The points -a-b-a- b-0-0 and a2- a b- q-0- b-0 are alwaysA- Vertices of a parallelogramB- lie on a circleC- CollinearD- vertices of a rectangle

The correct option is C CollinearThe equation of line joining the points ( a, b) and (a,b) is y= ba x So, clearly this line will satisfies (0,0) and (a^2,ab) ...

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Standard XII

Mathematics

Condition for Two Lines to Be Parallel

The points -a...

Question

The points $(- a, - b), (a, b), (0, 0)$ and $(a^{2}, a b), a \neq 0, b \neq 0$ are always

Collinear

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Vertices of a parallelogram

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vertices of a rectangle

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lie on a circle

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Solution

The correct option is A Collinear
The equation of line joining the points
$(- a, - b) and (a, b)$ is
$y = \frac{b}{a} x$

So, clearly this line will satisfies
$(0, 0)$ and $(a^{2}, a b)$

Henc they are always collinear points.

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Similar questions

Q. Are the points

(- a, - b) (0, 0), (a, b)

and

(a^{2}, a b)

vertices of a rectangle or colinear?

A \equiv (0, b), B = (0, 0)

and

C = (a, 0)

are the vertices of

△ A B C . D, E, F

are the mid-points of the sides

B C, C A

and

A B

respectively. If

a^{2} + b^{2} = 20

then

Q. If three vertices of a rectangle are

(0, 0), (a, 0)

and

(0, b)

, length of each diagonal is

5

and the perimeter is

14

, then the area of the rectangle is

Q. Show that

O (0, 0), C (a, 0), B (a, b), C (0, b)

are the vertices of a rectangle. If

P (x, y)

is a point in the coordinates plane, then prove using formula that

P Q^{2} + P B^{2} = P A^{2} + P C^{2}

Q. The coordinates of the point which is equidistant from the vertices of the triangle formed by the points O(0, 0), A(a, 0) and B(0, b), are _________.

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The points (−a,−b),(a,b),(0,0) and (a2,ab),a≠0,b≠0 are always

The correct option is A CollinearThe equation of line joining the points (−a,−b) and (a,b) is y=bax So, clearly this line will satisfies (0,0) and (a2,ab) Henc they are always collinear points.

The points $(- a, - b), (a, b), (0, 0)$ and $(a^{2}, a b), a \neq 0, b \neq 0$ are always

The correct option is A Collinear
The equation of line joining the points
$(- a, - b) and (a, b)$ is
$y = \frac{b}{a} x$

So, clearly this line will satisfies
$(0, 0)$ and $(a^{2}, a b)$

Henc they are always collinear points.