Question

Find the exact distance between the line 6x-y=3 and the point (6,2). Show your work. Explain your answer.

Answer 1

Answer:

The distance of point ( 6 , 2 ) from line  6 x - y = 3 is  $\frac{31}{\sqrt{37} }$  unit .

Step-by-step explanation:

Given as :

The equation of line is 6 x - y = 3

And The points is ( 6 , 2 )

Let The distance between the line and points is d unit

So, The distance of point from the line = $\frac{\begin{vmatrix}ax & +b y & + c\end{vmatrix}}{\sqrt{a^{2}+b^{2}}}$

Or, d = $\frac{\begin{vmatrix}6\times x & +(-1)\times y & + (-3)\end{vmatrix}}{\sqrt{6^{2}+(-1)^{2}}}$

Or, d =  $\frac{\begin{vmatrix}6\times 6 & +(-1)\times 2 & + (-3)\end{vmatrix}}{\sqrt{6^{2}+(-1)^{2}}}$

Or, d = $\frac{36 - 2 - 3}{\sqrt{37} }$

∴    d = $\frac{31}{\sqrt{37} }$  unit

Hence The distance of point ( 6 , 2 ) from line  6 x - y = 3 is  $\frac{31}{\sqrt{37} }$  unit .   Answer

Answer 2

Answer:

Using the distance formula, you can solve to show that the distance between  6x-y= 3  = 37/ sqrt 37, or  sqrt 37

Step-by-step explanation: