Question

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.

Answer 1

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

\[\therefore\] 2a + 11b − 5 = 0 

\[\Rightarrow b = \frac{5 - 2a}{11} . . . \left( i \right)\]

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

\[d_1 = \left| \frac{24a + 7b - 20}{\sqrt{{24}^2 + 7^2}} \right| = \left| \frac{24a + 7b - 20}{25} \right|\]

\[ \Rightarrow d_1 = \left| \frac{24a + 7 \times \frac{5 - 2a}{11} - 20}{25} \right| \left[ \text { from } (1) \right]\]

\[ \Rightarrow d_1 = \left| \frac{50a - 37}{55} \right|\]

Similarly,

\[d_2 = \left| \frac{4a - 3b - 2}{\sqrt{3^2 + \left( - 4 \right)^2}} \right| = \left| \frac{4a - 3 \times \frac{5 - 2a}{11} - 2}{5} \right|\]

\[ \Rightarrow d_2 = \left| \frac{44a - 15 + 6a - 22}{11 \times 5} \right| \left[ \text { from } (1) \right]\]

\[ \Rightarrow d_2 = \left| \frac{50a - 37}{55} \right|\]

∴ d₁ = d₂