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Find the equation of the straight line joining the point $(4,1)$ to the foot of the perpendicular drawn from the point $(3,2)$ on line $2x-3y=1$.

My Approach: Given equation: $$2x-3y=1$$ Slope of this line is $\frac {2}{3}$ Now the equation of the line perpendicular to this and passing through $(3,2)$ is $$y-2=\frac {-3}{2} (x-3)$$ $$ 2y-4=-3x+9$$ $$3x+2y-13=0$$

Now, please help me to continue.

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You want to find the point where $2x-3y=1$ and $3x+2y-13=0$ cross so:

$2x-3y=1\Leftrightarrow y=\frac{2}{3}x-\frac{1}{3}$

$3x+2y-13=0\Leftrightarrow y=-\frac{3}{2}x+\frac{13}{2}$

We are searching for the $x$ such that:

$\frac{2}{3}x-\frac{1}{3}=-\frac{3}{2}x+\frac{13}{2}$

giving $x=\frac{41}{13}$ so the point of intersection is $\left ( \frac{41}{13},\frac{23}{13} \right )$

Considering that the line we're looking for can be written as $y=ax+b$, all we need is to solve the system of equations:

$ \left\{\begin{matrix} \frac{23}{13}=\frac{41}{13}a+b\\ 4=a+b \end{matrix}\right.$

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Now that you have the equation of the perpendicular line, you can find the coordinates of the point that lies at the intersection of your two lines.

Let's name it $C = (x_C, y_C)$. Then, $C$ satisfies the two following conditions : $$ 2x_C -3y_C = 1 \quad \mbox{and} \quad 3x_C+2y_C -13 = 0 $$ which are just the mathematical way of saying that $C$ is on the two lines. We solve the system : $$ C = (41/13, \ 23/13) $$

Now, you only need to compute the slope of the line between $C$ and $(4,1)$ and deduce the equation of the straight line from here.

The slope : $\frac{1-23/13}{4-41/13} = -\frac{10}{11}$.

So the equation is of the form : $y = -10/11 x + k$ where $k$ is a constant. Knowing that $(4,1)$ is on the line, we plug in the coordinates to get $k$ : $1 = -10/11 \times 4 + k \implies k = 51/11$

In the end, you should find something of the form $y = -\frac{10}{11}x + \frac{51}{11}$

Construction

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You have two lines; the foot of the perpendicular is the point where those lines intersect. So you need to solve the system $$2x-3y=1, \quad 3x+2y-13=0,$$ giving some point $(a,b)$. You can then find the equation of the line passing through $(4,1)$ and $(a,b)$.

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