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**Hint:**We first try to express the formula for ratio division of two arbitrary points. The ratio of division between them is $m:n$ which gives $\left( x,y \right)\equiv \left( \dfrac{mc+an}{m+n},\dfrac{md+bn}{m+n} \right)$. We also find the midpoint theorem as $\left( x,y \right)\equiv \left( \dfrac{c+a}{2},\dfrac{d+b}{2} \right)$. We put the values for $A\equiv \left( 2,3 \right);B\equiv \left( 8,11 \right)$ to get the midpoint of AB.

**Complete step-by-step solution:**

We need to find the midpoint of AB with points $A\equiv \left( 2,3 \right);B\equiv \left( 8,11 \right)$.

We first find the formula for ratio division of two points. We take two arbitrary points.

They are $\left( a,b \right);\left( c,d \right)$. The ratio of division between them is $m:n$.

Then if the required point is $\left( x,y \right)$, we can say $\left( x,y \right)\equiv \left( \dfrac{mc+an}{m+n},\dfrac{md+bn}{m+n} \right)$.

Now for midpoints the ratio of division is $1:1$.

We put the values of m and n to get $\left( x,y \right)\equiv \left( \dfrac{c+a}{2},\dfrac{d+b}{2} \right)$.

So, the particular coordinates are half of the sum of the respective points.

Now, we find the midpoint of AB with points $A\equiv \left( 2,3 \right);B\equiv \left( 8,11 \right)$.

The replacement will be $\left( a,b \right)\equiv \left( 2,3 \right);\left( c,d \right)\equiv \left( 8,11 \right)$ for the theorem $\left( x,y \right)\equiv \left( \dfrac{c+a}{2},\dfrac{d+b}{2} \right)$

The midpoint will be $\left( x,y \right)\equiv \left( \dfrac{2+8}{2},\dfrac{3+11}{2} \right)$.

We complete the addition to get $2+8=10,3+11=14$.

Then we divide the added values by 2 to get $\dfrac{10}{2}=5,\dfrac{14}{2}=7$.

**The required midpoints of the segment AB with points $A\equiv \left( 2,3 \right);B\equiv \left( 8,11 \right)$ will be $\left( 5,7 \right)$.**

**Note:**The midpoint of a segment and its two end-points is similar. We actually need the end points to get the midpoint. The ratio value being negative indicates its direction. The division can be internal and external both.

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