Question

1. find the coordinates of the mid - point of the segment with the given endpoints. s(4, - 1) and t(6,0)
2. l(4,2) and p(0,2)
3. h(-5,5) and i(7,3)
4. g(-2,-8) and h(-3,-12)

the endpoints of two segments are given. find each segment length. tell whether the segments are congruent.

1. ab: a(2,6), b(0,3) cd: c(-1,0), d(1,3)
2. rs: r(5,4), s(0,4) tu: t(-4,-3), u(-1,1)
3. kl: k(-4,13), l(-10,6) mn: m(-1,-2), n(-1,-11)
4. op: o(6,-2), p(3,-2) qr: q(5,2), r(1,9)
5. find the length of the hypotenuse of the right triangle.

10.
11.

Explanation:

1. For finding the mid - point of a segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$:

The mid - point formula is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.

• Question 1: S(4, - 1) and T(6,0)
• Step 1: Calculate the x - coordinate of the mid - point
• Use the formula for the x - coordinate of the mid - point $x_m=\frac{x_1 + x_2}{2}$. Here, $x_1 = 4$, $x_2=6$. So, $x_m=\frac{4 + 6}{2}=\frac{10}{2}=5$.
• Step 2: Calculate the y - coordinate of the mid - point
• Use the formula for the y - coordinate of the mid - point $y_m=\frac{y_1 + y_2}{2}$. Here, $y_1=-1$, $y_2 = 0$. So, $y_m=\frac{-1+0}{2}=-\frac{1}{2}$.
• The mid - point is $(5,-\frac{1}{2})$.
• Question 2: L(4,2) and P(0,2)
• Step 1: Calculate the x - coordinate of the mid - point
• $x_m=\frac{4 + 0}{2}=\frac{4}{2}=2$.
• Step 2: Calculate the y - coordinate of the mid - point
• $y_m=\frac{2 + 2}{2}=\frac{4}{2}=2$.
• The mid - point is $(2,2)$.
• Question 3: H(-5,5) and I(7,3)
• Step 1: Calculate the x - coordinate of the mid - point
• $x_m=\frac{-5 + 7}{2}=\frac{2}{2}=1$.
• Step 2: Calculate the y - coordinate of the mid - point
• $y_m=\frac{5+3}{2}=\frac{8}{2}=4$.
• The mid - point is $(1,4)$.
• Question 4: G(-2,-8) and H(-3,-12)
• Step 1: Calculate the x - coordinate of the mid - point
• $x_m=\frac{-2+( - 3)}{2}=\frac{-2-3}{2}=-\frac{5}{2}$.
• Step 2: Calculate the y - coordinate of the mid - point
• $y_m=\frac{-8+( - 12)}{2}=\frac{-8 - 12}{2}=\frac{-20}{2}=-10$.
• The mid - point is $(-\frac{5}{2},-10)$.

2. For finding the length of a line segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$:

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

• Question 5: $\overline{AB}$ with $A(2,6)$ and $B(0,3)$
• Step 1: Substitute the values into the distance formula
• $d=\sqrt{(0 - 2)^2+(3 - 6)^2}=\sqrt{(-2)^2+( - 3)^2}=\sqrt{4 + 9}=\sqrt{13}$.
• Question 6: $\overline{RS}$ with $R(5,4)$ and $S(0,4)$
• Step 1: Substitute the values into the distance formula
• $d=\sqrt{(0 - 5)^2+(4 - 4)^2}=\sqrt{(-5)^2+0^2}=\sqrt{25}=5$.
• Question 7: $\overline{KL}$ with $K(-4,13)$ and $L(-10,6)$
• Step 1: Substitute the values into the distance formula
• $d=\sqrt{(-10+4)^2+(6 - 13)^2}=\sqrt{(-6)^2+( - 7)^2}=\sqrt{36 + 49}=\sqrt{85}$.
• Question 8: $\overline{OP}$ with $O(6,-2)$ and $P(3,-2)$
• Step 1: Substitute the values into the distance formula
• $d=\sqrt{(3 - 6)^2+(-2+2)^2}=\sqrt{(-3)^2+0^2}=\sqrt{9}=3$.
• Question 9: Right - triangle with legs 5 and 12
• Step 1: Use the Pythagorean theorem $c=\sqrt{a^2 + b^2}$
• Here, $a = 5$, $b = 12$. So, $c=\sqrt{5^2+12^2}=\sqrt{25 + 144}=\sqrt{169}=13$.
• Question 10: Right - triangle with legs 33 and 56
• Step 1: Use the Pythagorean theorem $c=\sqrt{a^2 + b^2}$
• Here, $a = 33$, $b = 56$. So, $c=\sqrt{33^2+56^2}=\sqrt{1089+3136}=\sqrt{4225}=65$.
• Question 11: Right - triangle with legs 40 and 42
• Step 1: Use the Pythagorean theorem $c=\sqrt{a^2 + b^2}$
• Here, $a = 40$, $b = 42$. So, $c=\sqrt{40^2+42^2}=\sqrt{1600 + 1764}=\sqrt{3364}=58$.

Answer:

1. $(5,-\frac{1}{2})$
2. $(2,2)$
3. $(1,4)$
4. $(-\frac{5}{2},-10)$
5. $\sqrt{13}$
6. $5$
7. $\sqrt{85}$
8. $3$
9. $13$
10. $65$
11. $58$