Question

find the distance of the point (4 1) from line 3x-4y-9=0​

Answer 1

Answer:

√2 unit

Step-by-step explanation:

3x-4y-9=0

3x=4y+9

x=(4y+9)/3

if y=0

then x=(4×0+9)/3

x=9/3

x=3

point will be (3,0)

let this point be A and (4,1) be B

AB=√(4-3)^2+(1-0)^2

AB=√1+1

AB=√2 unit

Answer 2

Given,

Equation of line, $\[3x - 4y - 9 = 0\]$

Point $\[ = \left( {4,1} \right)\]$

Solution,

Consider the equation of line is $\[ax + by + c = 0\]$ . The distance of this line from the point $\[\left( {{x_1},{y_1}} \right)\]$

$\[ = \frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}\]$

Calculate the distance.

$\[ = \frac{{3x - 4y - 9}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} }}\]$

$\[ = \frac{{3 \times 4 - 4 \times \left( { 1} \right) - 9}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} }}\]$

$\[ = \frac{{12 - 4 - 9}}{{\sqrt {9 + 16} }}\]$

$\[ = \frac{{ - 1}}{5}\]$

Distance is always positive.

$\[ = 0.2unit\]$

Hence the distance is $\[ 0.2unit\]$