Q.Find the distance of the point $(2, 5)$ from the line $3x - 4y + 8 = 0$. Also find the distance between the parallel lines $3x - 4y + 8 = 0$ and $3x - 4y - 7 = 0$.

[!TLDR] Apply $d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ for the point, and $d=\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}$ for the two parallel lines. [!ANSWER] Distance o …

Distance of $(2,5)$ from $3x-4y+8=0$ (here $A=3,B=-4,C=8$): $d=\dfrac{|3(2)-4(5)+8|}{\sqrt{3^2+(-4)^2}}=\dfrac{|6-20+8|}{\sqrt{25}}=\dfrac{|-6|}{5}=\dfrac{6}{5}$. Distance between the parallel lines $3x-4y+8=0$ ($C_1=8$) and $3x-4y-7=0$ ($C_2=-7$), which share the same $A=3,B=-4$: $d=\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}=\dfra …

For the point, substitute directly into the point-to-line distance formula. For the two lines, confirm they share the same $A,B$ (so they are genui …

Forgetting the absolute value and reporting a negative distance, or using $\sqrt{A^2+B^2}$ with the wrong $A,B$ pulled from the second line instead of confirming …