Question

please solve the math and give its solution

Answer 1

$\underline{\mathsf{\mathfrak{SOLUTION}}}$

$\textsf{From the point ( 4 , 3 ) perpendicular is dropped on both axes .}$

On the x - axis the perpendicular will fall on ( 4 , 0 )

By Distance formula :

$\mathsf{p=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\mathsf{\implies p=\sqrt{(4-4)^2+(3-0)^2}}$

$\mathsf{\implies p=\sqrt{0^2+3^2}}$

$\mathsf{\implies p=\sqrt{9}}$

$\mathsf{\implies p=3}$

On the y-axis the perpendicular will fall on ( 0 , 3 )

Again by distance formula :

$\mathsf{q=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\mathsf{\implies q=\sqrt{(4-0)^2+(3-3)^2}}$

$\mathsf{\implies q=\sqrt{4^2+0^2}}$

$\mathsf{\implies q=\sqrt{16}}$

$\mathsf{\implies q=4}$

$\textsf{ p = 3 and q = 4 }$

$\textsf{L.C.M of p and q is 12}$

$\mathsf{4p=12}$

$\mathsf{3q=12}$

$\mathsf{\implies 4p=3q}$

$\textsf{\underline{\underline{\underline{\huge{ANSWER}}}}}$

$\underline{\bf{OPTION\:\:C}}$

$\textsf{Hope it helps :-)}$

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