Which is the distance between the point with the coordinates (-2, 3) and the line with the equation 6x-y=-3

Answer:[tex]\frac{12}{\sqrt{37}}[/tex] units. Step-by-step explanation:The perpendicular distance from a point ([tex]x_{1},y_{1}[/tex]) from a given straight line ax + by + c = 0 is given by the formula  [tex]\frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2} + b^{2}}}[/tex]. So, the distance from the point with coordinates (-2,3) to the straight line with equation 6x - y = - 3 i.e. 6x - y + 3 = 0 will be  [tex]\frac{|6(-2) - 1(3) + 3|}{\sqrt{6^{2} + (- 1)^{2}}} = \frac{12}{\sqrt{37} }[/tex] units. (Answer)