Question

If (2, 2p + 2) is the mid point of (3p , 4) and (-2 , 2q). The value of p and q are​

Answer 1

Given,

The co-ordinate of the starting-point of the line segment$\[ = 3p,4\]$

The co-ordinate of the end-point of the line segment$\[ = - 2,2q\]$

The co-ordinate of the mid-point of the line segment$\[ = 2,2p + 2\]$

Solution,

Formula used, the co-ordinates of the mid-point of a line segment$\[ = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\]$

Calculate the value of p.

$\[\begin{array}{l} \Rightarrow \frac{{3p - 2}}{2} = 2\\ \Rightarrow 3p - 2 = 4\\ \Rightarrow 3p = 6\\ \Rightarrow p = 2\end{array}\]$

Calculate the value of q.

$\[\begin{array}{l} \Rightarrow \frac{{4 + 2q}}{2} = \left( {2 \times 2} \right) + 2\\ \Rightarrow 2 + q = 4 + 2\\ \Rightarrow q = 4\end{array}\]$

Hence, the value of p and q are $\[2\,{\rm{and}}\,4\]$ respectively.

Answer 2

Answer:

The value of p and q are​ $2,4$

Step-by-step explanation:

Given:

The midpoint as $(2,2p+2)$

The points are $(3p,4),(-2,2q)$

We need to find  the values of p and q

The midpoint formula for the points $(x_1,y_1),(x_2,y_2)$ is $(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )$

So by applying the formula we get as

$(2,2p+2)=(\frac{3p-2}{2} ,\frac{4+2q}{2} )$

By equating x-coordinates we get

$\frac{3p-2}{2} =2\\\\3p-2=4\\\\3p=6\\\\p=2$

By equating y-coordinate and substituting the p-value we get as

$\frac{2q+4}{2}=2(2)+2 \\\\2q+4=12\\\\2q=8\\\\q=4$