Question

What is the distance between point (-3, 1) and point (4, 2) rounded to the nearest tenth? 8 units

Answer 1

The distance between the point (-3,1) and (4,2) is 7.1 units

Solution:

The distance between two points
`\left(x_(1), y_(1)\right),\left(x_(2), y_(2)\right)` is given as

`\text { Distance } = \sqrt{\left(x_(2)-x_(1)\right)^(2) + \left(y_(2)-y_(1)\right)^(2)}` ---- eqn 1

Where
`x_(1) y_(1) \text { and } x_(2) y_(2)` are the coordinates of the given points

From question, given that the points are (-3, 1) and (4, 2)

Hence we get
`x_(1) = -3 ; y_(1) = 1 ;x_(2) = 4 ; y_(2) = 2`

By using eqn 1, we get the distance between the two points as,

`\text { Distance } = \sqrt{(4-(-3))^(2) + (2-1)^(2)}`

`= \sqrt{(4+3)^(2)+1}`

`= \sqrt{7^(2)+1} = √(49+1) = √(50) = 7.071`

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

Answer 2

50

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