Question

Derive the formula for the perpendicular distance of a point (x,y)from the line Ax+By+C=0

Answer 1

Answer:

Perpendicular distance

Step-by-step explanation:

Answer 2

Given:

point Q( x,y) and the line ax+by+c=0.

To find:

the perpendicular distance from the point Q( x,y) to the line ax+by+c=0.

PROOF:

• let d be the perpendicular distance from the point Q( x,y) to the line ax+by+c=0.

• let n be a vector normal to the line that starts from point P(x1, y1)

• the distance d is the orthogonal projection of the vector PQ.

              d=  ∥PQ∥ cos∅.

• Now, multiply both the numerator and the denominator of the right hand side of the equation by the magnitude of the normal vector n

             d=  ∥n ∥∥PQ∥cos∅/ ∥n ∥

• We know from the definition of dot product that∥PQ∥ ∥n∥cos∅ just means the dot product of the vector PQ and the normal vector n :

         d= PQ.n / ∥n∥

     PQ=(x  − x1  ,y − y1).

So

       PQ⋅n = (x − x1  ,y − y1 )⋅(a,b)

               =a(x − x1  ) + b(y − y1 ).

And we know that,

         ∥n∥ =$\sqrt{a^{2}+b^{2} }$

thus

          d=  ∣a(x −x1 )+ b(y −y1  )∣/$\sqrt{a^{2}+b^{2} }$

​           =∣a(x)−a(x1) + b(y )−b(y1)∣/$\sqrt{a^{2}+b^{2} }$

• From the equation of the line we have c = -a(x1) - b(y)

                 d=  ∣a(x )+ b(y)+c∣/ $\sqrt{a^{2}+b^{2} }$.