Find the distance from the point [tex](6, 4)[/tex] to the line [tex]y=x+4[/tex].

Answer:[tex]\large \boxed{3\sqrt{2}}[/tex]Step-by-step explanation:1. Express the line in standard form. ax + by + c = 0 [tex]\begin{array}{rcl}y & = & x + 4\\-x + y - 4 & = & 0\\\end{array}[/tex]2. Calculate the distance The formula for the distance d from a point (x, y) and the line is: [tex]d = \dfrac{|ax + by + c|}{\sqrt{a^{2} + b^{2}}}[/tex]Insert the values: a = -1; b = 1; c = -4; x = 6; y = 4 [tex]\begin{array}{rcl}d &= &\dfrac{|(-1)(6) + 1\times4 + (-4)|}{\sqrt{(-1)^{2} + 1^{2}}}\\\\& = & \dfrac{|-6 + 4 - 4|}{\sqrt{1 + 1}}\\\\& = & \dfrac{|-6|}{\sqrt{2}}\\\\& = &\dfrac{6}{\sqrt{2}}\\\\& = & \dfrac{3\times\sqrt{2}\times\sqrt{2}}{\sqrt{2}}\\\\& = & \mathbf{3{\sqrt{2}}}\end{array}\\\text{The distance from the point to the line is $\large \boxed{\mathbf{3\sqrt{2}}}$}[/tex]