The points whose abscissa and ordinate have different signs will lie in
(a) I and II quadrants
(b) II and III quadrants
(c) I and III quadrants
(d) II and IV quadrants
Solution:
Point which has abscissa and ordinate having different signs will lie in second and fourth quadrants. (d)
The points (-5, 2) and (2, -5) lie in
(a) same quadrant
(b) II and III quadrants respectively
(c) II and IV quadrants respectively
(d) IV and II quadrants respectively
Solution:
Points (-5, 2) and (2, -5) lie in second and fourth quadrants respectively. (b)
If P (-1,1), Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4) are plotted on the graph paper, then point(s) in the fourth quadrant are
(a) P and T
(b) Q and R
(c) S only
(d) P and R
Solution:
Points P (-1, 1), Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4) are plotted on graph The points in 4th quadrant are Q and R (b)
More Solutions:
- Three vertices of a rectangle are A (2, -1), B (2, 7), and C (4, 7).
- What is the area of the parallelogram?
- Also find slope and y-intercept of these lines.
- Draw the graph of the equation 3x – 4y = 12.
- The simultaneous equations: 2x – 3y = 7; x + 6y = 11.
- System of equations graphically: x – 2y – 4 = 0, 2x + .y – 3 = 0.
- Using a scale of l cm to 1 unit for both the axes.
- The lines : 8y – 3x + 7 = 0, 2x-y + 4 = 0 and 5x + 4y = 29.
- Find the area of the triangle formed by these lines.
- Find the abscissa of the other end.