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प्रश्न
Plot the following points and write the name of the figure obtained by joining them in order:
P(– 3, 2), Q(– 7, – 3), R(6, – 3), S(2, 2)
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उत्तर
Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(– 3, 2) lies in II quadrant, Q(– 7, – 3) lies in III quadrant, R(6, – 3) lies in IV quadrant and S(2, 2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.
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संबंधित प्रश्न
Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.
| x | -2 | -1 | 0 | 1 | 3 |
| y | 8 | 7 | -1.25 | 3 | -1 |
Plot the following point on the graph paper:
(0, 0)
Plot the following point on the graph paper:
(7, −4)
Plot the following point on the graph paper:
(−3, 2)
Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.
Plot the following point in a graph sheet.
J(8, −4)
Plot the following points and check whether they are collinear or not:
(0, 0), (2, 2), (5, 5)
Points A(5, 3), B(– 2, 3) and D(5, – 4) are three vertices of a square ABCD. Plot these points on a graph paper and hence find the coordinates of the vertex C.
Plot the points P(1, 0), Q(4, 0) and S(1, 3). Find the coordinates of the point R such that PQRS is a square.
Plot the points A(1, – 1) and B(4, 5). Extend this line segment and write the coordinates of a point on this line which lies outside the line segment AB.
