Answer:
50
Step-by-step explanation:
You have to apply the distance formula:
(x,y)
(-3,1)
(4,2)
√(x2-x1)²+(y2-y1)² = √(4+3)²+(2-1)² = √7²+1² = √49 + 1 = 50
Answer:
The distance between the point (-3,1) and (4,2) is 7.1 units
Solution:
The distance between two points [tex]\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)[/tex] is given as
[tex]\text { Distance } = \sqrt{\left(x_{2}-x_{1}\right)^{2} + \left(y_{2}-y_{1}\right)^{2}}[/tex] ---- eqn 1
Where [tex]x_{1} y_{1} \text { and } x_{2} y_{2}[/tex] are the coordinates of the given points
From question, given that the points are (-3, 1) and (4, 2)
Hence we get [tex]x_{1} = -3 ; y_{1} = 1 ;x_{2} = 4 ; y_{2} = 2[/tex]
By using eqn 1, we get the distance between the two points as,
[tex]\text { Distance } = \sqrt{(4-(-3))^{2} + (2-1)^{2}}[/tex]
[tex] = \sqrt{(4+3)^{2}+1}[/tex]
[tex] = \sqrt{7^{2}+1} = \sqrt{49+1} = \sqrt{50} = 7.071[/tex]
Rounding to the nearest tenth, we get the answer as 7.1
Hence the distance between the point (-3,1) and (4,2) when rounded to nearest tenth is 7.1
