We need to find a point [tex]E[/tex] which divides the line [tex]CD[/tex] in the ratio of [tex]4:3[/tex]
The point [tex]E[/tex] is at [tex]E(10,6)[/tex]
The two points of [tex]CD[/tex] are [tex]C(2,2)[/tex] and [tex]D(16,9)[/tex]
The ratio of [tex]CD:DE[/tex] is [tex]4:3[/tex] and [tex]E[/tex] is in between [tex]CD[/tex].
So, [tex]E[/tex] divides the line internally.
The required formula is
[tex]\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\\ =\left(\dfrac{4\times 16+3\times 2}{4+3},\dfrac{4\times 9+3\times 2}{4+3}\right)\\ =(10,6)[/tex]
The required point is [tex]E(10,6)[/tex].
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