Mindblown: a blog about philosophy.
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Let us move our study towards a line. Now, let us look into dividing line segment into two parts, mainly unequal divisions in a ratio m:n
On a line PQ or AB ,there could be make two divisions: internal and external.
In internal divisions, a line is separated into segments by points, say Z, that lies on PQ line segment
Internal divisions
In external division, a line segment is separated by a point on PQ produced beyond its length, ie. the point of division Z lies outside the line segment PQ
External division
Now let us try to find the location of this New point in the case we are having the location coordinates of end points of line segment AB or PQ, which is going to be divided..
Internal division
Let a line segment AB with A at (x1, y1) and B at (x2, y2). Suppose a point which divides the line internally into two sections
Then for internal division, the steps to locate the point which divide the line AB :
Let us say that the coordinates of point of division Z is (x ,y).
Drop a perpendicular line from B till it joins the YN line to form triangle(YNB).
Also, drop a perpendicular line from A till it joins the AM line to forming a triangle(AMY).
Let AY division of line AB be equal to ‘m’ and YB division of line between equal to ‘n’
In triangles, let’s try to prove similarity between triangles
$ϴ_1$ = $ϴ_2$
$ϴ_3$ = $ϴ_4$
Angle(AMZ) = Angle(ZNB)
$ϴ_3$ = $ϴ_4$
Angle(AMZ) = Angle(ZNB)
Since three angles of triangle are equal with each other
Hence triangle(AMZ) and triangle(ZNB) are similar by AAA similarity
Now, we can do our maths :
$\mathrm{\frac{AZ}{ZB} = \frac{ZM}{BN}}$ ….Since triangles are similar)
$\mathrm{\frac{m}{n} = \frac{y_2-y}{y-y_1}}$
On cross multiply we get
$\mathrm{m×(y – y_1)}$ = $\mathrm{n(y_2 -y)}$
$\mathrm{m×y – m×y_1}$ = $\mathrm{ny_2 -n×y}$
Rearranging the terms to get,
$\mathrm{m×y + n×y}$ = $\mathrm{ny_2 +m×y_1}$
$\mathrm{(m + n)× y}$ = $\mathrm{ny_2 +my_1}$
$\mathrm{y}$ = $\mathrm{\frac{m×y_1 + n×y_2}{m +n}}$
$\mathrm{\frac{m}{n} = \frac{y_2-y}{y-y_1}}$
On cross multiply we get
$\mathrm{m×(y – y_1)}$ = $\mathrm{n(y_2 -y)}$
$\mathrm{m×y – m×y_1}$ = $\mathrm{ny_2 -n×y}$
Rearranging the terms to get,
$\mathrm{m×y + n×y}$ = $\mathrm{ny_2 +m×y_1}$
$\mathrm{(m + n)× y}$ = $\mathrm{ny_2 +my_1}$
$\mathrm{y}$ = $\mathrm{\frac{m×y_1 + n×y_2}{m +n}}$
Now we can do also do the process to get x -coordinate of point of division Y
we can do our maths :
$\frac{AZ}{ZB}$ = $\frac{AN}{ZN}$. …(Since triangles are similar)
$\frac{m}{n}$ = $\frac{x_2-x}{x-x_1}$
On cross multiply we get
$\mathrm{m(x – x_1)}$ = $\mathrm{n(x_2 -x)}$
$\mathrm{mx – mx_1}$ = $\mathrm{nx_2 -nx}$
$\frac{m}{n}$ = $\frac{x_2-x}{x-x_1}$
On cross multiply we get
$\mathrm{m(x – x_1)}$ = $\mathrm{n(x_2 -x)}$
$\mathrm{mx – mx_1}$ = $\mathrm{nx_2 -nx}$
Rearranging the terms to get,
$\mathrm{mx + nx}$ = $\mathrm{nx_2 +mx_1}$
$\mathrm{(m + n)x}$ = $\mathrm{nx_2 +mx_1}$
$\mathrm{x}$ = $\mathrm{\frac{nx_2 +mx_1}{m +n}}$
$\mathrm{mx + nx}$ = $\mathrm{nx_2 +mx_1}$
$\mathrm{(m + n)x}$ = $\mathrm{nx_2 +mx_1}$
$\mathrm{x}$ = $\mathrm{\frac{nx_2 +mx_1}{m +n}}$
Now, we shall look over external divisions
External division
We will look on subsequent day (in this same post , I will update.. before onset of 1 Apeil)