Coordinate Geometry (Perpendicular bisector)

(CIE 0606/2021/m/12/Q3)
The line $AB$ is such that the points $A$ and $B$ have coordinates $(-4, 6)$ and $(2, 14)$ respectively. 

(a) The point $C$, with coordinates $(7, a)$ lies on the perpendicular bisector of $AB$. Find the value of $a.$ [4]

(b) Given that the point $D$ also lies on the perpendicular bisector of $AB,$ find the coordinates of $D$ such that the line $AB$ bisects the line $CD.$ [2]

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*********math solution*************

(a)      Gradient of $AB=\dfrac{14-6}{2-(-4)}=\dfrac{8}{6}=\dfrac 43.$

Gradient of line perpendicular to $AB=-\dfrac 34.$

Mid-point of $AB=M=\left(\dfrac{-4+2}{2},\dfrac{6+14}{2}\right)=(-1,10)$

Equation of line perpendicular to $AB$ is $y-10=-\dfrac 34(x+1)$.

Since $(7, a)$ lies on the perpendicular bisector of $AB, a-10=-\dfrac 34(7+1).$

Thus $a=4.$  Thus $C=(7,4)$.

 (b)  Since $M$ is also the mid-point of  $CD$, $(-1,10)=\left(\dfrac{7+x}{2},\dfrac{4+y}{2}\right)$. 

Hence $7+x=-2,4+y=20\Rightarrow x=-9,y-16.$

$\therefore D=(-9,16)$.


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