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In the diagram, C and D are located such that AB is divided into three equal parts. What are the coordinates of C and D?

In the diagram C and D are located such that AB is divided into three equal parts What are the coordinates of C and D class=

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Step-by-step Answer:

Topic: Points of division

There are scary looking formulas that can be used, but it is much easier to calculate by reasoning.

Given : A(-3,6), B(6,-3)

Solution:

The idea is to subdivide the DIFFERENCE in coordinates into thirds, and add onto that of A.  We choose A as the starting point, but method works equally well if we chose B.

Difference in coordinates (delta) between A & B is then

delta(Bx-Ax, By-Ay)

=(6-(-3), -3-6)

=delta(9, -9)

One third of difference (for point C)

=delta/3 = (3,-3)

So coordinates of point C

= A(-3,6)+(3,-3)

= C(0,3)

Two thirds of difference (for point D)

= (2/3)delta = (6, -6)

Coordinates for point D

= A(-3,6)+(6,-6)

= D(3,0)

If you prefer to use formulas, it would be

New coordinates = (Xa+(Xb-Xa)*k, Ya+(Yb-Ya)*k)

where

Xa,Xb = x-coordinates of points A & B respectively.

Ya,Yb = y-coordinates of points A & B respectively.

k=ratio (usually less than 1)

Here

k for point C = 1/3

k for point D = 2/3

Answer:

Coordinate of C is: (0,3)

and Coordinate of D is: (3,0)

Step-by-step explanation:

We know that if a point C(x,y) divides the given line segment A(a,b)B(c,d) into ratio of m:n

then the coordinates of points C are:

[tex]x=\dfrac{m\times c+n\times a}{m+n},\ y=\dfrac{m\times d+n\times b}{m+n}[/tex]

  • Point C cuts the line segment AB in the ratio 1:2.

Here A(a,b)=A(-3,6)

and B(c,d)=B(6,-3)

This means that the coordinate of Point C are:

[tex]x=\dfrac{1\times 6+2\times (-3)}{1+2},\ y=\dfrac{1\times (-3)+2\times 6}{1+2}\\\\i.e.\\\\x=\dfrac{6-6}{3},\ y=\dfrac{-3+12}{3}\\\\i.e.\\\\x=0,\ y=\dfrac{9}{3}\\\\i.e.\\\\x=0,\ y=3[/tex]

Hence, the coordinates of Point C are: (0,3)

  • Similarly Point D cuts the line AB in the ratio 2:1

Hence, the coordinates of point D is calculated by:

[tex]x=\dfrac{2\times (6)+1\times (-3)}{1+2},\ y=\dfrac{2\times (-3)+1\times 6}{1+2}\\\\i.e.\\\\x=\dfrac{12-3}{3},\ and\ y=\dfrac{-6+6}{3}\\\\i.e.\\\\x=\dfrac{9}{3},\ y=\dfrac{0}{3}\\\\i.e.\\\\x=3,\ y=0[/tex]

Hence, the coordinate of Point D is: (3,0)

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