Question

If A (x,O)B(0,Y)and C (1,1)are in collinear then find the value of 1/x+1/y​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\red{ \huge{ \underline{ \overline{ \mid{ \bold{ \purple{ \: \: SOLUTION\: \: \red{ \mid}}}}}}}}$

$\underline{ \underline{ \bold{ \: \: GIVEN \: \: \: : }}} \to \\ \\ \bold{The \: \: points \: \: A(x, \: 0) \: \: ;\: B \: (0 ,\: y) \: \: and \: \: \: C(1 ,\: 1)} \\ \bold{are \: \: in \: \: collinear.}$

$\underline{ \underline{ \bold{ \: \: TO \: \: FIND \: \: : }}} \to \: \: \: \: \: \bold{Value \: \: of \: \: \frac{1}{x} + \frac{1}{y} = ?}$

$\star \: \: \: \underline{\bold{ \: We \: \: know \: \: that \: \:} } \to \\ \\ \bold{If \: \: the \: \: points \: \: (a_1 \:, b_1) \:; \: (a_2 \:, b_2) \: \: and \: \: (a_3 \:, b_3)} \\ \bold{are \: \: in \: \: collinear \: \: then \: the \: \:are \: \: of \: \: the \: \: triaagle } \\ \bold{made \: \: by \: \: these \: \: points \: \: is \: \: equal \: \: to \: \: 0.} \\ \\ \bold{ i.e,} \\ \\ \boxed{\bold{ \frac{1}{2} [a_1(b_2 - b_3) + a_2(b_3 - b_1) + a_3(b_1 - b_2)] = 0}}$

$\underline{ \bold{ \: \: Suppose \: \: here \: \: }} \\ \\ \bold{a_1 = x \: \:, \: b_1 = 0} \\ \\ \bold{a_2 = 0 \: \: ,\: b_2 = y} \\ \\ \bold{a_3 = 1 \: \:, \: b_3 = 1}$

$\underline{ \bold{ \: \: Using \: \: the \: \: above \: \: formula \: \: }}$

$\bold{ \frac{1}{2} [x(y - 1) + 0(1 - 0) + 1(0 - y)] = 0} \\ \\ \bold{ = > x(y - 1) + 0 \times 1+ 1( - y) = 0} \\ \\ \bold{ = > xy - x + 0 - y = 0} \\ \\ \bold{ = > - x - y + xy = 0} \\ \\ \bold{ = > - x - y = - xy} \\ \\ \bold{ = > x + y = xy} \\ \\ \bold{ = > \frac{x + y}{xy} = \frac{xy}{xy} } \\ \\ \bold{ = > \frac{x}{xy} + \frac{y}{xy} = 1 } \\ \\ \bold{ = > \frac{1}{y} + \frac{1}{x} = 1 } \\ \\ \bold{ = > \boxed{ \bold{ \: \frac{1}{x} + \frac{1}{y} = 1 \: }}}$