Show that the path of a moving point such that its distances from two lines $3 x-2 y=5$ and $3 x+2 y=5$ are equal is a straight line.
Show that the path of a moving point such that its distances from two lines $3 x-2 y=5$ and $3 x+2 y=5$ are equal is a straight line.
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.
Similar Questions
A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is
A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is
Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :
Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :
- [IIT 2000]
The four points whose co-ordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form :
The four points whose co-ordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form :
Find the area of the triangle formed by the line $y-x=0, x+y=0$ and $x-k=0$.
Find the area of the triangle formed by the line $y-x=0, x+y=0$ and $x-k=0$.
The points $(1, 3)$ and $(5, 1)$ are the opposite vertices of a rectangle. The other two vertices lie on the line $y = 2x + c,$ then the value of c will be
The points $(1, 3)$ and $(5, 1)$ are the opposite vertices of a rectangle. The other two vertices lie on the line $y = 2x + c,$ then the value of c will be
- [IIT 1981]