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दर्शाइए कि एक गतिमान बिंदु, जिसकी दो रेखाओं $3 x-2 y=5$ और $3 x+2 y=5$ से दूरीयाँ समान है, का पथ एक रेखा है।
Solution
Given lines are
${3x – 2y = 5}$……$(1)$
and ${3x + 2y = 5}$…..$(2)$
Let $(h, k)$ is any point, whose distances from the lines $(1) $ and $(2)$ are equal. Therefore
$\frac{{|3h – 2k – 5|}}{{\sqrt {9 + 4} }} = \frac{{|3h + 2k – 5|}}{{\sqrt {9 + 4} }}$
${\text{or }}|3h – 2k – 5| = |3h + 2k – 5|$
${{\text{which gives }}3h – 2k – 5 = 3h + 2k – 5{\text{ or }} – (3h – 2k – 5)}$
${ = 3h + 2k – 5}$
Solving these two relations we get $k=0$ or $h=\frac{5}{3} .$ Thus, the point $(h, k)$ satisfies the equations $y=0$ or $x=\frac{5}{3},$ which represent straight lines. Hence, path of the point equidistant from the lines $(1)$ and $(2)$ is a straight line.
