We are given two straight lines:

$L_1: ax + by - 30 = 0$ $L_2: 5x - 2y + 4 = 0$

They are perpendicular to each other and intersect at a point on the y-axis. We are asked to find the values of $a$ and $b$.

Step 1: Condition for Perpendicularity

For two lines to be perpendicular, the product of their slopes must equal $-1$.

The general form of a line is $Ax + By + C = 0$, and the slope of the line is $-\frac{A}{B}$.

• For $L_1: ax + by - 30 = 0$, the slope is $-\frac{a}{b}$.
• For $L_2: 5x - 2y + 4 = 0$, the slope is $-\frac{5}{-2} = \frac{5}{2}$.

Using the condition for perpendicularity: $\left(-\frac{a}{b}\right) \times \frac{5}{2} = -1$

Simplifying: $\frac{a}{b} \times \frac{5}{2} = 1 \quad \Rightarrow \quad \frac{5a}{2b} = 1 \quad \Rightarrow \quad 5a = 2b \quad \Rightarrow \quad b = \frac{5a}{2}$

Step 2: Condition for Intersection on the y-axis

The lines intersect on the y-axis, which means that at the point of intersection, $x = 0$.

Substitute $x = 0$ into both equations.

For $L_1: ax + by - 30 = 0$, when $x = 0$: $b(0) + by - 30 = 0 \quad \Rightarrow \quad by = 30 \quad \Rightarrow \quad y = \frac{30}{b}$

For $L_2: 5x - 2y + 4 = 0$, when $x = 0$: $5(0) - 2y + 4 = 0 \quad \Rightarrow \quad -2y = -4 \quad \Rightarrow \quad y = 2$

Thus, the intersection point is $y = 2$.

Step 3: Solving for $a$ and $b$

We now know that $y = \frac{30}{b} = 2$, so: $\frac{30}{b} = 2 \quad \Rightarrow \quad b = \frac{30}{2} = 15$

Now, substitute $b = 15$ into the equation $b = \frac{5a}{2}$: $15 = \frac{5a}{2} \quad \Rightarrow \quad 30 = 5a \quad \Rightarrow \quad a = 6$

Final Answer:

The values of $a$ and $b$ are: $a = 6, \quad b = 15$

Would you like more details or have any further questions?

Here are some related questions for further practice:

1. What is the equation of the line perpendicular to $L_2$ that passes through a given point?
2. How do you find the point of intersection between two lines?
3. How can we confirm if two lines are parallel?
4. What is the geometric interpretation of the slope of a line?
5. How do you find the distance between a point and a line?

Tip: When lines intersect at the y-axis, you can directly substitute $x = 0$ to simplify the equations.