Chapter 1.3, Problem 11CYU

To determine

To find the coordinates of missing point.

Answer to Problem 11CYU

Hence, coordinates of G are $(-4,12)$ .

Explanation of Solution

Given information:

  $F(1,3.5),J(6,-2)$

F is the midpoint of $JG$ .

Formula used:

  $x_m^{}=\frac{x_1^{}+x_2^{}}{2}$ For x coordinate of midpoint.

  $y_m^{}=\frac{y_1^{}+y_2^{}}{2}$ For y coordinate of midpoint.

Calculation:

Let the coordinates of G be $(x_2^{},y_2^{})$

Let coordinates of J be $(x_1^{},y_1^{})$ and

Coordinates of F be $(x_m^{},y_m^{})$

According to question, P is the midpoint of $JG$

So for x coordinate of E ,

  $1=\frac{6+x_2^{}}{2}$ ,

Solving,

  $1(2)=6+x_2^{}$

Now,

  $2-6=x_2^{}$

Hence, x coordinate of G is -4.

For y coordinate of G ,

  $5=\frac{-2+y_2^{}}{2}$ ,

Solving,

  $5(2)=-2+y_2^{}$

Now,

  $10+2=y_2^{}$

Hence, y coordinate of G is 12.

Hence, coordinates of E are $(-4,12)$ .