Question

If the three points (a,0), (0, b) and (1,1) are collinear then let us show that
1 by a plus 1 by b is equal to 1​

Answer 1

$\bigstar \underline{\underline{\mathbb{GIVEN:-}}}$

• Three points (a,0);(0,b);&(1,1) are collinear.

$\bigstar \underline{\underline{\mathbb{TO\ PROVE:-}}}$

• $\sf \frac{1}{a} +\frac{1}{b} =1$

$\bigstar \underline{\underline{\mathbb{PROOF:-}}}$

We need to know:

❥︎$\\ \boxed{\sf \orange{Area\ of\ triangle=\frac{1}{2} \left|\begin{array}{ccc}\sf x_1&\sf y_1&1\\\sf x_2&\sf y_2&1\\\sf x_3&\sf y_3&1\end{array}\right| }}$

❥ As the points are collinear , area of triangle should be equal to zero.

Acc to question:

$\mapsto \sf x_1=a,x_2=0,x_3=1\\\\\mapsto \sf y_1=0,y_2=b,y_3=1$

Substituiting the values in the formula ,

$:\implies \sf Area =\frac{1}{2} \left|\begin{array}{ccc}a&0&1\\0&b&1\\1&1&1\end{array}\right|=0$

$:\implies \sf \frac{1}{2} [a(b-1)+0(1-0)+1(0-b)]=0\\\\:\implies \sf \frac{1}{2}(ab-a+0-b)=0 \\\\:\implies \sf ab-a-b=0\\\\:\implies ab-(b+a)=0\\\\:\implies b+a=ab$

Dividing both sides by 'ab' we get :

$:\implies \sf \frac{a}{ab} +\frac{b}{ab} =\frac{ab}{ab} \\\\:\implies \sf \orange{ \boxed{\boxed{\sf \frac{1}{a} +\frac{1}{b} =1 }}}$

Hence proved !

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ღ Know More :-

$\\\\\sf If\ A(x_1,y_1);B(x_2,y_2)&C(\sf x_3,y_3) \ are\ three\ points,$

For the points to be collinear:

❥ They should lie on same straight line.

❥ Slope of line AB = Slope of AC

$\sf i.e.,\ \frac{y_1-y_2}{x_1-x_2} =\frac{y_1-y_3}{x_1-x_3}$

❥ Sum of lengths of any two line segments should be equal to length of remaining line segment .

i.e., AB +BC=AC

      AB+AC=BC

       AC+BC=AB

❥ And to find length of line segment,we use the distance formula.

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HOPE IT HELPS !

Answer 2

Answer:

$\huge\rm{Question}$

$\rm \: If \: the \: points \: (a,0) \: (0,b) \: and \: (1,1) \: are \: collinear, \: then \: show \: that \: \frac{1}{a} + \frac{1}{b}[\tex][tex]\rm \: x_{1} (y_{2}- y_{3})+ x_{2}( y_{3}- y_{1})+ x_{3}( y_{1}- y_{2})=0$$\rm \: x_{1}, y_{1}→(a,0) \\ \rm \: x_{2} y_{2}=0,b \\ \rm \: x_3y_3=(1,1)$$\rm \: →a(b-1)+0+1(0-b)=0 \\ \rm \: →ab-a-b=0 \\ \rm \: →ab=a+b \\ \rm \: →1= \frac{a+b}{ab} \\ \rm \: → \frac{a}{ab} + \frac{b}{ab} = 1 \\ \\ \rm→ \frac{1}{a} + \frac{1}{b} =1 \: (proved)$