## Calculus 8th Edition

$\dfrac{13}{5}$
Scalar projection distance formula: $\dfrac{|ax_1+by_1+c|}{\sqrt {a^2+b^2}}$ Distance from the point $(-2,3)$ to the line $3x-4y+5=0$ Thus, $\dfrac{|ax_1+by_1+c|}{\sqrt {a^2+b^2}}=\dfrac{|3(-2)+-4(3)+5|}{\sqrt {3^2+(-4)^2}}$ $=\dfrac{|-6-12+5|}{\sqrt {9+16}}$ $=\dfrac{13}{5}$