Consider ABC as the right angles triangle where ∠C = 90°

Here infinity such lines are present.

m is the slope of AC

So the slope of BC =\(- \dfrac{1}{m}\)

Equation of AC –

y – 3 = m (x – 1)

By cross multiplication

x – 1 = \( \dfrac{1}{m}\) (y – 3)

Equation of BC –

y – 1 = \( - \dfrac{1}{m}\) (x + 4)

By cross multiplication

x + 4 = – m (y – 1)

By considering values of m we get

If m = 0,

So we get

y – 3 = 0, x + 4 = 0

If m = ∞,

So we get

x – 1 = 0, y – 1 = 0 we get x = 1, y = 1

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