if the sum of the distances of a point from two

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is-

A. square

B. circle

C. straight line

D. two intersecting lines

Answer : square

Solution :
Let the two perpendicular lines be the co-ordinate axes.
Let $$\left( {x,\,y} \right)$$ be the point sum of whose distances from two axes is 1 then we must have
$$\left| x \right| + \left| y \right| = 1\,\,\,\,{\text{or}}\,\, \pm x \pm y = 1$$
These are the four lines $$x+y=1, \,x-y=1, \,-x+y=1, \,-x-y=1$$
Any two adjacent sides are perpendicular to each other. Also each line is equidistant from origin. Therefore figure formed is a square.

Releted MCQ Question on
Geometry >> Straight Lines

Releted Question 1

The points $$\left( { - a, - b} \right),\left( {0,\,0} \right),\left( {a,\,b} \right)$$ and $$\left( {{a^2},\,ab} \right)$$ are :

A. Collinear

B. Vertices of a parallelogram

C. Vertices of a rectangle

D. None of these

View Answer

Releted Question 2

The point (4, 1) undergoes the following three transformations successively.
(i) Reflection about the line $$y =x.$$
(ii) Translation through a distance 2 units along the positive direction of $$x$$-axis.
(iii) Rotation through an angle $$\frac{p}{4}$$ about the origin in the counter clockwise direction.
Then the final position of the point is given by the coordinates.

A. $$\left( {\frac{1}{{\sqrt 2 }},\,\frac{7}{{\sqrt 2 }}} \right)$$

B. $$\left( { - \sqrt 2 ,\,7\sqrt 2 } \right)$$

C. $$\left( { - \frac{1}{{\sqrt 2 }},\,\frac{7}{{\sqrt 2 }}} \right)$$

D. $$\left( {\sqrt 2 ,\,7\sqrt 2 } \right)$$

View Answer

Releted Question 3

The straight lines $$x + y= 0, \,3x + y-4=0,\,x+ 3y-4=0$$ form a triangle which is-

A. isosceles

B. equilateral

C. right angled

D. none of these

View Answer

Releted Question 4

If $$P = \left( {1,\,0} \right),\,Q = \left( { - 1,\,0} \right)$$ and $$R = \left( {2,\,0} \right)$$ are three given points, then locus of the point $$S$$ satisfying the relation $$S{Q^2} + S{R^2} = 2S{P^2},$$ is-

A. a straight line parallel to $$x$$-axis

B. a circle passing through the origin

C. a circle with the centre at the origin

D. a straight line parallel to $$y$$-axis

View Answer

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is-

Releted MCQ Question on
Geometry >> Straight Lines

The points $$\left( { - a, - b} \right),\left( {0,\,0} \right),\left( {a,\,b} \right)$$ and $$\left( {{a^2},\,ab} \right)$$ are :

The straight lines $$x + y= 0, \,3x + y-4=0,\,x+ 3y-4=0$$ form a triangle which is-

If $$P = \left( {1,\,0} \right),\,Q = \left( { - 1,\,0} \right)$$ and $$R = \left( {2,\,0} \right)$$ are three given points, then locus of the point $$S$$ satisfying the relation $$S{Q^2} + S{R^2} = 2S{P^2},$$ is-

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If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is-

Releted MCQ Question on Geometry >> Straight Lines

The points $$\left( { - a, - b} \right),\left( {0,\,0} \right),\left( {a,\,b} \right)$$ and $$\left( {{a^2},\,ab} \right)$$ are :

The straight lines $$x + y= 0, \,3x + y-4=0,\,x+ 3y-4=0$$ form a triangle which is-

If $$P = \left( {1,\,0} \right),\,Q = \left( { - 1,\,0} \right)$$ and $$R = \left( {2,\,0} \right)$$ are three given points, then locus of the point $$S$$ satisfying the relation $$S{Q^2} + S{R^2} = 2S{P^2},$$ is-

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