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.