Question

bplz plz send me answer for my exam​

Answer 1

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

$\bold{If \: \: (x \: ,\: y) \: ;\: (a \:, \: 0) \: \: and \: \: (0 \:, \: b) \: \: are \: \: collinear} \\ \bold{points \: ;\: prove \: \: that \: \: \frac{x}{a} + \frac{y}{b} = 1.}$

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

$\underline{ \underline{ \bold{ \: \: GIVEN \: \: }}} \to \\ \\ \bold{(x \: ,\: y) \:; \: ( a \: ,\: 0) \: \: and \: \: (0 \:, \: b) \: \: are \: \: collinear \: \: points.} \\ \\ \underline{\underline{\bold{ \: \: TO \: \: PROVE \: \: }}} \to \: \: \: \bold{ \frac{x}{a} + \frac{y}{b} = 1}$

$\underline{ \bold{ \: \: We \: \: know \: \: that \: \: }} \\ \\ \bold{If \: \: the \: \: points \: \: (x_1 \: ,\: y_1) \:; \: (x_2 \: ,\: y_2) \: \: and} \\ \bold{(x_3 \: ,\: y_3) \: \: points \: \: are \: \: collinear \:; \: then} \\ \bold{the \: \: area \: \: of \: \: the \: \: triangle \: \: made \: \: by} \\ \bold{these \: \: points \: \: will \: \: be \: \: 0.} \\ \\ \bold{i.e} \\ \\ \boxed{\bold{ \frac{1}{2}[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] = 0}}$

$\bold{Suppose \: \: here,} \\ \\ \bold{(x_1 \: \: y_1) = (x \: ,\: y)} \\ \\ \bold{(x_2 \:, \: y_2) = (a \: ,\: 0)} \\ \\ \bold{(x_3 \: \: y_3) = (0 \:, \: b)} \\ \\ \\ \underline{ \bold{ \: \: Using \: \: the \: \: above \: \: formula \: \: }} \\ \\ \bold{ \frac{1}{2} [x(0 - b) + a(b - y) + 0(y - 0)] = 0} \\ \\ \bold{ \Longrightarrow \: \: [x \times ( - b) + ab - ay + 0] = 0} \\ \\ \bold{ \Longrightarrow \: \: - xb + ab - ay = 0} \\ \\ \bold{ \Longrightarrow \: \: - xb - ay = - xy} \\ \\ \bold{ \Longrightarrow \: \: - (xb + ay) = - xy} \\ \\ \bold{ \Longrightarrow \: \: xb + ay = ab} \\ \\ \bold{ \Longrightarrow \: \: \frac{xb + ay}{ab} = \frac{ \cancel{ab}}{ \cancel{ab}} } \\ \\ \bold{ \Longrightarrow \: \: \frac{x \: \cancel{b}}{a \: \cancel{b}} + \frac{ \cancel{a} \: y}{ \cancel{a} \: b} = 1} \\ \\ \bold{ \therefore \: \: \underline{ \: \: \: \: \frac{x}{a} + \frac{y}{b} = 1 \: \: \: \: }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\bold{ \: \: Hence \: \: Proved. \: \: }}$

Answer 2

$\huge\mathbb\red{QUESTION}$

$\begin{lgathered}\bold{If \: \: (x \: ,\: y) \: ;\: (a \:, \: 0) \: \: and \: \: (0 \:, \: b) \: \: are \: \: collinear} \\ \bold{points \: ;\: prove \: \: that \: \: \frac{x}{a} + \frac{y}{b} = 1.}\end{lgathered}$

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

$\begin{lgathered}\underline{ \underline{ \bold{ \: \: GIVEN \: \: }}} \to \\ \\ \bold{(x \: ,\: y) \:; \: ( a \: ,\: 0) \: \: and \: \: (0 \:, \: b) \: \: are \: \: collinear \: \: points.} \\ \\ \underline{\underline{\bold{ \: \: TO \: \: PROVE \: \: }}} \to \: \: \: \bold{ \frac{x}{a} + \frac{y}{b} = 1}\end{lgathered}$

$\begin{lgathered}\underline{ \bold{ \: \: We \: \: know \: \: that \: \: }} \\ \\ \bold{If \: \: the \: \: points \: \: (x_1 \: ,\: y_1) \:; \: (x_2 \: ,\: y_2) \: \: and} \\ \bold{(x_3 \: ,\: y_3) \: \: points \: \: are \: \: collinear \:; \: then} \\ \bold{the \: \: area \: \: of \: \: the \: \: triangle \: \: made \: \: by} \\ \bold{these \: \: points \: \: will \: \: be \: \: 0.} \\ \\ \bold{i.e} \\ \\ \boxed{\bold{ \frac{1}{2}[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] = 0}}\end{lgathered}$

$\begin{lgathered}\bold{Suppose \: \: here,} \\ \\ \bold{(x_1 \: \: y_1) = (x \: ,\: y)} \\ \\ \bold{(x_2 \:, \: y_2) = (a \: ,\: 0)} \\ \\ \bold{(x_3 \: \: y_3) = (0 \:, \: b)} \\ \\ \\ \underline{ \bold{ \: \: Using \: \: the \: \: above \: \: formula \: \: }} \\ \\ \bold{ \frac{1}{2} [x(0 - b) + a(b - y) + 0(y - 0)] = 0} \\ \\ \bold{ \Longrightarrow \: \: [x \times ( - b) + ab - ay + 0] = 0} \\ \\ \bold{ \Longrightarrow \: \: - xb + ab - ay = 0} \\ \\ \bold{ \Longrightarrow \: \: - xb - ay = - xy} \\ \\ \bold{ \Longrightarrow \: \: - (xb + ay) = - xy} \\ \\ \bold{ \Longrightarrow \: \: xb + ay = ab} \\ \\ \bold{ \Longrightarrow \: \: \frac{xb + ay}{ab} = \frac{ \cancel{ab}}{ \cancel{ab}} } \\ \\ \bold{ \Longrightarrow \: \: \frac{x \: \cancel{b}}{a \: \cancel{b}} + \frac{ \cancel{a} \: y}{ \cancel{a} \: b} = 1} \\ \\ \bold{ \therefore \: \: \underline{ \: \: \: \: \frac{x}{a} + \frac{y}{b} = 1 \: \: \: \: }}\end{lgathered}$

$\underline{\bold{ \: \: Hence \: \: Proved. \: \: }}$