Find coordinates of the foot of perpendicular and equation of perpendicular drawn from the point $$(2, 3)$$ to the line $$y=3x-4$$.
$$\\ The\quad given\quad equation\quad of\quad the\quad line\quad is\quad \\ y=3x-4\quad \quad \longrightarrow \left( 1 \right) \\ Slope\quad of\quad the\quad given\quad line\quad is\quad { m }_{ 1 }=3\\ suppose\quad the\quad equation\quad of\quad the\quad line\quad perpendicular\quad to\quad the\quad given\quad line\quad is\\ y={ m }_{ 2 }x+c\quad \quad \longrightarrow \left( 2 \right) \\ Since\quad both\quad the\quad lines\quad are\quad perpendicular\quad to\quad each\quad other\quad so\quad we\quad have;\\ { m }_{ 1 }{ m }_{ 2 }=-1\\ \Rightarrow 3{ m }_{ 2 }=-1\\ \Rightarrow { m }_{ 2 }=-\dfrac { 1 }{ 3 } \\ From\left( 2 \right) we\quad have,\quad y=-\dfrac { 1 }{ 3 } x+c\\ Since\quad the\quad line\quad passes\quad through\quad the\quad point\quad \left( 2,3 \right) ,we\quad have\\ 3=-\dfrac { 2 }{ 3 } +c\\ \Rightarrow c=\dfrac { 11 }{ 3 } \\ \therefore equation\quad of\quad the\quad perpendicular\quad to\quad the\quad given\quad line\quad is\\ y=-\dfrac { 1 }{ 3 } x+\dfrac { 11 }{ 3 } \\ \Rightarrow 3y=-x+11\quad \quad \longrightarrow \left( 3 \right) \\ Solving\quad equation\quad \left( 1 \right) and\left( 3 \right) \quad we\quad have\quad x=\dfrac { 23 }{ 11 } ,y=\dfrac { 25 }{ 11 } \\ Hence,the\quad foot\quad and\quad equation\quad of\quad the\quad perpendicular\quad is\quad \left( \dfrac { 23 }{ 11 } ,\dfrac { 25 }{ 11 } \right) and\quad 3y=-x+11\quad respectively.$$
Let $$a$$ and $$b$$ be any two numbers satisfying $$\dfrac {1}{a^2}+\dfrac {1}{b^2}=\dfrac {1}{4}$$. Then, the foot of perpendicular from the origin on the variable line, $$\dfrac {x}{a}+\dfrac {y}{b}=1$$, lies on :
The foot of the perpendicular drawn from the origin, on the line, $$3x + y = \lambda (\lambda \neq 0)$$ is $$P$$. If the line meets x-axis at $$A$$ and y-axis at $$B$$, then the ratio $$BP : PA$$ is
The foot of the perpendicular from the point $$(3, 4)$$ on the line $$3x-4y+5=0$$ is:
The ends $$A, B$$ of a straight line segment of constant length $$c$$ slide upon the fixed rectangular axes $$OX$$ & $$OY$$ respectively. If the rectangle $$OAPB$$ be completed then, the locus of the foot of the perpendicular drawn from $$P$$ to $$AB$$ is ?
A straight line passes through a fixed point $$(h,k)$$. Find the locus of feet of the perpendiculars, drawn from the origin to it.
If $$p$$ and $$p'$$ be the lengths of perpendiculars from origin to the lines $$x \sec \theta-y \cos\theta=a$$ and $$x \cos \theta-y \sin \theta=a \cos 2\theta$$ respectively, then prove that $$4p^{ 2 }+p' ^{ 2 }=a^{ a }$$.
Prove that the product of the perpendiculars from the point $$ \left[ \pm \sqrt { \left( { a }^{ 2 }-{ b }^{ 2 } \right) } ,0 \right]$$ to the line $$\dfrac{ x }{ a }\cos \theta+\dfrac{ y }{ b }\sin \theta=1$$ is $$b^{ 2 }$$.
If $$x \cos \alpha + y \sin \alpha = p$$ where, $$p=\dfrac{ \sin^{ 2 }\alpha }{ \cos \alpha }$$ be a straight line, prove that perpendiculars $$p_{ 1 },p_{ 2 }$$ and $$ p_{ 3 }$$ on this line from the points $$(m^{ 2 }, 2m), (mm,m+m)$$ and $$(m^{ 2 },2m)$$ respectively are in geometrical progression.
Find the co-ordinates of the foot of the perpendicular drawn from the point$$(2,3)$$to the line $$y=3x+4$$
Let $$\triangle ABC$$ be a triangle with $$AB = AC$$. If $$D$$ is the midpoint of $$BC$$, $$E$$ is the foot of the perpendicular drawn from $$D$$ to $$AC$$ and $$F$$ the mid-point of $$DE$$. Then $$AF$$ is perpendicular to