Answer:
The distance of point ( 6 , 2 ) from line 6 x - y = 3 is [tex]\frac{31}{\sqrt{37} }[/tex] 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 = [tex]\frac{\begin{vmatrix}ax & +b y & + c\end{vmatrix}}{\sqrt{a^{2}+b^{2}}}[/tex]
Or, d = [tex]\frac{\begin{vmatrix}6\times x & +(-1)\times y & + (-3)\end{vmatrix}}{\sqrt{6^{2}+(-1)^{2}}}[/tex]
Or, d = [tex]\frac{\begin{vmatrix}6\times 6 & +(-1)\times 2 & + (-3)\end{vmatrix}}{\sqrt{6^{2}+(-1)^{2}}}[/tex]
Or, d = [tex]\frac{36 - 2 - 3}{\sqrt{37} }[/tex]
∴ d = [tex]\frac{31}{\sqrt{37} }[/tex] unit
Hence The distance of point ( 6 , 2 ) from line 6 x - y = 3 is [tex]\frac{31}{\sqrt{37} }[/tex] unit . Answer
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:
