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?
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?
Show 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.
Updated On: Apr 28, 2026
- $(\frac{3}{4}, \frac{1}{2})$
- $(\frac{1}{2}, \frac{3}{4})$
- $(\frac{1}{4}, \frac{3}{2})$
- $(\frac{3}{2}, \frac{1}{2})$
- (2, 4)
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The Correct Option is A
Solution and Explanation
Step 1: Analysis
Divide the given condition $3a+2b+4c=0$ by 4 to make the constant term $c$. $\frac{3}{4}a + \frac{2}{4}b + c = 0 \implies a(\frac{3}{4}) + b(\frac{1}{2}) + c = 0$.
Step 2: Conclusion
Comparing this with $ax + by + c = 0$, we find $x = \frac{3}{4}$ and $y = \frac{1}{2}$. Thus, $(\frac{3}{4}, \frac{1}{2})$ lies on the line. Final Answer: (A)
Divide the given condition $3a+2b+4c=0$ by 4 to make the constant term $c$. $\frac{3}{4}a + \frac{2}{4}b + c = 0 \implies a(\frac{3}{4}) + b(\frac{1}{2}) + c = 0$.
Step 2: Conclusion
Comparing this with $ax + by + c = 0$, we find $x = \frac{3}{4}$ and $y = \frac{1}{2}$. Thus, $(\frac{3}{4}, \frac{1}{2})$ lies on the line. Final Answer: (A)
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