Show that the perpendiculars let fall from any point on the straight line 2x + 11y − 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x − 3y − 2 = 0 are equal to each other.

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Solution

et P(a, b) be any point on 2x + 11y − 5 = 0

$∴$ 2a + 11b − 5 = 0

$\Rightarrow b = \frac{5 - 2 a}{11} . . . (i)$

Let d₁ and d₂ be the perpendicular distances from point P
on the lines 24x + 7y = 20 and 4x − 3y − 2 = 0, respectively.

$d_{1} = |\frac{24 a + 7 b - 20}{\sqrt{24^{2} + 7^{2}}}| = |\frac{24 a + 7 b - 20}{25}| \Rightarrow d_{1} = |\frac{24 a + 7 \times \frac{5 - 2 a}{11} - 20}{25}| [from (1)] \Rightarrow d_{1} = |\frac{50 a - 37}{55}|$

Similarly,

$d_{2} = |\frac{4 a - 3 b - 2}{\sqrt{3^{2} + {(- 4)}^{2}}}| = |\frac{4 a - 3 \times \frac{5 - 2 a}{11} - 2}{5}| \Rightarrow d_{2} = |\frac{44 a - 15 + 6 a - 22}{11 \times 5}| [from (1)] \Rightarrow d_{2} = |\frac{50 a - 37}{55}|$

∴ d₁ = d₂

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