Mathematics MISCELLANEOUS

Length of the Perpendicular :

The length of the perpendicular from a point `(x_1, y_1)` to a line `ax + by+ c = 0` is

`|(ax_1+by_1+c)/(sqrt(a^2+b^2)|`

Proof :

The line `ax + by + c = 0` meets `x`-axis at `A(-c/a,0)` and `y`-axis at `B(0,-c/b)`

Let `P(x_1, y_1)` be the point. Draw `PN bot AB`.

Now, area of `Delta PAB`

`1/2|x_1(0+c/b)-c/a(-c/b-y_1)+0(y_1-0)|`

`1/2|(cx_1)/b+(cy_1)/a+c^2/(ab)|=|(ax_1+by_1+c)c/(2ab)|`.......................(i)

Also, area of `Delta PAB`

`=1/2 AB xx PN`

`=1/2sqrt(c^2/a^2+c^2/b^2) xx PN`

`= c/(2ab)sqrt(a^2+b^2) xx PN`....................(ii)

From equation (i) and (ii), we get

`|(ax_1+by_1+c)c/(2ab)|=c/(2ab) sqrt(a^2+b^2)xx PN`

`=> PN=(|ax_1+by_1+c|)/(sqrt(a^2+b^2))`

Distance Between Two Parallel Lines :

Let `(x_1 , y_1)` be any point on the line `ax + by +c_2 = 0`

Distance of point `(x_1, y_1)` from the line `ax+ by +c_1 = 0` is

`p=(|ax_1+by_1+c|)/(sqrt(a^2+b^2))`

Now point `(x_1, y_1)` lies on `ax + by+c_2 = 0` then

`ax_1 +by_1 +c_2 = 0`

`=> ax_1 + by_1 = -c_2`

`=> p=(|c_1-c_2|)/(sqrt(a^2+b^2))`

Area of Parallelogram :

Area of the llgm whose `4` sides are as shown in the fig. using

`A = p_1 p_2 cosec theta` is given by

`|((c_1-c_2) (d_1-d_2))/(m_1-m_2)|=((p_1p_2)/(sin theta))`

Position of a Point w.r.t a Line :

If the points `P(x_1, y_1)` and `Q(x_1 ,y_2)` lies on the same side of the line `Ax + Bx + C=0` then the expressions

`Ax_1 + By_1 +C` and `Ax_2+ By_2 + C` will be of the same sign and if `P(x_1, y_1)` and `Q(x_2 ,y_2 )` are on the opposite side of the line
then the expressions `Ax_1 + By_1 + C` and `Ax_2 + By_2 + C` will be of opposite sign.