Question

Find the distance of the point (3,-5) from the line 3x-4y-26=0

• A. 3/5
• B. 5/3
• C. 3
• D. 5

Answer 1

Answer:

A. 3/5

3x-4y-26=0

$(x_1,y_1)=(3,-5)$

$Ax+By+C=0$

$A=3,B=-4,C=-26$

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

$d=\frac{|3\times3+-4\times-5-26|}{\sqrt{3^2+(-4)^2}}$

$=\frac{|9+20-26|}{5}=\frac{3}{5}$