Question

Let $ax+by+c=0$ be the equation of a straight line such that $3a+2b+4c=0$. Which one of the following points lies on the line?

• A. $(\frac{3}{4}, \frac{1}{2})$
• B. $(\frac{1}{2}, \frac{3}{4})$
• C. $(\frac{1}{4}, \frac{3}{2})$
• D. $(\frac{3}{2}, \frac{1}{2})$
• E. (2, 4)

Hint:

To find a point on a line from a linear combination of coefficients, normalize the equation so the constant term matches the standard form.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a condition that the coefficients of a general line equation must satisfy. We need to find a specific point \( (x, y) \) that lies on this line regardless of the specific values of a, b, and c (as long as they satisfy the condition). This means the point's coordinates must make the line equation \( ax + by + c = 0 \) look like the given condition.
Step 2: Key Formula or Approach:
The given condition is \( 3a + 2b + 4c = 0 \).
The equation of the line is \( ax + by + c = 0 \).
We need to manipulate the condition so that it matches the form of the line equation.
Step 3: Detailed Explanation:
Start with the condition:
\[ 3a + 2b + 4c = 0 \] Our goal is to make this equation resemble \( ax + by + c = 0 \). The term 'c' in the line equation has a coefficient of 1. Let's achieve this in our condition by dividing the entire equation by 4 (assuming \( c \neq 0 \)).
\[ \frac{3a}{4} + \frac{2b}{4} + \frac{4c}{4} = \frac{0}{4} \] \[ \frac{3}{4}a + \frac{1}{2}b + c = 0 \] Now, let's rearrange the terms to match the \( ax + by + c = 0 \) format:
\[ a\left(\frac{3}{4}\right) + b\left(\frac{1}{2}\right) + c = 0 \] By comparing this equation with the general line equation \( ax + by + c = 0 \), we can see that if we substitute \( x = \frac{3}{4} \) and \( y = \frac{1}{2} \), the line equation becomes the given condition.
This means that the point \( \left(\frac{3}{4}, \frac{1}{2}\right) \) lies on the line for any a, b, c that satisfy the condition.
Step 4: Final Answer:
The point that lies on the line is \( \left(\frac{3}{4}, \frac{1}{2}\right) \).