Question

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


points​

Answer 1

$\huge\underline\blue{SolutioN:-}$

• The perpendicular distance between the point (3,4) and the line 3x+4y=5 is 4 units

$\underline\purple{Theory:}$

• The length perpendicular from a point $\sf\:(x_1,y_1)$

to a line ax+by+c=0 is

$\rm=|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$

$\underline{\pink{Step \: by \: step \: explanation:}}$

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

$\green{Then ,}$

$\underline{\red{Perpendicular \: distance}}$

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

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

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

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

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

$\sf\implies\:P=4$

Answer 2

\huge\underline\blue{SolutioN:-}

SolutioN:−

• The perpendicular distance between the point (3,4) and the line 3x+4y=5 is 4 units

\underline\purple{Theory:}

Theory:

• The length perpendicular from a point \sf\:(x_1,y_1)(x

1

,y

1

)

to a line ax+by+c=0 is

\rm=|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}=∣

a

2

+b

2

ax

1

+by

1

+c

\underline{\pink{Step \: by \: step \: explanation:}}

Stepbystepexplanation:

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

\green{Then ,}Then,

\underline{\red{Perpendicular \: distance}}

Perpendiculardistance

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

3

2

+4

2

3(3)+4(4)−5

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

9+16

9+16−5

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

9+16

9+16−5

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

25

25−5

\sf\implies\:P=|\dfrac{20}{5}⟹P=∣

5

20

\sf\implies\:P=4⟹P=4