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The midpoint P of the line segment joining the points A (-10, 4) and B (-2, 0) lies on the line segment joining the points C (-9, -4) and D (-4, y). Find the ratio in which P divides CD and also, find the value of y.

Answer Verified Verified
Hint: Find the coordinates of the midpoint P of the line segment AB. Then use the section formula of line segment CD for the abscissa of point P to find the ratio in which P divides CD. Use the section formula of line segment CD for the ordinate of point P to find y.



Complete step-by-step answer:

From section formula, if P (x, y) divides the line segment joining \[C({x_3},{y_3})\] and \[D({x_4},{y_4})\] in the ratio m:n, then:
\[x = \dfrac{{m{x_4} + n{x_3}}}{{m + n}};{\text{ }}y = \dfrac{{m{y_4} + n{y_3}}}{{m + n}}{\text{ }}.........{\text{(2)}}\]
Substituting equation (1) in equation (2) and using coordinates of C and D, we get:
\[ - 6 = \dfrac{{m( - 4) + n( - 9)}}{{m + n}}{\text{ }}..........{\text{(3)}}\]
\[ 2 = \dfrac{{m(y) + n( - 4)}}{{m + n}}{\text{ }}...........{\text{(4)}}\]
Simplifying equation (3) to get the ratio in which P divided CD, we get:
\[ - 6 = \dfrac{{ - 4m - 9n}}{{m + n}}{\text{ }}\]
\[ - 6(m + n) = - 4m - 9n\]
\[ - 6m - 6n = - 4m - 9n\]
\[ - 6m + 4m = - 9n + 6n\]
\[ - 2m = - 3n\]
\[\dfrac{m}{n}{\text{ = }}\dfrac{3}{2}{\text{ }}..........{\text{(5)}}\]
Simplifying equation (4) to obtain the value of y, we get:
\[2 = \dfrac{{my - 4n}}{{m + n}}\]
\[2(m + n) = my - 4n\]
\[2m + 2n = my - 4n\]
Gathering terms containing m on RHS and terms containing n on LHS, we get:
\[4n + 2n = my - 2m\]
\[6n = m(y - 2)\]
Divide both sides by n, to get:
\[6 = \dfrac{m}{n}(y - 2)\]
Substituting equation (5) in the above equation, we get:
\[6 = \dfrac{3}{2}(y - 2)\]
Multiply both sides by \[\dfrac{2}{3}\] and simplify.
\[\dfrac{2}{3} \times 6 = y - 2\]
\[4 = y - 2\]
\[y = 6\]
Hence, the value of y is 6
Therefore, P divides CD in the ratio 3:2 and the value of y is 6.

Note: The possibility for mistake is writing the section formula for points \[C({x_3},{y_3})\] and \[D({x_4},{y_4})\] wrongly as \[x = \dfrac{{m{x_3} + n{x_4}}}{{m + n}};{\text{ }}y = \dfrac{{m{y_3} + n{y_4}}}{{m + n}}\] instead of \[x = \dfrac{{m{x_4} + n{x_3}}}{{m + n}};{\text{ }}y = \dfrac{{m{y_4} + n{y_3}}}{{m + n}}\] . You might also think, it is impossible to find three variables from two equations but you are just finding the ratio between m and n and then the value of y, which requires only two equations.
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