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Q.
The coordinates of the foot of perpendicular from the point $(2, 3)$ on the line $y = 3x + 4$ is given by
Straight Lines
Solution:
Let the coordinates of foot of perpendicular be $A(\alpha, \beta)$ from the point $P(2, 3)$ (say) on line $y = 3x + 4$
Slope of $AP$, $m_{1}=\frac{\beta-3}{\alpha-2}$
Slope of given line, $m_{2}=3$
Since, both are perpendicular.
$\therefore m_{1}\times m_{2}=-1$
$\Rightarrow \frac{\beta-3}{\alpha-2} \times3=-1$
$\Rightarrow 3\beta=-\alpha+11\quad\ldots\left(i\right)$
Also, the point $A\left(\alpha, \beta\right)$ is on the given line.
So, $A \left(\alpha, \beta\right)$ satisfy the equation of the line.
$\therefore \beta=3\alpha+4\quad\ldots\left(ii\right)$
On solving $\left(i\right)$ and $\left(ii\right)$, we get $\alpha=-\frac{1}{10}$,
$\beta=\frac{37}{10}$
So, coordinates of foot of perpendicular is $\left(\frac{-1}{10}, \frac{37}{10}\right)$.