Solveeit Logo
Login

Question

Question: If the points \((a,b),(a',b')\) and \((a - a',b - b')\) are collinear, then \( 1)ab' = a'b \\\...

If the points (a,b),(a′,b′)(a,b),(a',b') and (a−a′,b−b′)(a - a',b - b') are collinear, then
1)ab′=a′b 2)ab=a′b′ 3)aa′=bb′ 4)a2+b2  1)ab' = a'b \\\ 2)ab = a'b' \\\ 3)aa' = bb' \\\ 4){a^2} + {b^2} \\\

Explanation

Solution

Hint : Here, we will use the concepts of the collinear points and take the determinant for the points and by expanding the determinant and simplification of the equation gives the required resultant expression.

Complete step-by-step answer :
Given that three points are collinear means when we take the determinant of these three points it will be zero.
\left| {\begin{array}{*{20}{c}} a&b;&1 \\\ {a'}&{b'}&1 \\\ {a - a'}&{b - b'}&1 \end{array}} \right| = 0
Expand the above determinant –
⇒a[b′−(b−b′)]−b[a′−(a−a′)]+1[a′(b−b′)−(b′(a−a′)]=0\Rightarrow a[b' - (b - b')] - b[a' - (a - a')] + 1[a'(b - b') - (b'(a - a')] = 0
Open the brackets in the above expression, when there is negative sign outside the bracket then the sign of the terms inside the bracket changes. Positive term becomes negative and the negative term becomes positive but when there is a positive sign outside the bracket then the sign of the terms inside the bracket remains the same when brackets open.
⇒a[b′−b+b′)]−b[a′−a+a′)]+1[a′b−a′b′−b′a+a′b′]=0\Rightarrow a[b' - b + b')] - b[a' - a + a')] + 1[a'b - a'b' - b'a + a'b'] = 0
Similarly multiply the term inside the bracket –
⇒ab′−ab+ab′−ba′+ab−a′b+a′b−a′b′−b′a+a′b′=0\Rightarrow ab' - ab + ab' - ba' + ab - a'b + a'b - a'b' - b'a + a'b' = 0
Combine like terms together in the above expression –
⇒ab′+ab′‾−ab+ab‾−ba′−b′a‾−a′b+a′b‾−a′b′+a′b′‾=0\Rightarrow \underline {ab' + ab'} - \underline {ab + ab} - \underline {ba' - b'a} - \underline {a'b + a'b} - \underline {a'b' + a'b'} = 0
Like terms with the same value and the opposite sign cancels each other.
⇒ab′+ab′‾−ba′−b′a‾=0\Rightarrow \underline {ab' + ab'} - \underline {ba' - b'a} = 0
When you combine two terms with the negative sign add both the terms and then give negative sign to the resultant value.
⇒2ab′−2ba′=0\Rightarrow 2ab' - 2ba' = 0
Move one term to the opposite side, when you move any term from one side to another side then the sign of the term also changes. Negative term becomes positive and vice-versa.
⇒2ab′=2ba′\Rightarrow 2ab' = 2ba'
Common multiples from both the sides of the equation cancels each other.
⇒ab′=ba′\Rightarrow ab' = ba'
Therefore, the correct answer is Option (1).

Note : Remember expansion of determinant can also be done by first reducing the determinant’s value and if possible, by row transformation making the elements of the row to be zero and then opening the determinant. Be careful about the sign convention while simplifying remember minus minus will be plus.