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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