If `Q(h,k)` is the foot of the perpendicular of `P(x_(1),y_(1))` on the line `ax+by+c=0` then prove that `(h-x_(1)),a=(k-y_(1)),b=-(ax_(1)+by_(1)+c):(a^(2)+b^(2))`.

If Q(h,k) is the foot of the perpendicular from P(x_(1) ,y_(1)) on the line ax + by+ c=0 then Q(h,k) is

If Q(h,k) is the foot of the perpendicular from P(x_(1),y_(1)) on the line ax+by+c=0 then (h-x_(1))/(a)=(k-y_(1))/(b) =_____ .

A : The foot of the perpendicular from (3, 4) on the line 3x-4y+5=0 is (81//25, 92//25) R : If (h, k) is the foot of the perpendicular from (x_(1), y_(1)) to the line ax+by+c=0 then (h-x_(1))/(a)=(h-k_(1))/(b)=(-(ax_(1)+by_(1)+c))/(a^(2)+b^(2))

If (h,k) is the foot of the perpendicular from (x_(1),y_(1)) to the line ax+by+c=0, (a!=0,b!=0) then (h-x_(1))/(a)=(k-y_(1))/(b)=(-(ax_(1)+by_(1)+c))/(a^(2)+b^(2)) Find the foot of the perpendicular from (-1,3) to the line 5x-y=18

If Q(h,k) is the image of the point P(x_(1),y_(1)) w.r.to the straight line ax+by+c=0 then prove that (h-x_(1)):a=(k-y_(1)):b=-2(ax_(1)+by_(1)+c):a^(2)+b^(2) or (h-x_(1))/(a)=(k-y_(1))/(b)=(-2(ax_(1)+by_(1)+c))/(a^(2)+b^(2)) and find the image of (1,-2) w.r.t the straight line 2x-3y+5=0

Foot of the perpendicular from a point (x_(1),y_(1)) on the line ax + by + c= 0 is given by

Foot of the perpendicular from a point (x_(1),y_(1)) on the line ax + by + c= 0 is given by

A : The image of the origin with respect to the line x+y+1=0 is (-1, -1) R : If (h, k) is the image of (x_(1), y_(1)) with respect to the line ax+by+c=0 then (h-x_(1))/(a)=(h-k_(1))/(b)=(-2(ax_(1)+by_(1)+c))/(a^(2)+b^(2))