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Show that the points (a,0),(0,b) and (3a,-2b) are collinear. Also, find the equation of line containing them. |
Answer» Let `A(a,0),B(0,b) and C(3a,-2b)` be the given points. Then, the equation of the line AB is given by `(y-0)/(x-a)=(b-0)/(0-a) Rightarrow -ay=bx-ab` `Rightarrow bx+ay-ab=0` Thus, the equation of the line AB is bx+ay-ab=0…………(i) Putting x=3a and y-02b (i), we get `LHS=b.3a+a(-2b)-ab=3ab-2ab=0=RHS` Thus, the point C(3a,-2b) also lies on AB. Hence, the given points are collinear and the equation of hte line containing them is bx+ay-ab=0 |
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