Question

the length of the perpendicular from the point (a,3) on the line 3x+4y+5=0 is 4 find value of a

Answer 1

$The \: \: answer \: \: is \: \: given \: \: below \\ \\ RULE \\ \\ Let ,\: \: any \: \: straight\: \:line \: \: be \\ ax + by + c = 0 \\ and \: \: any \: \: point \: \: be \: \: (p,q). \\ \\ Then, \: \: the \: \: required \: \: distance \\ from \: \: the \: \: point \: \: (p,q) \: \: on \: \: \\ to \: \: the \: \: line \: \: ax + by + c = 0 \: \: be \\ = \frac{ ap + bq + c}{ \sqrt{ {a}^{2} + {b}^{2} } } \: \: units \\ \\ SOLUTION \\ \\ The \: \: given \: \: line \: \: is \\ 3x + 4y + 5 = 0 \\ \\ So, \: \: the \: \: distance \: \: of \: \: this \: \: line \: \: \\ from \: \: the \: \: point \: \: (a,3) \: \: is \\ = \frac{ 3a + 12 + 5 }{ \sqrt{ {3}^{2} + {4}^{2} } } \: \: units \\ \\ = \frac{ 3a + 17 }{ \sqrt{9 + 16} } \: \: units \\ \\ = \frac{3a + 17}{ \sqrt{25} } \: \: units \\ \\ = \frac{3a + 17}{5} \: \: units \\ \\ Now, \: \: give n\: \: that \\ \\ \frac{3a + 17}{5} = 4 \\ \\ or \: \: 3a + 17 = 20 \\ \\ or \: \: 3a = 20 - 17 \\ \\ or \: \: 3a = 3 \\ \\ So ,\: \: a = 1 \\ \\ Thank \: \: you \: \: fo r\: \: your \: \: question.$