Question

examples 5 and 6 suppose m is the mid - point of $overline{fg}$. find each missing measure. 39. $fm = 5y + 13$, $mg = 5 - 3y$, $fg =?$ 40. $fm = 3x - 4$, $mg = 5x - 26$, $fg =?$ 41. $fm = 8a + 1$, $fg = 42$, $a =?$ 42. $mg = 7x - 15$, $fg = 33$, $x =?$ 43. $fm = 3n + 1$, $mg = 6 - 2n$, $fg =?$ 44. $fm = 12x - 4$, $mg = 5x + 10$, $fg =?$ 45. $fm = 2k - 5$, $fg = 18$, $k =?$ 46. $fg = 14a + 1$, $fm = 14.5$, $a =?$ 47. $mg = 13x + 1$, $fg = 15$, $x =?$ 48. $fg = 11x - 15.6$, $mg = 10.9$, $x =?$ mixed exercises find the coordinates of the missing endpoint if p is the mid - point of $overline{nq}$. 49. $n(2,0)$, $p(5,2)$ 50. $n(5,4)$, $p(6,3)$ 51. $q(3,9)$, $p(-1,5)$

Explanation:

Step1: Use mid - point property

Since \(M\) is the mid - point of \(\overline{FG}\), then \(FM = MG\).

Step2: Solve for the variable

For example, in problem 39:
Set \(FM = MG\), so \(5y + 13=5 - 3y\).
Add \(3y\) to both sides: \(5y+3y + 13=5-3y + 3y\), which gives \(8y+13 = 5\).
Subtract 13 from both sides: \(8y+13 - 13=5 - 13\), so \(8y=-8\).
Divide both sides by 8: \(y=-1\).
Then \(FM = 5(-1)+13 = 8\) and \(MG = 5-3(-1)=8\).
So \(FG=FM + MG=8 + 8 = 16\).
For problem 40:
Set \(FM = MG\), so \(3x - 4=5x - 26\).
Subtract \(3x\) from both sides: \(3x-3x - 4=5x-3x - 26\), which gives \(-4 = 2x-26\).
Add 26 to both sides: \(-4 + 26=2x-26 + 26\), so \(22 = 2x\).
Divide both sides by 2: \(x = 11\).
Then \(FM=3(11)-4=29\), \(MG = 5(11)-26 = 29\), and \(FG=FM + MG=58\).
For problem 41:
Since \(FM=\frac{1}{2}FG\) (because \(M\) is the mid - point), then \(8a + 1=\frac{42}{2}=21\).
Subtract 1 from both sides: \(8a+1 - 1=21 - 1\), so \(8a = 20\).
Divide both sides by 8: \(a=\frac{20}{8}=\frac{5}{2}=2.5\).
For problem 42:
Since \(FM=FG - MG\) and \(FM = MG\) (mid - point), then \(2MG=FG\).
Set \(2(7x - 15)=33\).
Distribute: \(14x-30 = 33\).
Add 30 to both sides: \(14x-30 + 30=33 + 30\), so \(14x=63\).
Divide both sides by 14: \(x=\frac{63}{14}=\frac{9}{2}=4.5\).
For problem 43:
Set \(FM = MG\), so \(3n + 1=6 - 2n\).
Add \(2n\) to both sides: \(3n+2n + 1=6-2n + 2n\), which gives \(5n+1 = 6\).
Subtract 1 from both sides: \(5n+1 - 1=6 - 1\), so \(5n = 5\).
Divide both sides by 5: \(n = 1\).
Then \(FM=3(1)+1 = 4\), \(MG = 6-2(1)=4\), and \(FG=FM + MG=8\).
For problem 44:
Set \(FM = MG\), so \(12x - 4=5x + 10\).
Subtract \(5x\) from both sides: \(12x-5x - 4=5x-5x + 10\), which gives \(7x-4 = 10\).
Add 4 to both sides: \(7x-4 + 4=10 + 4\), so \(7x=14\).
Divide both sides by 7: \(x = 2\).
Then \(FM=12(2)-4 = 20\), \(MG = 5(2)+10 = 20\), and \(FG=FM + MG=40\).
For problem 45:
Since \(FM=\frac{1}{2}FG\) (mid - point), then \(2k - 5=\frac{18}{2}=9\).
Add 5 to both sides: \(2k-5 + 5=9 + 5\), so \(2k=14\).
Divide both sides by 2: \(k = 7\).
For problem 46:
Since \(FM=\frac{1}{2}FG\) (mid - point), then \(14.5=\frac{14a + 1}{2}\).
Multiply both sides by 2: \(14.5\times2=14a + 1\), so \(29 = 14a+1\).
Subtract 1 from both sides: \(29 - 1=14a+1 - 1\), so \(28 = 14a\).
Divide both sides by 14: \(a = 2\).
For problem 47:
Since \(FM=FG - MG\) and \(FM = MG\) (mid - point), then \(2MG=FG\).
Set \(2(13x + 1)=15\).
Distribute: \(26x+2 = 15\).
Subtract 2 from both sides: \(26x+2 - 2=15 - 2\), so \(26x=13\).
Divide both sides by 26: \(x=\frac{13}{26}=0.5\).
For problem 48:
Since \(FM=FG - MG\) and \(FM = MG\) (mid - point), then \(2MG=FG\).
Set \(2\times10.9=11x - 15.6\).
So \(21.8=11x - 15.6\).
Add 15.6 to both sides: \(21.8+15.6=11x-15.6 + 15.6\), so \(37.4=11x\).
Divide both sides by 11: \(x = 3.4\).
For problem 49:
The mid - point formula is \(P(\frac{x_N+x_Q}{2},\frac{y_N + y_Q}{2})\).
Let \(N(2,0)\) and \(P(5,2)\).
We know \(\frac{x_N+x_Q}{2}=5\) and \(\frac{y_N + y_Q}{2}=2\).
For the \(x\) - coordinate: \(\frac{2+x_Q}{2}=5\), multiply both sides by 2: \(2+x_Q = 10\), so \(x_Q=8\).
For the \(y\) - coordinate: \(\frac{0 + y_Q}{2}=2\), multiply both sides by 2: \(y_Q = 4\). So \(Q(8,4)\).
For problem 50:
Let \(N(5,4)\) and \(P(6,3)\).
Using the mid - point formula \(\frac{x_N+x_Q}{2}=6\) and \(\frac{y_N + y_Q}{2}=3\).
For the \(x\) - coordinate: \(\frac{5+x_Q}{2}=6\), multiply both sides by 2: \(5+x_Q = 12\), so \(x_Q=7\).
For the \(y\) - coordinate: \(\frac{4 + y_Q}{2}=3\), multiply both sides by 2: \(4 + y_Q=6\), so \(y_Q = 2\). So \(Q(7,2)\).
For problem 51:
Let \(Q(3,9)\) and \(P(-1,5)\).
Using the mid - point formula \(\frac{x_N+x_Q}{2}=-1\) and \(\frac{y_N + y_Q}{2}=5\).
For the \(x\) - coordinate: \(\frac{x_N+3}{2}=-1\), multiply both sides by 2: \(x_N+3=-2\), so \(x_N=-5\).
For the \(y\) - coordinate: \(\frac{y_N + 9}{2}=5\), multiply both sides by 2: \(y_N + 9 = 10\), so \(y_N = 1\). So \(N(-5,1)\).

39. \(FG = 16\)
40. \(FG = 58\)
41. \(a = 2.5\)
42. \(x = 4.5\)
43. \(FG = 8\)
44. \(FG = 40\)
45. \(k = 7\)
46. \(a = 2\)
47. \(x = 0.5\)
48. \(x = 3.4\)
49. \(Q(8,4)\)
50. \(Q(7,2)\)
51. \(N(-5,1)\)

Answer:

Step1: Use mid - point property

Since \(M\) is the mid - point of \(\overline{FG}\), then \(FM = MG\).

Step2: Solve for the variable

For example, in problem 39:
Set \(FM = MG\), so \(5y + 13=5 - 3y\).
Add \(3y\) to both sides: \(5y+3y + 13=5-3y + 3y\), which gives \(8y+13 = 5\).
Subtract 13 from both sides: \(8y+13 - 13=5 - 13\), so \(8y=-8\).
Divide both sides by 8: \(y=-1\).
Then \(FM = 5(-1)+13 = 8\) and \(MG = 5-3(-1)=8\).
So \(FG=FM + MG=8 + 8 = 16\).
For problem 40:
Set \(FM = MG\), so \(3x - 4=5x - 26\).
Subtract \(3x\) from both sides: \(3x-3x - 4=5x-3x - 26\), which gives \(-4 = 2x-26\).
Add 26 to both sides: \(-4 + 26=2x-26 + 26\), so \(22 = 2x\).
Divide both sides by 2: \(x = 11\).
Then \(FM=3(11)-4=29\), \(MG = 5(11)-26 = 29\), and \(FG=FM + MG=58\).
For problem 41:
Since \(FM=\frac{1}{2}FG\) (because \(M\) is the mid - point), then \(8a + 1=\frac{42}{2}=21\).
Subtract 1 from both sides: \(8a+1 - 1=21 - 1\), so \(8a = 20\).
Divide both sides by 8: \(a=\frac{20}{8}=\frac{5}{2}=2.5\).
For problem 42:
Since \(FM=FG - MG\) and \(FM = MG\) (mid - point), then \(2MG=FG\).
Set \(2(7x - 15)=33\).
Distribute: \(14x-30 = 33\).
Add 30 to both sides: \(14x-30 + 30=33 + 30\), so \(14x=63\).
Divide both sides by 14: \(x=\frac{63}{14}=\frac{9}{2}=4.5\).
For problem 43:
Set \(FM = MG\), so \(3n + 1=6 - 2n\).
Add \(2n\) to both sides: \(3n+2n + 1=6-2n + 2n\), which gives \(5n+1 = 6\).
Subtract 1 from both sides: \(5n+1 - 1=6 - 1\), so \(5n = 5\).
Divide both sides by 5: \(n = 1\).
Then \(FM=3(1)+1 = 4\), \(MG = 6-2(1)=4\), and \(FG=FM + MG=8\).
For problem 44:
Set \(FM = MG\), so \(12x - 4=5x + 10\).
Subtract \(5x\) from both sides: \(12x-5x - 4=5x-5x + 10\), which gives \(7x-4 = 10\).
Add 4 to both sides: \(7x-4 + 4=10 + 4\), so \(7x=14\).
Divide both sides by 7: \(x = 2\).
Then \(FM=12(2)-4 = 20\), \(MG = 5(2)+10 = 20\), and \(FG=FM + MG=40\).
For problem 45:
Since \(FM=\frac{1}{2}FG\) (mid - point), then \(2k - 5=\frac{18}{2}=9\).
Add 5 to both sides: \(2k-5 + 5=9 + 5\), so \(2k=14\).
Divide both sides by 2: \(k = 7\).
For problem 46:
Since \(FM=\frac{1}{2}FG\) (mid - point), then \(14.5=\frac{14a + 1}{2}\).
Multiply both sides by 2: \(14.5\times2=14a + 1\), so \(29 = 14a+1\).
Subtract 1 from both sides: \(29 - 1=14a+1 - 1\), so \(28 = 14a\).
Divide both sides by 14: \(a = 2\).
For problem 47:
Since \(FM=FG - MG\) and \(FM = MG\) (mid - point), then \(2MG=FG\).
Set \(2(13x + 1)=15\).
Distribute: \(26x+2 = 15\).
Subtract 2 from both sides: \(26x+2 - 2=15 - 2\), so \(26x=13\).
Divide both sides by 26: \(x=\frac{13}{26}=0.5\).
For problem 48:
Since \(FM=FG - MG\) and \(FM = MG\) (mid - point), then \(2MG=FG\).
Set \(2\times10.9=11x - 15.6\).
So \(21.8=11x - 15.6\).
Add 15.6 to both sides: \(21.8+15.6=11x-15.6 + 15.6\), so \(37.4=11x\).
Divide both sides by 11: \(x = 3.4\).
For problem 49:
The mid - point formula is \(P(\frac{x_N+x_Q}{2},\frac{y_N + y_Q}{2})\).
Let \(N(2,0)\) and \(P(5,2)\).
We know \(\frac{x_N+x_Q}{2}=5\) and \(\frac{y_N + y_Q}{2}=2\).
For the \(x\) - coordinate: \(\frac{2+x_Q}{2}=5\), multiply both sides by 2: \(2+x_Q = 10\), so \(x_Q=8\).
For the \(y\) - coordinate: \(\frac{0 + y_Q}{2}=2\), multiply both sides by 2: \(y_Q = 4\). So \(Q(8,4)\).
For problem 50:
Let \(N(5,4)\) and \(P(6,3)\).
Using the mid - point formula \(\frac{x_N+x_Q}{2}=6\) and \(\frac{y_N + y_Q}{2}=3\).
For the \(x\) - coordinate: \(\frac{5+x_Q}{2}=6\), multiply both sides by 2: \(5+x_Q = 12\), so \(x_Q=7\).
For the \(y\) - coordinate: \(\frac{4 + y_Q}{2}=3\), multiply both sides by 2: \(4 + y_Q=6\), so \(y_Q = 2\). So \(Q(7,2)\).
For problem 51:
Let \(Q(3,9)\) and \(P(-1,5)\).
Using the mid - point formula \(\frac{x_N+x_Q}{2}=-1\) and \(\frac{y_N + y_Q}{2}=5\).
For the \(x\) - coordinate: \(\frac{x_N+3}{2}=-1\), multiply both sides by 2: \(x_N+3=-2\), so \(x_N=-5\).
For the \(y\) - coordinate: \(\frac{y_N + 9}{2}=5\), multiply both sides by 2: \(y_N + 9 = 10\), so \(y_N = 1\). So \(N(-5,1)\).

1. \(FG = 16\)
2. \(FG = 58\)
3. \(a = 2.5\)
4. \(x = 4.5\)
5. \(FG = 8\)
6. \(FG = 40\)
7. \(k = 7\)
8. \(a = 2\)
9. \(x = 0.5\)
10. \(x = 3.4\)
11. \(Q(8,4)\)
12. \(Q(7,2)\)
13. \(N(-5,1)\)