Question

If the points (a1 b1), (a2 b2)and (a1-a2,b1-b2)are collinear then show that a1b2=a2b1

Answer 1

Here:

($x_{1}$ = $a_{1}$,  $y_{1}$ = $b_{1}$), ($x_{2}$ = $a_{2}$,  $y_{2}$ = $b_{2}$) and ($x_{3}$ = $a_{1}-a_{2}$,  $y_{3}$ = $b_{1}-b_{2}$)

We have to prove that $a_1b_2$ = $a_2b_1$.

Solution:

We know that,

The three points are collinear

$x_{1} (y_{2}-y_{3})+x_{2} (y_{3}-y_{1})+x_{3} (y_{1}-y_{2})$ = 0

Put ($x_{1}$ = $a_{1}$,  $y_{1}$ = $b_{1}$), ($x_{2}$ = $a_{2}$,  $y_{2}$ = $b_{2}$) and ($x_{3}$ = $a_{1}-a_{2}$,  $y_{3}$ = $b_{1}-b_{2}$), we get

∴ $a_{1}$( $b_{2}$ - $b_{1}+b_{2}$) + $a_{2}$( $b_{1}-b_{2}$ - $b_{1}$) + ($a_{1}-a_{2}$)($b_{1}$ - $b_{2}$) = 0

⇒ $a_{1}$(2$b_{2}$ - $b_{1}$) + $a_{2}$( - $b_{2}$) + ($a_{1}-a_{2}$)($b_{1}$ - $b_{2}$) = 0

⇒ 2$a_{1}$$b_{2}$ - $a_{1}$$b_{1}$ - $a_{2}$ $b_{2}$ + $a_{1}$$b_{1}$ - $a_{1}$$b_{2}$ - $a_{2}$ $b_{1}$ + $a_{2}$ $b_{2}$ = 0

⇒ 2$a_{1}$$b_{2}$  - $a_{1}$$b_{2}$ - $a_{2}$ $b_{1}$ = 0

⇒ $a_{1}$$b_{2}$  - $a_{2}$ $b_{1}$ = 0

⇒ $a_{1}$$b_{2}$  = $a_{2}$ $b_{1}$, proved.

Thus, $a_{1}$$b_{2}$  = $a_{2}$ $b_{1}$, proved.

Answer 2

Given :

Three points (a1 b1), (a2 b2) and

(a1-a2,b1-b2) are collinear

To Show : a1b2=a2b1

Solution :

•Area of triangle with its coordinates as ( x1, y1) , ( x2,y2) & ( x3, y3) is

Area = 1/2[ x1(y2-y3) + x2(y3-y1)

+x3(y1-y2) ]

•Since the given three points are collinear then area of the triangle formed by these points will be zero

=> 0 = 1/2[a1(b2-(b1-b2)) +

a2(b1-b2-b1) + (a1-a2)(b1-b2) ]

0 = a1(2b2-b1) + a2(-b2)

+ (a1-a2)(b1-b2)

0 = 2a1b2 - a1b1 -a2b2 + a1b1 - a1b2

- a2b1 + a2b2

0 = 2a1b2 - a1b2 - a2b1

a2b1 = a1b2

Hence proved