}}-1}\] and as both lines AC and \[x+2y=0\] are perpendicular then,
(slope of AC) \[\times \] (slope of \[x+2y=0\]) = -1
\[\Rightarrow \left( \dfrac{{{y}_{2}}+2}{{{x}_{2}}-1} \right)\times \left( \dfrac{-1}{2} \right)=-1\]
\[\Rightarrow {{y}_{2}}+2=2\left( {{x}_{2}}-1 \right)\Rightarrow 2{{x}_{\partial 2}}-{{y}_{2}}-4=0\] - (4)
Solving equation (2) and equation (3) we have,
\[{{x}_{1}}-{{y}_{1}}+13=0\]
\[\underline{{{x}_{1}}+{{y}_{1}}+1=0}\]
\[2{{x}_{1}}+14=0\]
\[{{x}_{1}}=\dfrac{-14}{2}=-7\]
\[\Rightarrow {{x}_{1}}=-7\] and \[{{y}_{1}}=6\].
Similarly solving equation (2) and equation (4) we have,
\[\begin{align}
  & {{x}_{2}}+2{{y}_{2}}-3=0 \\
 & 2{{x}_{2}}-{{y}_{2}}-4=0 \\
\end{align}\]
Multiplying equation (2) by 2 and substituting from equation (4) we get,
\[\begin{align}
  & 2{{x}_{2}}+4{{y}_{2}}-6=0 \\
 & \underline{\begin{align}
  & 2{{x}_{2}}-{{y}_{2}}-4=0 \\
 & ++ \\
\end{align}} \\
 & +5{{y}_{2}}-2=0\Rightarrow +5{{y}_{2}}-2=0 \\
\end{align}\]
\[\Rightarrow {{y}_{2}}=\dfrac{+2}{5}\] and \[{{x}_{2}}=\dfrac{11}{5}\].
Thus co – ordinates of C are \[\left( \dfrac{11}{5},\dfrac{2}{5} \right)\].
Then equation of line BC is
\[\begin{align}
  & y-6=\dfrac{\dfrac{2}{5}-6}{\dfrac{11}{5}+7}\left( x+7 \right) \\
 & \Rightarrow y-6=\dfrac{-28}{46}\left( x+7 \right) \\
 & \Rightarrow 14x+23y-40=0 \\
\end{align}\]
\[\therefore \] Equation of line BC is \[14x+23y-40=0\].

Note:
The key point to note here in this question is that while calculating the equation of line BC this formula is used. If \[B=\left( \alpha ,\beta \right)\] and \[C=\left( \alpha ',\beta ' \right)\] then BC = \[y-\alpha =\dfrac{\beta '-\beta }{\alpha '-\alpha }\left( x-\alpha \right)\].
Another point to note here is that slope of both lines \[x+2y=0\] and \[x-y+5=0\] is important and are used as \[\dfrac{dy}{dx}=m\] = slope of line.