Question

find the perpendicular distance between the (3, 4) and the lines 3x + 4y =5​

Answer 1

Answer:

Formula |ax1+by1+c|/root(a^2+b^2). By applying it you get the answer as 20/5=4units

Answer 2

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

$\sf\:P=|\dfrac{3(3)+4(4)-5}{\sqrt{3^2+4^2}}|$

$\sf\implies\:P=|\dfrac{9+16-5}{\sqrt{9+16}}|$

$\sf\implies\:P=|\dfrac{9+16-5}{\sqrt{9+16}}|$

$\sf\implies\:P=|\dfrac{25-5}{\sqrt{25}}|$

$\sf\implies\:P=|\dfrac{20}{5}|$

$\sf\implies\:P=4$