Question

What is the distance between the point (2, 7) and the
point (‒1, 3) in the xy-coordinate plane?
F. √17
G. 4√2
H. 5
J. 5√2
K. 6

Answer 1

We know that :-

Distance between the two points (a,b) and (c,d) :-

$\underline{ \boxed{ \large \rm \: \sqrt{ {(c - a)}^{2} + {(d - b)}^{2} } }}$

Now , we have to find out distance between points (2, 7) and (‒1, 3).

$\rm \longrightarrow \large \sqrt{ {( - 1 - 2)}^{2} + {( 3 - 7)}^{2} } \\ \\ \\ \rm \longrightarrow \large \sqrt{ {( -3)}^{2} + {( - 4)}^{2} } \\ \\ \\ \rm \longrightarrow \large \sqrt{ 9+ 16 } \\ \\ \\ \rm \longrightarrow \large \sqrt{ 25 } \\ \\ \\ \rm \longrightarrow \large \sqrt{ 5 \times 5}\\ \\ \\ \rm \longrightarrow \large 5 \: units$

Therefore, distance between points (2, 7) and (‒1, 3) = 5 units

Answer :

Option H. 5 is correct

Answer 2

Answer:

Distance between two points (x₁ , y₁) and (x₂ , y₂) can be calculated by :

$\bf\dag\:\:\large\underline{\boxed{\sf\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}}$

⠀⠀⠀⌬ (x₁ , y₁) = (2 , 7)

⠀⠀⠀⌬ (x₂ , y₂) = (-1 , 3)

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\Longrightarrow\sf Distance= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\\\\\:\Longrightarrow\sf Distance=\sqrt{(-1-2)^2 + (3-7)^2}\\\\\\:\Longrightarrow\sf Distance=\sqrt{(-3)^2 + (-4)^2}\\\\\\:\Longrightarrow\sf Distance=\sqrt{9 + 16}\\\\\\:\Longrightarrow\sf Distance=\sqrt{25}\\\\\\:\Longrightarrow\sf Distance = 5$

⠀

$\therefore\:\underline{\textsf{Distance between following points is h) \textbf{5}}}.$