Find the projection of the point $(-6,4)$ onto the line $4x-5y+3=0$
I can find the distance between the point and the line, but I do not think it can help
2026-04-24 12:59:29.1777035569
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Projection of a point on a line
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I assume you're talking about the orthogonal projection. So let's find the equation of the perpendicular to the given line through the given point. The equation of a perpendicular is of the form
$$5x+4y+c=0$$
It passes through $(-6,4)$ so $c=30-16=14$. The projection (orthogonal) is the intersection of the two lines
$$\begin{cases}4x-5y+3=0\\5x+4y+14=0\end{cases}$$
The intersection point is $(-2,-1)$
One way to solve this problem is to first find the line that goes through $(-6,4)$ and is perpendicular to $4x-5y+3=0$. Since the slope of the given line is $m_0=\frac{4}{5}$, the slope of a line perpendicular to $4x-5y+3=0$ will be $m_1=-\frac{5}{4}$. Now you have a point $(-6,4)$ and a slope $m_1=-\frac{5}{4}$, so it is simple to find the line defined by that point and slope.
The projection of $(-6,4)$ to $4x-5y+3=0$ will then be the intersection of the given line and the perpendicular line found.