Question

Derive the formula to find the distance of the point (x1,y1)from the line ax+by+c=0

Answer 1

Consider a point $(x_{1},y_{1})$ in two dimensional coordinate plane. As we have to find distance of point  $(x_{1},y_{1})$ from the line ,ax+by+c=0.

The shortest distance between line and point is Perpendicular distance.

Draw perpendicular from point  $(x_{1},y_{1})$ on the line ,ax+by+c=0.

Equation of any line perpendicular to , ax + by +c=0 is bx - a y +k=0

it passes through  $(x_{1},y_{1})$.

$b x_{1}- a y_{1}+k=0\\\\ k=-b x_{ 1}+a y_{1}$

So,equation of  line will be , b x- a y + $-b x_{ 1}+a y_{1}$=0---(1)

and , a x + by + c=0----(2)

We will find the point of intersection of two lines.

(1) × a - (2) × b gives

→ -y (a²+b²)=  b c +  $ab x_{ 1}-a²y_{1}$

$y=\frac{ b c + ab x_{ 1}-a^2y_{1}}{ -(a^2+b^2)}$

(1) × b + (2) × a gives

x(a²+b²)=-ac+$b^2 x_{ 1}-ab y_{1}$

x= $\frac{-ac+b^2 x_{ 1}-ab y_{1}}{a^2+b^2}$

Distance between two points  $(x_{1},y_{1})$ and [x= $\frac{-ac+b^2 x_{ 1}-ab y_{1}}{a^2+b^2}$,$y=\frac{ b c + ab x_{ 1}-a^2y_{1}}{ -(a^2+b^2)}$]  is given by distance formula:\frac{ b c + ab x_{ 1}-a^2y_{1}}{

which is =$\sqrt{(A-P)^2+(B-Q)^2}$

between two points (A,B) and (P,Q).

= $\sqrt{(x_{1}- \frac{-ac+b^2 x_{ 1}-ab y_{1}}{a^2+b^2})^2+(y_{1}+\frac{ b c + ab x_{ 1}-a^2y_{1}}{ (a^2+b^2)})^2}\\\\  \sqrt{\frac{a^2(ax_{1}+by_{1}+c)^2+b^2(ax_{1}+by_{1}+c)^2}{(a^2+b^2)^2}}=\frac{(ax_{1}+by_{1}+c)}{\sqrt{a^2+b^2}}$

Answer 2

Answer:

Consider a point (x_{1},y_{1})(x

1

,y

1

) in two dimensional coordinate plane. As we have to find distance of point (x_{1},y_{1})(x

1

,y

1

) from the line ,ax+by+c=0.

The shortest distance between line and point is Perpendicular distance.

Draw perpendicular from point (x_{1},y_{1})(x

1

,y

1

) on the line ,ax+by+c=0.

Equation of any line perpendicular to , ax + by +c=0 is bx - a y +k=0

it passes through (x_{1},y_{1})(x

1

,y

1

) .

\begin{gathered}b x_{1}- a y_{1}+k=0\\\\ k=-b x_{ 1}+a y_{1}\end{gathered}

bx

1

−ay

1

+k=0

k=−bx

1

+ay

1

So,equation of line will be , b x- a y + -b x_{ 1}+a y_{1}−bx

1

+ay

1

=0---(1)

and , a x + by + c=0----(2)

We will find the point of intersection of two lines.

(1) × a - (2) × b gives

→ -y (a²+b²)= b c + ab x_{ 1}-a²y_{1}abx

1

−a²y

1

y=\frac{ b c + ab x_{ 1}-a^2y_{1}}{ -(a^2+b^2)}y=

−(a

2

+b

2

)

bc+abx

1

−a

2

y

1

(1) × b + (2) × a gives

x(a²+b²)=-ac+b^2 x_{ 1}-ab y_{1}b

2

x

1

−aby

1

x= \frac{-ac+b^2 x_{ 1}-ab y_{1}}{a^2+b^2}

a

2

+b

2

−ac+b

2

x

1

−aby

1

Distance between two points (x_{1},y_{1})(x

1

,y

1

) and [x= \frac{-ac+b^2 x_{ 1}-ab y_{1}}{a^2+b^2}

a

2

+b

2

−ac+b

2

x

1

−aby

1

,y=\frac{ b c + ab x_{ 1}-a^2y_{1}}{ -(a^2+b^2)}y=

−(a

2

+b

2

)

bc+abx

1

−a

2

y

1

is given by distance formula:\frac{ b c + ab x_{ 1}-a^2y_{1}}{

which is =\sqrt{(A-P)^2+(B-Q)^2}

(A−P)

2

+(B−Q)

2

between two points (A,B) and (P,Q).

= \begin{gathered}\sqrt{(x_{1}- \frac{-ac+b^2 x_{ 1}-ab y_{1}}{a^2+b^2})^2+(y_{1}+\frac{ b c + ab x_{ 1}-a^2y_{1}}{ (a^2+b^2)})^2}\\\\ \sqrt{\frac{a^2(ax_{1}+by_{1}+c)^2+b^2(ax_{1}+by_{1}+c)^2}{(a^2+b^2)^2}}=\frac{(ax_{1}+by_{1}+c)}{\sqrt{a^2+b^2}}\end{gathered}

(x

1

−

a

2

+b

2

−ac+b

2

x

1

−aby

1

)

2

+(y

1

+

(a

2

+b

2

)

bc+abx

1

−a

2

y

1

)

2

(a

2

+b

2

)

2

a

2

(ax

1

+by

1

+c)

2

+b

2

(ax

1

+by

1

+c)

2

=

a

2

+b

2

(ax

1

+by

1

+c)