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Question

If $$A$$ and $$B$$ are two points having coordinates $$(-2,-2)$$ and $$(2,-4)$$ respectively, find the coordinates of $$P$$ such that $$AP=\cfrac{3}{7}AB$$.


Solution


Given:
$$AP=\cfrac { 3 }{ 7 } AB$$

$$\dfrac{AP}{AB}=\dfrac{3}{7}$$

$$\dfrac{AP}{AP+BP}=\dfrac{3}{3+4}$$

$$\dfrac{AP}{BP}=\dfrac{3}{4}$$

Hence, the line joining the points $$A(-2,-2)$$ and $$B(2,-4)$$ is divided by $$P$$ in ratio $$3:4$$.

$$P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$$

Substituting the values, $$x_1=-2,y_1=-2, \ x_2=2,y_2=-4$$ in the section formula.



$$P=\left( \cfrac { 3(2)+4(-2) }{ 3+4 } ,\cfrac { 3(-4)+4(-2) }{ 3+4}  \right)$$

$$P=\left( \cfrac { -2 }{ 7 } ,\cfrac { -20 }{ 7 }  \right) $$

Mathematics

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