#### 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.