Question

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.

Answer 1

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 lie on a line segment they must satisfy the equation of line segment i.e. $x+y-10=0$ , so, 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,
$\dfrac{3+k}{2}+5-10=0$
$\Rightarrow \dfrac{3+k}{2}-5=0$
$\Rightarrow 3+k+2\left( -5 \right)=0$
$\Rightarrow 3+k-10=0\Rightarrow k-7=0$
$\Rightarrow k=7$
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.