Question

If (a,0), (0, b) and (p, q) are collinear. Prove:
$\dfrac{p}{a} + \dfrac{q}{b} = 1$

Answer 1

Hint: We will use the concept that area of a triangle formed by three collinear points is 0. The formula for area of triangle in coordinate geometry is given as:
Area of $\Delta = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$ where, $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$ are the coordinates of the vertices of the triangle.

Complete step by step solution:
Now, we are given three points (a, 0), (0, b) and (p, q). Let these three points be the three vertices $A({x_1},{y_1}),B({x_2},{y_2})$and $C({x_3},{y_3})$. then,
$ ({x_1},{y_1}) = (a,0) \\
  ({x_2},{y_2}) = (0,b) \\
  ({x_3},{y_3}) = (p,q) \\ $.
Now, area of $\Delta ABC = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$
Now, putting the values of the coordinates of A, B and C we will get the area of $\Delta ABC$:
$ \Delta ABC = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right| \\
\Rightarrow 0 = \dfrac{1}{2}\left| {a(b - q) + 0(q - 0) + p(0 - b)} \right| \\
\Rightarrow 0 = \dfrac{1}{2}\left| {ab - aq + 0 - 0 + 0 - pb} \right| \\
\Rightarrow 0 = \dfrac{1}{2}\left| {ab - aq - bp} \right| \\
\Rightarrow 0 = \left| {ab - aq - bp} \right| \\
\Rightarrow ab = aq + bp \\ $
Now dividing the both sides by ab, we will get:
$ \Rightarrow \dfrac{{ab}}{{ab}} = \dfrac{{aq}}{{ab}} + \dfrac{{bp}}{{ab}} \\
\Rightarrow 1 = \dfrac{q}{b} + \dfrac{p}{a} \\
\Rightarrow \dfrac{p}{a} + \dfrac{q}{b} = 1 \\ $
Hence proved that $\dfrac{p}{a} + \dfrac{q}{b} = 1$.

Note: The coordinates and their order has to be taken very carefully in the formula for calculating the area of Triangle.In the formula for Area of $\Delta = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$.
If instead of ${x_1}$ in the first term in the bracket, we take ${x_2}$ for the x coordinates and similarly if we change the order of the y-ordinates anywhere in the formula then the formula will become invalid.