MHB *find eq of perpendicular line so that the area of the triangle is 8

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To find the equation of a perpendicular line that encloses a triangle with an area of 8 alongside the line y=4x-2, the base and height of the triangle must be calculated. The base is expressed as B=(4x-2)+2+(1/4)x, simplifying to B=(17/4)h, where h is the height. The area formula, Area=(1/2)BH, leads to the equation (b+2)(4b+2)/30=8 for calculating b. The intersection of the two line equations is used to determine the height, H=(4b+2)/15. Ultimately, there are two possible triangles that satisfy the conditions of the problem.
karush
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for the line y=4x-2 there is one perpendicular line of which will enclose a triangle on the lines and the values of the y-axis whose area is 8. What is the equation of this line?

Well, I chose the x value to be the height of the triangle and that would make the base $$B=(4x-2)+2+\frac{1}{4}x$$ or just $B=\frac{17}{4}h$ if $h=$ the $x$ value.

just seeing if I am going the right direction with this seem more complicated than is should be. got to be a slam dunk method...

I got a weird answer for $h=\frac{8}{\sqrt{17}}$ and $b=2\sqrt{17}$
 
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karush said:
View attachment 3268
for the line y=4x-2 there is one perpendicular line of which will enclose a triangle on the lines and the values of the y-axis whose area is 8. What is the equation of this line?

Well, I chose the x value to be the height of the triangle and that would make the base $$B=(4x-2)+2+\frac{1}{4}x$$ or just $B=\frac{17}{4}h$ if $h=$ the $x$ value.

just seeing if I am going the right direction with this seem more complicated than is should be. got to be a slam dunk method...

I got a weird answer for $h=\frac{8}{\sqrt{17}}$ and $b=2\sqrt{17}$

The area of a triangle is given from the formula $$\text{ Area }=\frac{1}{2} B H$$

You will find the part of the basis $B$ that is over the $x$-axis, by setting at the equation $y=\frac{x}{4}+b$, $x=0$. So, it is equal to $b$.
The part of $B$ under the $x$-axis is equal to $2$.

Therefore, $B=b+2$.

You will find the height $H$ by finding the intersection between the two line equations:
$$\frac{x}{4}+b=4x-2 \Rightarrow x=\frac{4b+2}{15}$$

Therefore, $H=\frac{4b+2}{15}$.

So, to calculate $b$ we have to solve the following:

$$B \cdot H=8 \Rightarrow \frac{b+2}{2} \frac{4b+2}{15}=8$$
 
Karush,
Actually, there are 2 such triangles.
r0ozlf.png
 
ok I posted this problem also on Linkedin since it had over 1000 views
but there was an image with this problem which was from a SAT pdf
but I couldn't find it but apparently the image is not necessary to solve it
Anyway
 
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