The sides of a triangle are on the lines $2x+3y+4=0$, $ \ \ x-y+3=0$, and $5x+4y-20=0$. Find the equations of the altitudes of the triangle.
Should I find the vertices first? Or is there a direct way? Actually, I tried finding the vertices, using the substitution method but I find it hard to turn it into an equation.
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$\begingroup$Plan: For the given equation of line as $$ax + by + c = 0 \tag{1}$$ you can find the general equation of perpendicular lines as $$bx - ay + d = 0 \tag{2}$$for some $d\in \mathbb R$. Now, you just need to find the values of $d$.
In order to find $d$, find the vertex point (the intersection of the other two sides than $(1)$) and plug that into $(2)$ (because the perpendicular line should pass through that vertex). Since $a$ and $b$ are known values you can derive the value of $d$ from there.
Solution: Take the third equation, for example. $$5x+4y-20=0 \tag{1}$$Then, the general equation of perpendicular lines is $$4x-5y+d = 0 \tag{2}$$Now, we will find the intersection of the other two lines than $(1)$.$$\begin{cases} 2x + 3y + 4 = 0 \\ x - y + 3 = 0\end{cases} \implies (x,y) = \left(-\frac{13}{5} , \frac{2}{5}\right)$$Now plug this into $(2)$ to get$$4\left(-\frac{13}{5}\right) - 5 \left(\frac{2}{5}\right) + d = 0 \\ d= \frac{62}{5}$$Thus, the equation of altitude to the side which is given by $(1)$ is $$\boxed{4x-5y + \frac{62}{5} = 0}$$ You can find the other two altitudes with the same method.
$\endgroup$ 3 $\begingroup$$$r:2x+3y+4=0,\;s:x-y+3=0,\;t:5x+4y-20=0$$The set of all lines passing through the intersection point $r\cap s$ is$$\mathcal{F}:a(2x+3y+4)+b(x-y+3)=0\tag{1}$$$$\mathcal{F}:(2a+b)x+(3a-b)y+4a+3b=0$$To find the one perpendicular to $t:5x+4y-20=0$ we use the condition$$5(2a+b)+4(3a-b)=0\to b=-22a$$plug this value in the equation $(1)$ and get
$a(2x+3y+4)-22a(x-y+3)=0$
simplify $a$ and get the first height equation$$20 x-25 y+62=0$$In a similar way we can get the other two equations$$27x-18y+46=0;\;7x+7y-16=0$$
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