Respuesta :
Answer:
see explanation
Step-by-step explanation:
Find the coordinates of c using the section formula
[tex]x_{c}[/tex] = [tex]\frac{(3(1)+(2(6)}{2+3}[/tex] = [tex]\frac{3+12}{5}[/tex] = 3
[tex]y_{c}[/tex] = [tex]\frac{(3(4))+(2(-1))}{2+3}[/tex] = [tex]\frac{12-2}{5}[/tex] = 2
coordinates of c = (3, 2)
Similarly to find the coordinates of d
[tex]x_{d}[/tex] = [tex]\frac{x(2(1))+(3(3))}{3+2}[/tex] = [tex]\frac{2+9}{5}[/tex] = [tex]\frac{11}{5}[/tex]
[tex]y_{d}[/tex] = [tex]\frac{(2(4))+(3(2))}{3+2}[/tex] = [tex]\frac{8+6}{5}[/tex] = [tex]\frac{14}{5}[/tex]
coordinates of d = ( [tex]\frac{11}{5}[/tex], [tex]\frac{14}{5}[/tex])
Answer: The co-ordinates of point C are (3, 2) and the co-ordinates of the point D are [tex]\left(\dfrac{11}{5},\dfrac{14}{5}\right).[/tex]
Step-by-step explanation: Given that the endpoints of a line segment AB are A(1,4) and B(6,-1).
We are to find the co-ordinates of a point C that divides the line segment AB in the ratio 2 : 3.
Also, to find the co-ordinates of the point D that divides the line segment AC in the ratio 3 : 2.
We know that
the co-ordinate of a point that divides a line segment with endpoints (p. q) and (r, s) in the ratio m : n is given by
[tex]\left(\dfrac{mr+np}{m+n},\dfrac{ms+nq}{m+n}\right).[/tex]
Therefore, the co-ordinates of point C will be
[tex]C\\\\\\=\left(\dfrac{2\times 6+3\times 1}{2+3},\dfrac{2\times (-1)+3\times 4}{2+3}\right)~~~~~~~~~~[\textup{here, m : n = 2 : 3}]\\\\\\=\left(\dfrac{12+3}{5},\dfrac{-2+12}{5}\right)\\\\\\=\left(\dfrac{15}{5},\dfrac{10}{5}\right)\\\\=(3,2).[/tex]
And, the co-ordinates of the point D will be
[tex]D\\\\\\=\left(\dfrac{3\times 3+2\times 1}{3+2},\dfrac{3\times 2+2\times 4}{3+2}\right)~~~~~~~~~~[\textup{here, m : n = 3 : 2}]\\\\\\=\left(\dfrac{9+2}{5},\dfrac{6+8}{5}\right)\\\\\\=\left(\dfrac{11}{5},\dfrac{14}{5}\right).[/tex]
Thus, the co-ordinates of point C are (3, 2) and the co-ordinates of the point D are [tex]\left(\dfrac{11}{5},\dfrac{14}{5}\right).[/tex]
