The foot of the perpendicular from (2, 3) upon the line 4x - 5y + 8 = 0 is

Question

Prepp Team · Accepted Answer

Finding the Foot of the Perpendicular

We are asked to find the coordinates of the foot of the perpendicular from a given point (2, 3) to the line 4x - 5y + 8 = 0. The foot of the perpendicular is the point on the line where a perpendicular line segment from the given point meets the line.

Method to Find the Foot of the Perpendicular

Let the given point be $P(x_1, y_1)$ and the equation of the line be $ax + by + c = 0$. The coordinates of the foot of the perpendicular, let's call it $Q(x_2, y_2)$, can be found using the formula:

\[ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2} \]

Applying the Formula

In this problem:

Given point $P(x_1, y_1) = (2, 3)$
Given line equation: $4x - 5y + 8 = 0$
Comparing the line equation with $ax + by + c = 0$, we have $a = 4$, $b = -5$, and $c = 8$.

Now, substitute these values into the formula to find the foot of the perpendicular $(x_2, y_2)$:

\[ \frac{x_2 - 2}{4} = \frac{y_2 - 3}{-5} = -\frac{4(2) + (-5)(3) + 8}{4^2 + (-5)^2} \]

Let's calculate the value of the fraction on the right side:

\[ -\frac{4(2) + (-5)(3) + 8}{4^2 + (-5)^2} = -\frac{8 - 15 + 8}{16 + 25} = -\frac{1}{41} \]

So the formula becomes:

\[ \frac{x_2 - 2}{4} = \frac{y_2 - 3}{-5} = -\frac{1}{41} \]

Now, we can find $x_2$ and $y_2$ by equating the first and second parts of the formula to the calculated value:

Finding $x_2$

\[ \frac{x_2 - 2}{4} = -\frac{1}{41} \] \[ x_2 - 2 = 4 \times \left(-\frac{1}{41}\right) \] \[ x_2 - 2 = -\frac{4}{41} \] \[ x_2 = 2 - \frac{4}{41} \] \[ x_2 = \frac{2 \times 41 - 4}{41} = \frac{82 - 4}{41} = \frac{78}{41} \]

Finding $y_2$

\[ \frac{y_2 - 3}{-5} = -\frac{1}{41} \] \[ y_2 - 3 = -5 \times \left(-\frac{1}{41}\right) \] \[ y_2 - 3 = \frac{5}{41} \] \[ y_2 = 3 + \frac{5}{41} \] \[ y_2 = \frac{3 \times 41 + 5}{41} = \frac{123 + 5}{41} = \frac{128}{41} \]

Therefore, the foot of the perpendicular from the point (2, 3) to the line 4x - 5y + 8 = 0 is $\left(\frac{78}{41}, \frac{128}{41}\right)$. This point is the foot of the perpendicular onto the given line.

This calculation confirms the coordinates of the foot of the perpendicular.