Question

Find the coordinates of foot of perpendicular from the point $( - 1,3)$ to the line
$3x - 4y - 16 = 0$.

Answer 1

Hint – Find the slope of the line perpendicular to the given line then apply the condition of perpendicular lines for slopes that is ${m_1}{m_2} = - 1$.
We know the foot of perpendicular lies on a line perpendicular to line $3x - 4y - 16 = 0$.
Slope of the line joining $( - 1,3)$ and $(a,b)$, is ${m_1} = \dfrac{{b - 3}}{{a + 1}}\,\,\,\,\,\,\,\,\,...({\text{i}})$
Slope of the line $3x - 4y - 16 = 0$ or $y = \dfrac{3}{4}x - 4$, ${m_2} = \,\dfrac{3}{4}$
Since the two lines are perpendicular ,

$${m_1}{m_2} = - 1 \\\ {m_1}\, = \dfrac{{ - 4}}{3}\,\,\,\,\,...({\text{ii}}) \\\$$

From {\text{(i) & (ii)}} we know ,

$$\dfrac{{b - 3}}{{a + 1}} = \dfrac{{ - 4}}{3}\, \\\ 3b - 9 = - 4a - 4 \\\ 4a + 3b - 5 = 0\,\, \\\$$

On multiplying ${\text{4}}$ to above equation we get,
$16a + 12b - 20 = 0\,\,\,\,\,...({\text{iii}})$
The point ${\text{(}}a,b{\text{)}}$ lies on the line $3x - 4y - 16 = 0$
$3a - 4b - 16 = 0$
On multiplying $3$ to the above equation we get,
${\text{9}}a - 12b - 48 = 0\,\,\,\,...({\text{iv}})$
Adding equation {\text{(iii) & (iv)}} we get,

$$25a = 68 \\\ a = \dfrac{{68}}{{25}} \\\$$

On putting the value of $a$ in ${\text{(iv)}}$ we get the value of $b$ as,
$b = \dfrac{{49}}{{25}}$
Hence the coordinates of foot of perpendicular is $\left( {\dfrac{{68}}{{25}},\dfrac{{49}}{{25}}} \right)$.
Note – In these type of question we have to find the slope of perpendicular line with the help of given line then we will get an equation which will pass through the given point and foot of perpendicular also and we know that the foot of perpendicular will pass through the given line. Then on getting two equations we can get those points by solving those equations.