
How do you find the midpoint of AB with points $A\equiv \left( 2,3 \right);B\equiv \left( 8,11 \right)$
Answer
444.9k+ views
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.
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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