Find the perpendicular distance between the (3, 4) and the lines 3x +

1.	Find the perpendicular distance between the (3, 4) and the lines 3x + 4y =5
Answer» Solution : The PERPENDICULAR distance between the POINT (3,4) and the line 3x+4y=5 is 4 units Theory : The length perpendicular from a point $\sf\:(x_1,y_1)$ to a line ax+by+c=0 is $\rm=\|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\|$ Step by step EXPLANATION: We have to Find the perpendicular distance between the point (3,4) and the line 3x+4y=5 Let the perpendicular distance be P , and point (3,4) and line 3x+4y-5=0 Then , Perpendicular distance $\sf\:P=\|\dfrac{3(3)+4(4)-5}{\sqrt{3^2+4^2}}\|$ $\sf\implies\:P=\|\dfrac{9+16-5}{\sqrt{9+16}}\|$ $\sf\implies\:P=\|\dfrac{9+16-5}{\sqrt{9+16}}\|$ $\sf\implies\:P=\|\dfrac{25-5}{\sqrt{25}}\|$ $\sf\implies\:P=\|\dfrac{20}{5}\|$ $\sf\implies\:P=4$