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))`.

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))

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))

The perpendicular distance of the plance ax + b y + c z + d = 0 from the point P(x_(0) , y_(0) , z_(0)) is (ax_(0)+by_(0)+cz_(0)+d)/sqrt(a^(2)+b^(2)+c^(2))

At the point (x_(1),y_(1)) on the curve x^(3)+y^(3)=3axy show that the tangent is (x_(1)^(2)-ay_(1))x+(y_(1)^(2)-ax_(1))y=ax_(1)y_(1)

The foot of the perpendicular from the point (1,2,3) to the line (x)/(2)=(y-1)/(3)=(z-1)/(3) is

A : The ratio in which the perpendicular through (4, 1) divides the line joining (2, -1), (6, 5) is 5:8 . R : The ratio in which the line ax+by+c=0 divides the line segment joining (x_(1), y_(1)), (x_(2), y_(2)) is (ax_(1)+by_(1)+c): -(ax_(2)+by_(2)+c) .

Three points P (h, k), Q(x_(1) , y_(1))" and " R (x_(2) , Y_(2)) lie on a line. Show that (h - x_(1)) (y_(2) - y_(1)) = (k - y_(1)) (x_(2) - x_(1)) .

If one of the lines given by the equation a_(1)x^(2)+2h_(1)xy+b_(1)y^(2)=0 coincides with one of the lines given by a_(2)x^(2)+2h_(2)xy+b_(2)y^(2)=0 and the other lines represented by them are perpendecular then prove that. h_(1)((1)/(a_(1))-(1)/(b_(1)))=h_(2)((1)/(a_(2))-(1)/(b_(2)))