Question

Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

Answer 1

A

Explanation:

Answer 2

`D = (25√(13))/(13)`

Explanation:

Given

`y = 4 - (2)/(3)x`

`(x_1,y_1) = (-2,-3)`

Required

Determine the distance

`y = 4 - (2)/(3)x`

Write the above equation in standard form:

`Ax + By + C = 0`

So, we have:

`(2)/(3)x+y - 4 = 0`

By comparison:

`A = (2)/(3)`
`B = 1` and
`C = -4`

The distance is calculated using:

`D = (|Ax_1 + By_1 + C|)/(√(A^2 + B^2))`

Where:

`(x_1,y_1) = (-2,-3)`

`A = (2)/(3)`
`B = 1` and
`C = -4`

This gives:

`D = \frac{\sqrt{((2)/(3))^2 + 1^2}}`

`D = \frac{\sqrt{(4)/(9) + 1}}`

Take LCM

`D = \frac(-4-9-12)/(3){\sqrt{(4+9)/(9)}}`

`D = \frac{\sqrt{(13)/(9)}}`

`D = |(-25)/(3)|/\sqrt{(13)/(9)}`

`D = (25)/(3)/\sqrt{(13)/(9)}`

Split the square root

`D = (25)/(3)/(√(13))/(√(9))`

Change / to *

`D = (25)/(3)*(√(9))/(√(13))`

`D = (25)/(3)*(3)/(√(13))`

`D = (25)/(√(13))`

Rationalize

`D = (25)/(√(13)) * (√(13))/(√(13))`

`D = (25√(13))/(13)`

Hence, the distance is:

`D = (25√(13))/(13)`