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NCERT Class XI Mathematics - Straight Lines - Solutions

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Question : 73 of 74
Marks: +1, -0
If sum of the perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.
Solution:  
The given equations of lines are
x + y – 5 = 0 ... (i)
3x – 2y + 7 = 0 ... (ii)
According to question,
⇒ ∣x+y−5∣12+12\frac{\left|x+y-5\right|}{\sqrt{1^2+1^2}} + ∣3x−2y+7∣32+22\frac{\left|3x-2y+7\right|}{\sqrt{3^2+2^2}} = 10 ⇒ ±(x+y−5)2\frac{\pm(x+y-5)}{\sqrt{2}} ± 3x−2y+713\frac{3x-2y+7}{\sqrt{13}} = 10 which shows a set of four lines.
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