Question

State and prove foot of perpendicular theorem

Answer 1

Answer:

Let ax+by+c=0 be a straight line. If a perpendicular line is drawn from any point on the plane to this straight line then the point of intersection of the given straight line and its perpendicular is called the foot of the corresponding perpendicular. But how to find the coordinates of the foot of the perpendicular drawn from a point to the given line? Let us derive the formula now.

Let ax+by+c=0 be the equation of straight line and assume that a perpendicular is drawn from a point (p,q) to this line and let the corresponding foot of the perpendicular be (h,k).

The slope of perpendicular line joining (p,q) and (h,k) is k-q/h-p

The slope of line ax+by+c=0 is –a/b.

Since the product of slopes of two perpendicular lines is -1,

k-q/h-p * -a/b = -1

h-p/a = k-q/b = m (say)

h-p/a = m and k-q/b = m

h = am+p and k = bm +q

Since (h,k) is a point on the straight line ax+by+c=0 we have

Ah+bk+c=0

a(am+p) + b(bm+q) + c=0

a2m+b2m + (ap+bq+c)=0

m = -(ap+bq+c)/(a2+b2)

Thus the formula for finding the foot of the perpendicular is,

h-p/a = k-q/b =-(ap+bq+c)/(a2+b2)

Answer 2

Answer:

The foot of perpendicular theorem

Let ax + by + c = 0 be the equation of straight line.

The slope of the perpendicular line joining (p, q) and (h, k) is k - q/h - p.

The slope of line ax + by + c = 0 is -a/b.

Since the product of slopes of two perpendicular lines is -1,

( k-q / h-p) * (-a/b ) = -1

h-p / a = k-q / b = m (says)

h-p/a = m and k - q/ b = m

h = am + p and k = bm+ q

since (h , k) is a point on the straight line ax +  by + c = 0 we have

ah + bk + c  = 0

a(am + p) + b( bm + q ) + c = 0

a²m + b²m + ( ap + bq + c) = 0

m = -( ap + bq + c ) / (a² + b²)

Thus the formula is

h - p / a = k-q / b

             = -(ap + bq + c ) / (a² + b²)