Question

if Q ( h,k ) is the foot of the perpendicular form P ( x1,y1 ) on the straight line
ax + by+c =0, then ( h-x1 ):a = ( k-y1 ):b = - ( ax1+by1+c ):( a²+b² )

Answer 1

Answer:

We know that foot of perpendicular from $(x_{1} ,y_{1})$ on straight line ax + by + c = 0$(x_{1} , y_{1})$

is (h,k) then:

$\frac{h-x_{1} }{a} = \frac{k-y_{1} }{b} = -\frac{ax_{1}+by_{1} +c }{a^{2} + b^{2} }$

Explanation:

We know that foot of perpendicular from $(x_{1} , y_{1})$ on line ax + by + c = 0

Formula is :

$\frac{x - x_{1} }{a} = \frac{y-y_{1} }{b} =-\frac{ax_{1} + by_{1} + c }{a^{2} + b^{2} }$

where (x,y) is foot of perpendicular of $(x_{1},y_{1} )$ from ax + by + c = 0

So here in question given to us is foot of perpendicular is (h,k) and we nedd to prove formula.

Putting value of of point(h,k) in question,

$\frac{h-x_{1} }{a} = \frac{k-y_{1} }{b} = -\frac{ax_{1}+by_{1} +c }{a^{2} + b^{2} }$

Hence proved