Let A = \[{\text{ }}\left( {3, - 4{\text{ }}} \right)\], B = \[\left( {1,2} \right)\]. Let P = \[\left( {2k - 1{\text{ }},{\text{ }}2k + 1{\text{ }}} \right)\]be a variable point such that PA + PB is the minimum . Then k is:

Hint: In order to solve the given question, we must have the knowledge of basics of straight line where we study about slope also which comes under coordinate geometry. The Slope of a line determines a number that measures its steepness or mathematically we can say that it is the change in y for a unit change in x along the line . When any two points on the line is given , the slope of a line can be calculated by the formula $ m = \dfrac{{y - {y_1}}}{{x - {x_1}}} $

Complete step by step solution:
In the question given , we need to find the minimum value of PA + PB when we are given the points of A, B and P respectively as –
Let A = \[{\text{ }}\left( {3, - 4{\text{ }}} \right)\]
B = \[\left( {1,2} \right)\]
 P = \[\left( {2k - 1{\text{ }},{\text{ }}2k + 1{\text{ }}} \right)\]
We can get the minimum value of PA + PB iff point P lies on the line AB or the joining points A and B .
So, we will equate the slopes and get our answer as follows by using the formula of slope = $ m = \dfrac{{y - {y_1}}}{{x - {x_1}}} $
\[
  \dfrac{{2k + 1 - 2}}{{2k - 1 - 1}} = \dfrac{{ - 4 - 2}}{{3 - 1}} \\
  \dfrac{{2k + 1}}{{2k - 2}} = \dfrac{{ - 6}}{2} \\
  \dfrac{{2k - 1}}{{2k - 2}} = - 3 \\
  2k - 1 = - 3(2k - 2) \\
  2k - = - 6k + 6 \;
 \]
To simplify we will do calculations adding and subtracting both the sides of the equation as per our requirement .
\[
  6k + 2k = 6 + 1 \\
  8k = 7 \\
  k = \dfrac{7}{8} \;
 \]
So, the correct answer is “ $ k = \dfrac{7}{8} $ ”.

Note: In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation .
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Always try to understand the mathematical statement carefully and keep things distinct .
Choose the options wisely , it's better to break the question and then solve part by part .
Cross check the answer and always keep the final answer simplified .