Find the equation of the straight line passing through the origin and

1.	Find the equation of the straight line passing through the origin and bisecting the portion of the line ax + by + c = 0 intercepted between the coordinate axes.
Answer» Concept Used: The equation of the line passing through the origin is y = mx To find: Equation of the straight line passing through the origin and bisecting the portion of a line intercepted between the coordinate axes. Assuming: The line ax + by + c = 0 meets the coordinate axes at A and B. Explanation: So, the coordinate of A and B are A(-\(\frac{c}{a}\),0) and B (0, -\(\frac{c}{a}\)) Now, The midpoint of AB is \(\Big(-\frac{c}{2a},-\frac{c}{2b}\Big)\) ∴ \(-\frac{c}{2b}\) = \( m\times-\frac{c}{2a}\) ⇒ m = \(\frac{a}{b}\) Hence, the equation of the required line is y = \(\frac{a}{b}x\) ⇒ ax – by = 0

Find the equation of the straight line passing through the origin and bisecting the portion of the line ax + by + c = 0 intercepted between the coordinate axes.

Answer»

Concept Used:

The equation of the line passing through the origin is y = mx

To find:

Equation of the straight line passing through the origin and bisecting the portion of a line intercepted between the coordinate axes.

Assuming:

The line ax + by + c = 0 meets the coordinate axes at A and B.

Explanation:

So, the coordinate of A and B are A(-\(\frac{c}{a}\),0) and B (0, -\(\frac{c}{a}\))

Now,

The midpoint of AB is \(\Big(-\frac{c}{2a},-\frac{c}{2b}\Big)\)

∴ \(-\frac{c}{2b}\) = \( m\times-\frac{c}{2a}\)

⇒ m = \(\frac{a}{b}\)

Hence, the equation of the required line is

y = \(\frac{a}{b}x\)

⇒ ax – by = 0