# Thread: Kindly help me for this triangle problem

1. ## Kindly help me for this triangle problem

The problem is attached. It is triangle and its peak is having value 1

the sides are at a and -a. The new point is on a/4. from which i have drawn perpendicular line

i want to know what is the value of that point on y-axis scale.
book says it is 0.733
but how???

2. The problem is attached. It is triangle and its peak is having value 1

the sides are at a and -a. The new point is on a/4. from which i have drawn perpendicular line

i want to know what is the value of that point on y-axis scale.
book says it is 0.733
but how???
Since I would consider this a typical proportional problem, I wonder that too?

If a can be any number, triangle height y can be 0<y<1 ?

3. Originally Posted by moonnightingale
The problem is attached. It is triangle and its peak is having value 1

the sides are at a and -a. The new point is on a/4. from which i have drawn perpendicular line

i want to know what is the value of that point on y-axis scale.
book says it is 0.733
but how???
The two known points on that line segment are

$(a, 0)$ and $(0, 1)$.

$m = \frac{1 - 0}{0 - a}$

$= \frac{1}{-a}$

$= -\frac{1}{a}$.

The $y$ intercept is $1$, so $c = 1$.

Therefore the line segment has equation

$y = -\frac{1}{a}\,x + 1$

$y = 1 - \frac{x}{a}$.

So if $x = \frac{a}{4}$

$y = 1 - \frac{\frac{a}{4}}{a}$

$= 1 - \frac{1}{4}$

$= \frac{3}{4}$.

4. one can do that?
isn't that like saying a=y=1 ?

so all you get is a mirror?
y = 1-x ?

I'm asking since I really don't know.

//====edit
aren't you supposed to plug that back to the equation? so that:
3/4 = -(1/a)x + 1

//====edit 2
wow, that was really stupid of me of course 3/4