Question

t%20%5Clarge%20Distance%20%20%5C%3A%20Formula%20%5C%3A%20%20%20%5C%5C%20%20%5Clarge%20%5Ctt%20and%20%5C%5C%20%5Clarge%5Ctt%20%20%5C%3A%20Coordinate%20%20%5C%3A%20Proof%20%5Cend%7Barray%7D%7D%7D" id="TexFormula1" title=" \boxed{\boxed{ \begin{array}{}\large\tt The \\ \tt \large Distance \: Formula \: \\ \large \tt and \\ \large\tt \: Coordinate \: Proof \end{array}}}" alt=" \boxed{\boxed{ \begin{array}{}\large\tt The \\ \tt \large Distance \: Formula \: \\ \large \tt and \\ \large\tt \: Coordinate \: Proof \end{array}}}" align="absmiddle" class="latex-formula">
Direction: Find the distance between each pair of points on the coordinate plane and write the solution.


1. A(3, 8) and B(3, -2)

2. O(5, -2) and K(7, -2)

3. L(4, -4) and M(-4, 6)

4. P(5, 1) and C(2, 3)

5. Q(-10, 9) and R(-2, 3)

$\boxed{\boxed{ \begin{array}{}\large\tt Nonsense + Useless \\ \large \tt= Report! \end{array}}}$
​

Answer 1

Distance formula

$\boxed{\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Now

#1

$\\ \tt\Rrightarrow AB=\sqrt{(3-3)^2+(-2-8)^2}=\sqrt{100}=10units$

#2

$\\ \tt\Rrightarrow OK=\sqrt{(5-7)^2+(-2+2)^2}=\sqrt{4}=2units$

#3

$\\ \tt\Rrightarrow LM=\sqrt{(-4-4)^2+(6+4)^2}=\sqrt{64+100}=\sqrt{164}\approx 13units$

#4

$\\ \tt\Rrightarrow PC=\sqrt{(5-2)^2+(1-3)^2}=\sqrt{9+4}=\sqrt{13}=3.2units$

#5

$\\ \tt\Rrightarrow QR=\sqrt{(-2+10)^2+(3-9)^2}=\sqrt{64+36}=\sqrt{100}=10units$

Answer 2

Answer:

Answer of all given questions are given below :

• 1) AB = 10 units
• 2) OK = 2 units
• 3) LM = 2√41 units
• 4) PC ≈ 3.6 units
• 5) QR = 10 units

Step-by-step explanation:

Here's the required formula to find distance between points :

$\star{\small{\underline{\boxed{\sf{\red{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}}}$

According to this formula, we'll solve all the given questions and find the distance between points.

1. A(3, 8) and B(3, -2)

Substituting all the given values in the formula to find the distance between points:

${\implies{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\implies{\small{\sf{AB = \sqrt{\Big(3 - 3 \Big)^{2} + \Big( - 2 - 8 \Big)^{2}}}}}}$

${\implies{\small{\sf{AB = \sqrt{\Big(0\Big)^{2} + \Big( - 10 \Big)^{2}}}}}}$

${\implies{\small{\sf{AB = \sqrt{\Big(0 \times 0\Big) + \Big( - 10 \times - 10 \Big)}}}}}$

${\implies{\small{\sf{AB = \sqrt{\big(0 + 100\big)}}}}}$

${\implies{\small{\sf{AB = \sqrt{100}}}}}$

${\implies{\sf{\underline{\underline{\purple{AB = 10}}}}}}$

Hence, the distance between points AB is 10 units..

$\begin{gathered}\end{gathered}$

2. O(5, -2) and K(7, -2)

Substituting all the given values in the formula to find the distance between points:

${\longrightarrow{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\Big(7 - 5\Big)^{2} + \Big( - 2 + 2 \Big)^{2}}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\Big(2\Big)^{2} + \Big(0 \Big)^{2}}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\Big(2 \times 2\Big) + \Big(0 \times 0 \Big)}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\big(4 + 0 \big)}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{4}}}}}$

${\longrightarrow{\sf{\underline{\underline{\pink{OK = 2}}}}}}$

Hence, the distance between points OK is 2 units.

$\begin{gathered}\end{gathered}$

3. L(4, -4) and M(-4, 6)

Substituting all the given values in the formula to find the distance between points:

${\longmapsto{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\Big( - 4 - 4\Big)^{2} + \Big( 6 + 4 \Big)^{2}}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\Big( - 8\Big)^{2} + \Big(10 \Big)^{2}}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\Big( - 8 \times - 8\Big) + \Big(10 \times 10\Big)}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\big( 64 + 100\big)}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{164}}}}}$

${\longmapsto{\sf{\underline{\underline{\orange{LM = 2 \sqrt{41} }}}}}}$

Hence, the distance between points LM is 2√41 units.

$\begin{gathered}\end{gathered}$

4. P(5, 1) and C(2, 3)

Substituting all the given values in the formula to find the distance between points:

${\dashrightarrow{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\Big(2 - 5 \Big)^{2} + \Big(3 - 1 \Big)^{2}}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\Big( - 3\Big)^{2} + \Big(2\Big)^{2}}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\Big( - 3 \times - 3\Big) + \Big(2 \times 2\Big)}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\big( 9 + 4\big)}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{13}}}}}$

${\dashrightarrow{\sf{\underline{\underline{\green{PC \approx 3.6}}}}}}$

Hence, the distance between points PC is 3.6 units.

$\begin{gathered}\end{gathered}$

5. Q(-10, 9) and R(-2, 3)

Substituting all the given values in the formula to find the distance between points:

${\twoheadrightarrow{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\Big( - 2 + 10 \Big)^{2} + \Big(3 - 9 \Big)^{2}}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\Big( 8 \Big)^{2} + \Big( - 6\Big)^{2}}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\Big( 8 \times 8 \Big)+ \Big( - 6 \times - 6\Big)}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\big(64 + 36\big)}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{100}}}}}$

${\twoheadrightarrow{\sf{\underline{\underline{\blue{QR = 10}}}}}}$

Hence, the distance between points QR is 10 units.

$\rule{300}{2.5}$