Question

t is the mid - point of su. find the missing point using the information below.

1. s(4,9) and t(2, - 10)
2. t(-4,2) and u(-5, - 8)
3. s(-9,9) and u(21, - 29)
4. u(2, - 7) and s(0, - 25)

Explanation:

1. Recall the mid - point formula:

• The mid - point formula between two points \(S(x_1,y_1)\) and \(U(x_2,y_2)\) is \(T(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).

1. For problem 28: Given \(S(4,9)\) and \(T(2,-10)\), find \(U(x,y)\)

• Step 1: Find the \(x\) - coordinate of \(U\)
• Using the \(x\) - coordinate part of the mid - point formula \(\frac{x_1 + x_2}{2}=x_m\), where \(x_1 = 4\), \(x_m = 2\).
• Substitute into the formula: \(\frac{4 + x}{2}=2\).
• Multiply both sides by 2: \(4 + x=4\).
• Subtract 4 from both sides: \(x = 0\).
• Step 2: Find the \(y\) - coordinate of \(U\)
• Using the \(y\) - coordinate part of the mid - point formula \(\frac{y_1 + y_2}{2}=y_m\), where \(y_1 = 9\), \(y_m=-10\).
• Substitute into the formula: \(\frac{9 + y}{2}=-10\).
• Multiply both sides by 2: \(9 + y=-20\).
• Subtract 9 from both sides: \(y=-29\).
• So \(U(0,-29)\).

1. For problem 29: Given \(T(-4,2)\) and \(U(-5,-8)\), find \(S(x,y)\)

• Step 1: Find the \(x\) - coordinate of \(S\)
• Using the \(x\) - coordinate part of the mid - point formula \(\frac{x + x_U}{2}=x_T\), where \(x_U=-5\), \(x_T=-4\).
• Substitute into the formula: \(\frac{x+( - 5)}{2}=-4\).
• Multiply both sides by 2: \(x - 5=-8\).
• Add 5 to both sides: \(x=-3\).
• Step 2: Find the \(y\) - coordinate of \(S\)
• Using the \(y\) - coordinate part of the mid - point formula \(\frac{y + y_U}{2}=y_T\), where \(y_U=-8\), \(y_T = 2\).
• Substitute into the formula: \(\frac{y+( - 8)}{2}=2\).
• Multiply both sides by 2: \(y - 8 = 4\).
• Add 8 to both sides: \(y = 12\).
• So \(S(-3,12)\).

1. For problem 30: Given \(S(-9,9)\) and \(U(21,-29)\), find \(T(x,y)\)

• Step 1: Find the \(x\) - coordinate of \(T\)
• Using the \(x\) - coordinate part of the mid - point formula \(x=\frac{x_S+x_U}{2}\), where \(x_S=-9\), \(x_U = 21\).
• Substitute into the formula: \(x=\frac{-9 + 21}{2}=\frac{12}{2}=6\).
• Step 2: Find the \(y\) - coordinate of \(T\)
• Using the \(y\) - coordinate part of the mid - point formula \(y=\frac{y_S+y_U}{2}\), where \(y_S = 9\), \(y_U=-29\).
• Substitute into the formula: \(y=\frac{9+( - 29)}{2}=\frac{9 - 29}{2}=\frac{-20}{2}=-10\).
• So \(T(6,-10)\).

1. For problem 31: Given \(U(2,-7)\) and \(S(0,-25)\), find \(T(x,y)\)

• Step 1: Find the \(x\) - coordinate of \(T\)
• Using the \(x\) - coordinate part of the mid - point formula \(x=\frac{x_S+x_U}{2}\), where \(x_S = 0\), \(x_U = 2\).
• Substitute into the formula: \(x=\frac{0 + 2}{2}=1\).
• Step 2: Find the \(y\) - coordinate of \(T\)
• Using the \(y\) - coordinate part of the mid - point formula \(y=\frac{y_S+y_U}{2}\), where \(y_S=-25\), \(y_U=-7\).
• Substitute into the formula: \(y=\frac{-25+( - 7)}{2}=\frac{-25 - 7}{2}=\frac{-32}{2}=-16\).
• So \(T(1,-16)\).

Answer:

1. \(U(0,-29)\)
2. \(S(-3,12)\)
3. \(T(6,-10)\)
4. \(T(1,-16)\)