Question

If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length

[1 mark]

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Answer 1

length was AOB triangle was 12 and I hope length was 90digry.............

Answer 2

$\mathfrak{ \red{question: - }} \\ \\ {If \: the \: graph \: of \: the \: equation \: 4x + 3y = 12 \: } \\ {cuts \: the \: coordinate \: axes \: at \: A \: andm B, \: then \: } \\ {hypotenuse \: of right \: triangle \: AOB \: is \: of \: length.}</p><p> \\ \\ \mathfrak{ \red{solution : - }} \\ \\ we \: know \: right \: angle \: triangle \: having \: a\: \\ hypotenuse. \\ \\ \mathfrak{ \red{given : -}} \\ equation \: is \: \: 4x + 3y = 12 \\ \\ \red{\textit{When the graph cuts x-axis, }} \\ \red{\textit{y coordinate becomes zero.}} \\ \\ = > y \: = 0 \\ \\ 4x + 3(0) = 12 \\ \\ = > 4x = 12 \\ \\ = > x = \frac{12}{4} \\ \\ = > x = 3 \: \: \: \: \: ..........(1) \\ \\ \\ \\ \red{\textit{so same here when the graph cuts y-axis, }} \\ \red{\textit{x coordinate becomes zero.}} \\ \\ = > x = 0 \\ \\ = > 4(0) + 3y = 12 \\ \\ = > 3y = 12 \\ \\ = > y = \frac{12}{3} \\ \\ = > y = 4 \: \: \: \: \: ...........(2) \\ \\ \: \textit{If OA=3 units \:and \: </p><p>OB=4 units.}</p><p> \: \: \\ \\ we \: know \: AO+ OB = AOB = AB \\ \\ \mathfrak{ \red{by \: pythagorus \: theorem}} \: \\ \\ = > {AB}^{2} = {AO}^{2} + {OB}^{2} \\ \\ = > {AB}^{2} = {3}^{2} + {4}^{2} \\ \\ = > {AB}^{2} = 9 + 16 \\ \\ = > {AB}^{2} = 25 \\ \\ = > {AB} = \sqrt{25} \\ \\ = > {AB} = 5 \: units \\ \\ \underline\purple{\: hence \: the \: length \: of \: hypotenious \: } \\ \underline\purple{ is \: 5 \: units.} \\ \\ \fcolorbox{red}{white}{i \: hope \: it \: helps \: you.}$