Question

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Answer 1

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Answer 2

1) Take a point C on the graph such that ABCD is a square i.e., all sides AB,BC,CD and AD are equal.

So, abscissa of C should be equal to abscissa of B that is, −2 and ordinate of C should be equal to ordinate of D that is, −4. Hence, the coordinates of C are (−2,–4).

The graph obtained by plotting the points A,B,C,D is given above.

NOTE : PICTURE 1 IS GIVEN.

2) (0,0),(−5,0),(0,−3) & (−5,−3)

Let the rectangle be OABC with the vertex O at the origin.

Let OA be the longer side.

∴ The co-ordinate of O is O(0,0).

Since one of the vertex is in the third quadrant and the longer side of length 5 units lie on the x-axis, the co-ordinate of A is A(−5,0).

Also, the rectangle lie in the III quadrant.

∴ The side OC will lie on the y-axis to the negative direction.

Then, the length of OC is 3 units.

∴ The co-ordinates of C will be C(0,−3).

Then obviously, the co-ordinates of B will be (−5,−3).

Here, OC∥AB and ∠AOC=90° , hence is a rectangle.

Therefore, the vertices of the rectangle are O(0,0),A(−5,0),B(−5,−3),C(0,−3).

PICTURE 2 IS GIVEN.

3) We plot the given points P(1,0),Q(4,0),S(1,3).

From S we draw a line parallel to X-axis and from Q we draw a line parallel to Y-axis.

Both the lines will meet at R.

From the graph find co-ordinates of R are (4,3)

PICTURE 3 IS GIVEN

4)

(i) We know that, the point whose abscissa is 0 will lie on Y-axis. So, the required points whose abscissa is 0 are A, L and O.

(ii) We know that, the point whose ordinate is 0 will lie on X-axis. So, the required points, whose ordinate is 0 are G,l and O.

(iii) Here, abscissa ‘-5’ is negative, which shows that point with abscissa -5 will lie in II and III quadrants. So, the required points whose abscissa is -5, are D and H.

Note: We know that, origin O is the intersection point of both axes. So, we can consider it on X-axis as well as on Y-axis.

5)In point A(1,−1), x - coordinate is positive and y - coordinative is negative. So it lies in IV quadrant. In point B(4,5), both coordinates are positive,so it lies in I quadrant. On plotting these point, we get the following graph.

(i) On joining the points A and B, we get the line segment AB. Now to find the coordinates of a point on this line segment between A and B draw a perpendicular to X - axis from x=2. [ Since, x=2 lies between A and B] say it intersect line segment AB at P. Now, draw a perpendicular to Y - axis from P, it intersects Y - axis at y=1. Thus we get points (2,1) which lie between line segment AB.

(ii) Extant the line segment AB. Now draw a perpendicular to X - axis from x=0, say it intersects extended line segment at Q on Y -axis at y=−3. Thus , we get the point Q(0,−7) which lies outside the line segment AB.

PICTURE 5 IS GIVEN.

ALL THE PICTURES ARE ATTACHED FOR BETTER UNDERSTANDING EXCEPT 4 AND ALL ARE ARRANGED IN ORDER.