Question

If (3,3/4)is the midpoint of the line segment joining the points (k,0) and (7,2/3),find the value of k.

Answer 1

Answer:

14

Step-by-step explanation:

Given : Midpoint : A(x,y)=(3,3/4)

           End points :$B (x_{1} ,y_{1})=(k,0)$

                               $C (x_{2} ,y_{2})=(7,\frac{2}{3})$

Solution:

Now use mid point formula :

$(x,y)=(\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2} )$

$(3,\frac{3}{4})=(\frac{k+7}{2} ,\frac{0+\frac{2}{3}}{2} )$

⇒$3=\frac{k+7}{2}$ and $\frac{3}{4}=\frac{0+\frac{2}{3}}{2}$

Thus for k

⇒$3=\frac{k+7}{2}$

⇒$21=k+7$

⇒$21-7=k$

⇒$14=k$

Hence the value of k is 14

Answer 2

Answer:  The required value of k is -1.

Step-by-step explanation:  Given that the point $\left(3,\dfrac{3}{4}\right)$ is the midpoint of the line segment joining the points (k, 0) and $\left(7,\dfrac{2}{3}\right)$.

We are to find the value of k.

We know that

the co-ordinates of the midpoint of the line segment joining the points (a, b) and (c, d) are given by

$\left(\dfrac{a+c}{2},\dfrac{b+d}{2}\right).$

Therefore, according to the given information, we get

$\dfrac{k+7}{2}=3\\\\\Rightarrow k+7=6\\\\\Rightarrow k=6-7\\\\\Rightarrow k=-1.$

Thus, the required value of k is -1.