This is a Test of Mathematics Solution Subjective 113 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Find the vertices of the two right angles triangles, each having area 18 and such that the point (2, 4) lies on the hypotenuse, and the other two sides are formed by the \(x\) and \(y\) axes.
Suppose the vertices are (a, 0) and (0, b). Clearly $ \frac{1}{2} ab = 18 $ or ab = 36.
Also the equation of the line through (0,a), (b,0) is $ \displaystyle{ \frac{x}{a} + \frac{y}{b} = 1 } $. Since we know that (2, 4) is on that line, there fore $ \displaystyle { \frac{2}{a} + \frac{4}{b} = 1 } $.
In this equation, lets replace $ a $ by $ \frac{36}{b} $. Hence we get $ \displaystyle { \frac{2}{\frac{36}{b}} + \frac{4}{b} = 1 } $ or $ \displaystyle { \frac{b}{18} + \frac{4}{b} = 1 } $.
Therefore we get a quadratic in b.
$ \displaystyle { b^2 -18b + 72 = 0 } $. We can simply middle term factorize this to find $ \displaystyle {(b-12)(b-6) = 0 } $. Thus b = 12 or 6 implying a = 3 or 6.
The other two vertices are: (0,12) and (3, 0) OR (0,6) and (6,0).
This is a Test of Mathematics Solution Subjective 113 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Find the vertices of the two right angles triangles, each having area 18 and such that the point (2, 4) lies on the hypotenuse, and the other two sides are formed by the \(x\) and \(y\) axes.
Suppose the vertices are (a, 0) and (0, b). Clearly $ \frac{1}{2} ab = 18 $ or ab = 36.
Also the equation of the line through (0,a), (b,0) is $ \displaystyle{ \frac{x}{a} + \frac{y}{b} = 1 } $. Since we know that (2, 4) is on that line, there fore $ \displaystyle { \frac{2}{a} + \frac{4}{b} = 1 } $.
In this equation, lets replace $ a $ by $ \frac{36}{b} $. Hence we get $ \displaystyle { \frac{2}{\frac{36}{b}} + \frac{4}{b} = 1 } $ or $ \displaystyle { \frac{b}{18} + \frac{4}{b} = 1 } $.
Therefore we get a quadratic in b.
$ \displaystyle { b^2 -18b + 72 = 0 } $. We can simply middle term factorize this to find $ \displaystyle {(b-12)(b-6) = 0 } $. Thus b = 12 or 6 implying a = 3 or 6.
The other two vertices are: (0,12) and (3, 0) OR (0,6) and (6,0).
