Question

find the equation of the line which passes through the point (-1,2) and whose distance from origin is one unit ​

Answer 1

$\textsf{\underline {\Large {Straight Lines}}}$ :

Given, Point = ( - 1, 2 )

x = - 1, y = 2

Equation of line,

$\boxed{\mathsf{(\:y\:-\:y_0 \:)\:=\:m(\:x\:-\:x_0\:)}}$

Here, m = Slope of the line

$\mathsf{(\:2\:-\:y_0 \:)\:=\:m(\:-1\:-\:x_0\:)}$

$\mathsf{(\:2\:-\:y_0 \:)\:=\:-m \:-mx_0}$

$\mathsf{(\:mx_0\:-\:y_0 \:)\:=\:-m \:-\:2}$

➡️ $\mathsf{\:mx_0\:-\:y_0\:+m \:+\:2\:=\:0}$ --> ( i )

Distance from origin, d = 1 unit

Coordinates of origin, $\mathsf{( \:x_1,\: y_1\:)}$ = ( 0, 0 )

Perpendicular distance formula,

$\boxed{\mathsf{d\:=\:{\dfrac{|Ax_{1} \:+\:By_{1} \:+\:C|} {\sqrt{{A} ^{2}\:+\:{B}^{2}}}}}}$

Here,

A = $\mathsf{(\:m\:)}$

B = $\mathsf{(\:-y\:)}$

C = $\mathsf{(\:m\:+\:2\:)}$

Putting these values,

1 = $\mathsf{d\:=\:{\dfrac{|(\:m\:)(0) \:+\:(\:-1\:)( 0 ) \:+\:(\:m\:+\:2\:)|} {\sqrt{{(\:m\:)} ^{2}\:+\:{(\:-1\:)}^{2}}}}}$

1 = $\mathsf{d\:=\:{\dfrac{(\:m\:+\:2\:)} {\sqrt{{m}^{2}\:+\:1}}}}}$

$\mathsf{\sqrt{{m} ^{2}\:+\:1} \:=\:m\:+\:2}$

$\mathsf{{m} ^{2}\:+\:1\:=\:{(m\:+\:2)}^{2}}$

$\mathsf{{m} ^{2}\:+\:1\:=\:{m} ^{2}\:+\:4\:+\:4m}$

$\mathsf{4m\:-\:3 \:= \:0}$

$\mathsf{4m\:= \:3}$

$\mathsf{m\:= \:{\dfrac{3}{4}}}$

Putting value of ' m' in equation ( i ),

Replacing $\mathsf{x_0\:by\:x\:and\:y_0\:by\:y}$

$\mathsf{\:mx_0\:-\:y_0\:+m \:+\:2\:=\:0}$

$\mathsf{\dfrac{3}{4}x\:-\:y\:+{\dfrac{3}{4}\:+\:2\:=\:0}}$

$\mathsf{\dfrac{3}{4}x\:-\:y\:{\dfrac{11}{4}}\:=\:0}$

➡️ Equation of line,

$\boxed{\mathsf{3x\:-\:4y\:+\:11\:=\:0}}$

Answer 2

Answer:

:

Given, Point = ( - 1, 2 )

x = - 1, y = 2

Equation of line,

Here, m = Slope of the line

➡️  --> ( i )

Distance from origin, d = 1 unit

Coordinates of origin,  = ( 0, 0 )

Perpendicular distance formula,

Here,

A =  

B =  

C =  

Putting these values,

1 =  

1 =  

Putting value of ' m' in equation ( i ),

Replacing  

➡️ Equation of line,

Step-by-step explanation: