1.

In what ratio does the point (\(\frac {-19}{3}, \frac {7}{3}\)) divide the line segment joining A(3, 7) and B(-11, 0)?(a) 1:2 (externally)(b) 1:2 (internally)(c) 2:1 (externally)(d) 2:1 (internally)The question was posed to me in an international level competition.The origin of the question is Geometry topic in portion Coordinate Geometry of Mathematics – Class 10

Answer»

Right answer is (d) 2:1 (internally)

For EXPLANATION I WOULD say: Let the ratio in which the point (\(\frac {-19}{3}, \frac {7}{3}\)) divides the LINE segment joining the POINTS A(3, 7) and B(-11, 0) be k:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(3, 7) and B(-11, 0) and the ratio is k:1

∴ x = \(\frac {k(-11)+1(3)}{k+1} = \frac {-11k+3}{k+1}\)

y = \(\frac {k(0)+1(7)}{k+1} = \frac {7}{k+1}\)

Since, the point is (\(\frac {-19}{3}, \frac {7}{3}\)).

∴ \(\frac {-19}{3} = \frac {-11k+3}{k+1}\)

-19(k + 1) = 3(-11k + 3)

-19k – 19 = -33k + 9

-19k + 33k = 19 + 9

14k = 28

k = \(\frac {28}{14}\) = 2

The ratio is 2:1.


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