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

[karush]

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 $b=2\sqrt{17}$

 

[mathmari]

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:

$$B \cdot H=8 \Rightarrow \frac{b+2}{2} \frac{4b+2}{15}=8$$

 

[johng1]

Karush,
Actually, there are 2 such triangles.

 

[karush]

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