Question 32

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Jambmaths question: 

Find the equation of the perpendicular bisector of the line joining P(2,-3) to Q(-5,1)

Option A: 

$8y-14x+13=0$

Option B: 

$8y-14x-13=0$

Option C: 

$8y+14x-13=0$

Option D: 

$8y+14x+13=0$

Jamb Maths Solution: 

$\begin{align}  & \text{Given }P=({{x}_{1}},{{y}_{1}})=(2,-3),\text{ }Q=({{x}_{2}},{{y}_{2}})=(-5,1) \\ & \text{Slope of a line}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\ & \text{Slope of line }PQ=\frac{1-(-3)}{-5-2} \\ & {{m}_{1}}=-\frac{4}{7} \\ & \text{Mid point of }PQ=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\ & =\left( \frac{2-5}{2},\frac{-3+1}{2} \right)=\left( \frac{-3}{2},-1 \right) \\ & \text{For two line to be perpendicular }{{m}_{1}}{{m}_{2}}=-1 \\ & -\frac{4}{7}\times {{m}_{2}}=-1 \\ & {{m}_{2}}=\frac{7}{4} \\ & \text{Equation of the line perpendicular } \\ & \text{(but also perpendicular bisector) to the line PQ} \\ & \text{will have a slope of }\frac{7}{4} \\ & \text{Using one point slope equation} \\ & (y-{{y}_{2}})={{m}_{2}}(x-{{x}_{2}}) \\ & [y-(-1)]=\tfrac{7}{4}[x-(-\tfrac{3}{2})] \\ & y+1=\tfrac{7}{4}[(x+\tfrac{3}{2})] \\ & 4y+4=7x+\tfrac{21}{2} \\ & 8y+8=14x+21 \\ & 8y-14x-13=0 \\\end{align}$

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