Question

If the points \((a,0),(0,b)\) and \((1,1)\) are collinear, then \(\frac{1}{a}+\frac{1}{b} =\)

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

The Correct Option is C

Given points: \((a,0)\), \((0,b)\), and \((1,1)\) 

Step 1: Condition for collinearity

Three points are collinear if the area of the triangle they form is zero.

The area of a triangle formed by three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:

\[ \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| = 0 \]

Step 2: Substitute given points

\[ \frac{1}{2} \left| a( b - 1 ) + 0( 1 - 0 ) + 1( 0 - b ) \right| = 0 \] \[ \frac{1}{2} \left| a(b - 1) - b \right| = 0 \] \[ a(b - 1) - b = 0 \] \[ ab - a - b = 0 \]

Step 3: Solve for \( \frac{1}{a} + \frac{1}{b} \)

\[ ab - a - b = 0 \Rightarrow ab = a + b \] \[ \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \] \[ = \frac{ab}{ab} = 1 \]

Final Answer: 1

Answer 2

To determine if the points \((a,0)\), \((0,b)\), and \((1,1)\) are collinear, they must satisfy the collinearity condition which states that the area of the triangle formed by these points should be zero. The formula for the area of a triangle given three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is:

\[\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\]

Applying the given points \((a,0)\), \((0,b)\), and \((1,1)\):

\[\text{Area}=\frac{1}{2}\left|a(b-1)+0(1-0)+1(0-b)\right|=0\]

Simplify:

\[\frac{1}{2}\left|ab-a-b\right|=0\]

This implies:

\[ab-a-b=0\]

Rearranging gives:

\[ab=a+b\]

Divide the entire equation by \(ab\):

\[\frac{a}{ab}+\frac{b}{ab}=1\]

Which simplifies to:

\[\frac{1}{b}+\frac{1}{a}=1\]

Hence, \(\frac{1}{a}+\frac{1}{b}=1\)