Answer:
The distance between line 6 x - y = - 3 and point (6,2) is[tex]\sqrt{37}[/tex]
Step-by-step explanation:
Given equation of line as :
6 x - y = - 3
And The points be ( 6 , 2 )
Let The distance between line and points = d
So , from point to line distance formula
d = [tex]\frac{\begin{vmatrix}ax & + by & + c\end{vmatrix}}{\sqrt{a^{2}+b^{2}}}[/tex]
Or, d = [tex]\frac{\begin{vmatrix}6x & - y & + 3\end{vmatrix}}{\sqrt{6^{2}+(-1)^{2}}}[/tex]
∵ points is ( 6 , 2 )
so, d = [tex]\frac{\begin{vmatrix}6\times 6 & + (-1)\times 2 & + 2\end{vmatrix}}{\sqrt{6^{2}+(-1)^{2}}}[/tex]
or, d = [tex]\frac{36-2+3}{\sqrt{37} }[/tex]
∴ d = [tex]\frac{37}{\sqrt{37} }[/tex]
I.e d = [tex]\sqrt{37}[/tex]
Hence The distance between line 6 x - y = - 3 and point (6,2) is[tex]\sqrt{37}[/tex] Answer
