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The problem asks us to find the value of \( m \) such that the two lines:

1. \( 3y + mx - 15 = 0 \)
2. \( y - 2mx + 5x - 3m + 4 = 0 \)

are perpendicular to each other.

### Step 1: Rewriting both equations in slope-intercept form

For two lines to be perpendicular, the product of their slopes must be \( -1 \). Let’s rewrite each equation in the form \( y = mx + b \) (slope-intercept form) to determine their slopes.

#### Line 1: \( 3y + mx - 15 = 0 \)
Solve for \( y \):
\[
3y = -mx + 15
\]
\[
y = -\frac{m}{3}x + 5
\]
So, the slope of the first line is \( -\frac{m}{3} \).

#### Line 2: \( y - 2mx + 5x - 3m + 4 = 0 \)
Combine like terms:
\[
y = (2m - 5)x + (3m - 4)
\]
So, the slope of the second line is \( 2m - 5 \).

### Step 2: Applying the perpendicularity condition

For the lines to be perpendicular, the product of their slopes must equal \( -1 \):
\[
\left(-\frac{m}{3}\right)(2m - 5) = -1
\]
Multiply both sides:
\[
-\frac{m}{3}(2m - 5) = -1
\]
Simplify:
\[
\frac{m(2m - 5)}{3} = 1
\]
Multiply both sides by 3 to eliminate the denominator:
\[
m(2m - 5) = 3
\]
Expand:
\[
2m^2 - 5m = 3
\]
Rearrange into a standard quadratic form:
\[
2m^2 - 5m - 3 = 0
\]

### Step 3: Solving the quadratic equation

Now, solve the quadratic equation \( 2m^2 - 5m - 3 = 0 \) using the quadratic formula:
\[
m = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}
\]
\[
m = \frac{5 \pm \sqrt{25 + 24}}{4}
\]
\[
m = \frac{5 \pm \sqrt{49}}{4}
\]
\[
m = \frac{5 \pm 7}{4}
\]
Thus, we have two possible solutions for \( m \):
\[
m = \frac{5 + 7}{4} = 3 \quad \text{or} \quad m = \frac{5 - 7}{4} = -\frac{1}{2}
\]

### Final Answer:
The possible values of \( m \) are \( 3 \) and \( -\frac{1}{2} \).

---

Do you need further details or clarifications on any part of the solution?

Here are 5 related questions:
1. How do we derive the slope-intercept form from a general linear equation?
2. What is the geometric interpretation of perpendicular lines in a coordinate plane?
3. How do you solve a quadratic equation using the quadratic formula?
4. What other methods could be used to solve \( 2m^2 - 5m - 3 = 0 \)?
5. How do different values of \( m \) affect the slopes and positions of lines?

**Tip:** When lines are perpendicular, their slopes multiply to give -1. This is a key relationship in many geometry problems.

The two lines 3y + mx - 15 = 0 and y - 2mx + 5x - 3m + 4 = 0 are perpendicular for some real number(s) m. What is m?

Discover how to find the value of m for which two lines are perpendicular using slope-intercept form and quadratic equations. This problem is suitable for high school students learning algebra and geometry concepts.

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