Question

example 4
find the coordinates of the missing endpoint if b is the midpoint of $overline{ac}$.

1. $c(-5,4),b(-2,5)$
2. $a(1,7),b(-3,1)$
3. $a(-4,2),b(6,-1)$
4. $c(-6,-2),b(-3,-5)$
5. $a(4,-0.25),b(-4,6.5)$
6. $c(\frac{5}{3},-6),b(\frac{8}{3},4)$

Explanation:

Step1: Recall mid - point formula

The mid - point formula between two points $A(x_1,y_1)$ and $C(x_2,y_2)$ is $B(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. If we know one endpoint and the mid - point, we can find the other endpoint. Let the coordinates of the missing endpoint be $(x,y)$. For the $x$ - coordinate: $\frac{x_1+x}{2}=x_m$ (where $x_m$ is the $x$ - coordinate of the mid - point), and for the $y$ - coordinate: $\frac{y_1 + y}{2}=y_m$ (where $y_m$ is the $y$ - coordinate of the mid - point). We can solve these equations for $x$ and $y$.

Step2: Solve for $x$ and $y$ in general

From $\frac{x_1+x}{2}=x_m$, we get $x = 2x_m-x_1$. From $\frac{y_1 + y}{2}=y_m$, we get $y=2y_m - y_1$.

Step3: Solve problem 33

Given $C(-5,4)$ and $B(-2,5)$. Let the coordinates of $A$ be $(x,y)$.
For the $x$ - coordinate: $x=2\times(-2)-(-5)=-4 + 5=1$.
For the $y$ - coordinate: $y=2\times5 - 4=10 - 4 = 6$. So the coordinates of $A$ are $(1,6)$.

Step4: Solve problem 34

Given $A(1,7)$ and $B(-3,1)$. Let the coordinates of $C$ be $(x,y)$.
For the $x$ - coordinate: $x=2\times(-3)-1=-6 - 1=-7$.
For the $y$ - coordinate: $y=2\times1 - 7=2 - 7=-5$. So the coordinates of $C$ are $(-7,-5)$.

Step5: Solve problem 35

Given $A(-4,2)$ and $B(6,-1)$. Let the coordinates of $C$ be $(x,y)$.
For the $x$ - coordinate: $x=2\times6-(-4)=12 + 4 = 16$.
For the $y$ - coordinate: $y=2\times(-1)-2=-2 - 2=-4$. So the coordinates of $C$ are $(16,-4)$.

Step6: Solve problem 36

Given $C(-6,-2)$ and $B(-3,-5)$. Let the coordinates of $A$ be $(x,y)$.
For the $x$ - coordinate: $x=2\times(-3)-(-6)=-6 + 6=0$.
For the $y$ - coordinate: $y=2\times(-5)-(-2)=-10 + 2=-8$. So the coordinates of $A$ are $(0,-8)$.

Step7: Solve problem 37

Given $A(4,-0.25)$ and $B(-4,6.5)$. Let the coordinates of $C$ be $(x,y)$.
For the $x$ - coordinate: $x=2\times(-4)-4=-8 - 4=-12$.
For the $y$ - coordinate: $y=2\times6.5-(-0.25)=13 + 0.25 = 13.25$. So the coordinates of $C$ are $(-12,13.25)$.

Step8: Solve problem 38

Given $C(\frac{5}{3},-6)$ and $B(\frac{8}{3},4)$. Let the coordinates of $A$ be $(x,y)$.
For the $x$ - coordinate: $x=2\times\frac{8}{3}-\frac{5}{3}=\frac{16 - 5}{3}=\frac{11}{3}$.
For the $y$ - coordinate: $y=2\times4-(-6)=8 + 6 = 14$. So the coordinates of $A$ are $(\frac{11}{3},14)$.

Answer:

1. $(1,6)$
2. $(-7,-5)$
3. $(16,-4)$
4. $(0,-8)$
5. $(-12,13.25)$
6. $(\frac{11}{3},14)$