# Does anyone have a derivation of the formula for the shortest distance to a line? (1 Viewer)

#### SB257426

##### Very Important User

View attachment 40165

I was doing a hunters hill high extension 2 trial and they asked to find the min value of |z|. Would I need to derive this formula or can i just use it in an exam?

#### SB257426

##### Very Important User

here is the formula...

##### done hsc yay
is this from point P to a line AB??? I thought we do that with projections?

#### SB257426

##### Very Important User
is this from point P to a line AB??? I thought we do that with projections?
yeah it is ....

i thought as well but they used that formula to find the min value of z

##### done hsc yay
yeah it is ....

i thought as well but they used that formula to find the min value of z
what is the context? send the question???

#### SB257426

##### Very Important User
yeah it is ....

i thought as well but they used that formula to find the min value of z
if u look at hunters hill ext 2 trial if its on here then have a look at 11 D)

#### synthesisFR

##### Well-Known Member
View attachment 40166

here is the formula...
iirc thats the formula for perpendicular distance. i forgot how to derive it or idk if we even did derive it, last time i did it was year ten.
usually for short dist i just do projections or dot product

#### Luukas.2

##### Well-Known Member
The perpendicular distance formula gives the shortest distance from a point $\bg_white P\left(x_0,\ \!y_0\right)$ to the line $\bg_white Ax + By + C = 0$, which is

$\bg_white d_{\perp} = \cfrac{\left|Ax_0 + By_0 + C\right|}{\sqrt{A^2 + B^2}}$

It was a part of the 2 unit syllabus until the change in 2020. Now the only way is to derive the formula before using it or use vectors / projections for MX1 and MX2 students.

It can be derived as follows: Let X and Y be the points on the line such that PX and PY are parallel to the x- and y-axes, respectively. Let Q be the point on the line that is closest to P. The result follows by setting PQ = d (the perpendicular distance sought) and finding the area of triangle XPY in two different ways.

#### Luukas.2

##### Well-Known Member
Another derivation: Solving the equations of the circle $\bg_white (x - x_0)^2 + (y - y_0)^2 = r^2$ and the line $\bg_white Ax + By + C = 0$ simultaneously should yield a quadratic equation, which has only one solution if the line is a tangent to the circle. By setting the discriminant to zero, the result should follow by making the radius (which is also the perpendicular distance sought) the subject.