If the sum of the distances of a point from t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the two perpendicular lines be the coordinate axes $X$ and $Y$. The distance of a point $P(x, y)$ from the $Y$-axis is $|x|$ and from the $X$-

Vedclass · Accepted Answer

Square

Vedclass · Answer

(A) Let the two perpendicular lines be the coordinate axes $X$ and $Y$. The distance of a point $P(x, y)$ from the $Y$-axis is $|x|$ and from the $X$-axis is $|y|$.Given that the sum of the distances is $1$,we have $|x| + |y| = 1$.If the point is in the first quadrant,$x > 0, y > 0$,so $x + y = 1$.If the point is in the second quadrant,$x  0$,so $-x + y = 1$.If the point is in the third quadrant,$x If the point is in the fourth quadrant,$x > 0, y These four equations represent the sides of a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,find its locus.

Similar Questions

$A$ straight line cuts off intercepts on the coordinate axes,the reciprocal of whose sum is $1/p$. Through which fixed point does this line always pass?

Let $P(2, 3)$,$Q(6, 0)$,and $R(\alpha, \beta)$ be three points in the $x-y$ plane such that $|PR + QR| + |PR - QR|$ is minimum. Then the value of $(\alpha - 2\beta)$ is:

$A$ rod of length $2l$ slides with its ends on two perpendicular lines. The locus of its mid-point is

$A$ straight line through the point of intersection of the lines $x+2y=4$ and $2x+y=4$ meets the coordinate axes at $A$ and $B$. The locus of the mid-point of $AB$ is

If $A(2,-3)$ and $B(-2,1)$ are two vertices of a triangle and the third vertex moves on the line $2x + 3y = 9$,then the locus of the centroid of the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module