If the midpoint of the line segment joining the points$A\left( {3,4} \right)$and $B\left( {k,6} \right)$ is $P\left( {x,y} \right)$ and $x + y - 10 = 0$, find the value of\[k\].

Hint: First of all we will use the formula of midpoint i.e. $H\left( {{x_3},{y_3}} \right) = \left( {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right),\left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)} \right)$, to find the value of midpoint $P\left( {x,y} \right)$ of points $A\left( {3,4} \right)$ and $B\left( {k,6} \right)$. Now, as all the points mentioned in the question P, A and B lie on a line segment, therefore, they must satisfy the equation of line segment i.e. $x + y - 10 = 0$. Hence, by substituting the values of x and y in the equation we will find the value of \[k\].

Complete step-by-step answer:
In question we are given that midpoint of points $A\left( {3,4} \right)$ and $B\left( {k,6} \right)$ is $P\left( {x,y} \right)$ and the equation of line segment is given as, $x + y - 10 = 0$. Figure will be like this.

We are asked to find the value of \[k\], so, first of all, the equation for midpoint $H = \left( {{x_3},{y_3}} \right)$of two points $D\left( {{x_1},{y_1}} \right)$ and $E\left( {{x_2},{y_2}} \right)$, can be given as,
${x_3} = \left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)$ and ${y_3} = \left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)$ or $H\left( {{x_3},{y_3}} \right) = \left( {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right),\left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)} \right)$
Here, ${x_1} = 3$, ${y_1} = 4$, ${x_2} = k$, ${y_2} = 6$, ${x_3} = x$, ${y_3} = y$, on substituting these values in above formula, we will get,
$P\left( {x,y} \right) = \left( {\left( {\dfrac{{3 + k}}{2}} \right),\left( {\dfrac{{4 + 6}}{2}} \right)} \right)$
$ \Rightarrow P\left( {x,y} \right) = \left( {\left( {\dfrac{{3 + k}}{2}} \right),5} \right)$
$ \Rightarrow x = \dfrac{{3 + k}}{2},\;y = 5$
As, the point P lies on line segment its value of x and y satisfies the equation, $x + y - 10 = 0$, so, on substituting the value of x and y we will get,
$ \Rightarrow \dfrac{{3 + k}}{2} + 5 - 10 = 0$
$ \Rightarrow \dfrac{{3 + k}}{2} - 5 = 0$
We will divide the complete equation by 2.
$ \Rightarrow 3 + k + 2\left( { - 5} \right) = 0$
$ \Rightarrow 3 + k - 10 = 0$
We will combine the like terms together.
$ \Rightarrow k - 7 = 0$
We will isolate the\[k\] in the equation and solve for\[k\].
$ \Rightarrow k = 7$
Thus, we can say that the value of \[k\] is 7.

Note: There are chances of students getting the wrong answer while substituting $B\left( {k,6} \right)$ directly in the equation $x + y - 10 = 0$. By doing this, we get the equation as $k + 6 - 10 = 0$. So, on solving we will get a value of \[k\] as $k = 4$ which leads to an incorrect answer. So, do not make this mistake.