Co-ordinate Geometry - Questions & Answers
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In equation 2x - 5 = 0; ordinate is :
(i) 0 (ii) 2 (iii) -5 (iv) 1
Step 2: Solving for x, we get 2x = 5, which means x = 2.5.
Step 3: This represents a straight line parallel to the y-axis, crossing the x-axis at 2.5.
Step 4: The abscissa (x-coordinate) of every point on this line is 2.5.
Step 5: The ordinate (y-coordinate) can be any real number.
Step 6: Note: There appears to be a misprint in the question options as none describe the correct ordinate.
(b) In equation y = 3 + 4x/5 , the independent variable is :
(i) y (ii) x (iii) 4x/5 (iv) 3 + 4x/5
Step 2: This means the value of y depends on whatever value we choose for x.
Step 3: Since x can be chosen freely, x is the independent variable.
Step 4: Therefore, the correct option is (ii) x.
(c) If (2x - 1, y + 5) = (3, 10), the values of x and y are :
(i) x = 4, y = 5 (ii) x = 8, y = 5 (iii) x = 2, y = 5 (iv) x = 1, y = 5
Step 2: Equating the first components: 2x - 1 = 3.
Step 3: Solving for x: 2x = 4, so x = 2.
Step 4: Equating the second components: y + 5 = 10.
Step 5: Solving for y: y = 10 - 5, so y = 5.
Step 6: Therefore, the correct option is (iii) x = 2, y = 5.
(d) (5, -7) belongs to :
(i) 1st quadrant (ii) 2nd quadrant (iii) 3rd quadrant (iv) 4th quadrant
Step 2: Check the sign of the ordinate (y-coordinate), which is -7 (negative).
Step 3: A point with coordinates (+, -) lies in the fourth quadrant.
Step 4: Therefore, the correct option is (iv) 4th quadrant.
(e) Abscissa of a point is the solution of equation 3x - 2 = 7 and its ordinate is the solution of equation 8 - 3y = 2. The point is :
(i) (3, 2) (ii) (-3, 2) (iii) (3, -2) (iv) (-3, -4)
Step 2: 3x = 7 + 2, so 3x = 9, which gives x = 3.
Step 3: Find the ordinate by solving 8 - 3y = 2.
Step 4: -3y = 2 - 8, so -3y = -6, which gives y = 2.
Step 5: The coordinates of the point are (3, 2).
Step 6: Therefore, the correct option is (i) (3, 2).
2. For each equation given below; name the dependent and independent variables.
(i) y = 4/3 x - 7 (ii) x = 9y + 4 (iii) x = (5y + 3) / 2 (iv) y = 1/7 (6x + 5)
Step 2: For (ii) x = 9y + 4, 'x' is the subject of the formula. Dependent variable is x, independent variable is y.
Step 3: For (iii) x = (5y + 3) / 2, 'x' is the subject of the formula. Dependent variable is x, independent variable is y.
Step 4: For (iv) y = 1/7 (6x + 5), 'y' is the subject of the formula. Dependent variable is y, independent variable is x.
3. Plot the following points on the same graph paper :
(i) (8, 7) (ii) (3, 6) (iii) (0, 4) (iv) (0, -4) (v) (3, -2) (vi) (-2, 5) (vii) (-3, 0) (viii) (5, 0) (ix) (-4, -3)
Step 2: For (i) (8, 7), move 8 units right on the x-axis and 7 units up.
Step 3: For (ii) (3, 6), move 3 units right on the x-axis and 6 units up.
Step 4: For (iii) (0, 4), the point lies directly on the y-axis, 4 units up.
Step 5: For (iv) (0, -4), the point lies directly on the y-axis, 4 units down.
Step 6: For (v) (3, -2), move 3 units right on the x-axis and 2 units down.
Step 7: For (vi) (-2, 5), move 2 units left on the x-axis and 5 units up.
Step 8: For (vii) (-3, 0), the point lies directly on the x-axis, 3 units left.
Step 9: For (viii) (5, 0), the point lies directly on the x-axis, 5 units right.
Step 10: For (ix) (-4, -3), move 4 units left on the x-axis and 3 units down.
4. Find the values of x and y if :
(i) (x - 1, y + 3) = (4, 4)
(ii) (3x + 1, 2y - 7) = (9, -9)
(iii) (5x - 3y, y - 3x) = (4, -4)
Step 2: Solving gives x = 5 and y = 1.
Step 3: For (ii), equate corresponding components: 3x + 1 = 9 and 2y - 7 = -9.
Step 4: Solving gives 3x = 8 (so x = 8/3) and 2y = -2 (so y = -1).
Step 5: For (iii), equate components: 5x - 3y = 4 and -3x + y = -4.
Step 6: Multiply the second equation by 3 to get -9x + 3y = -12.
Step 7: Add it to the first equation: (5x - 9x) + (-3y + 3y) = 4 - 12.
Step 8: This simplifies to -4x = -8, so x = 2.
Step 9: Substitute x = 2 into -3x + y = -4 to get -6 + y = -4, so y = 2. (Final values: x = 2, y = 2).
5. Use the graph given below, to find the co-ordinates of the point(s) satisfying the given conditions :
(i) the abscissa is 2.
(ii) the ordinate is 0.
(iii) the ordinate is 3.
(iv) the ordinate is -4.
(v) the abscissa is 5.
(vi) the abscissa is equal to the ordinate.
(vii) the ordinate is half of the abscissa.
Step 2: Point A is at (2, 2). Point B is at (5, 0).
Step 3: Point C is at (-4, 2). Point D is at (4, -4).
Step 4: Point E is at (6, 3). Point F is at (-3, -2).
Step 5: Point G is at (5, -2). Point H is at (4, 5).
Step 6: Answer for (i) abscissa is 2: Point A (2, 2).
Step 7: Answer for (ii) ordinate is 0: Point B (5, 0).
Step 8: Answer for (iii) ordinate is 3: Point E (6, 3).
Step 9: Answer for (iv) ordinate is -4: Point D (4, -4).
Step 10: Answer for (v) abscissa is 5: Points B (5, 0) and G (5, -2).
Step 11: Answer for (vi) abscissa equals ordinate: Point A (2, 2).
Step 12: Answer for (vii) ordinate is half of abscissa: Point E (6, 3) because 3 = 6 / 2.
6. State, true or false :
(i) The ordinate of a point is its x-co-ordinate.
(ii) The origin is in the first quadrant.
(iii) The y-axis is the vertical number line.
(iv) Every point is located in one of the four quadrants.
(v) If the ordinate of a point is equal to its abscissa; the point lies either in the first quadrant or in the second quadrant.
(vi) The origin (0, 0) lies on the x-axis.
(vii) The point (a, b) lies on the y-axis if b = 0.
Step 2: (ii) False. The origin lies on both the x-axis and y-axis, not inside any quadrant.
Step 3: (iii) True. The y-axis represents the vertical axis.
Step 4: (iv) False. Points can also lie exactly on the x-axis or y-axis, which are borders, not quadrants.
Step 5: (v) False. If abscissa equals ordinate (+,+ or -,-), the point lies in the 1st or 3rd quadrant.
Step 6: (vi) True. The origin (0,0) lies on both axes, so it lies on the x-axis.
Step 7: (vii) False. If b=0, the point (a, 0) lies on the x-axis. It would lie on the y-axis if a=0.
7. In each of the following, find the co-ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation :
(i) 3 - 2x = 7 ; 2y + 1 = 10 - 2 1/2 y.
(ii) 2a/3 - 1 = a/2 ; (15 - 4b)/7 = (2b - 1)/3 .
(iii) 5x - (5 - x) = 1/2 (3 - x) ; 4 - 3y = (4 + y)/3
Step 2: Solve 2y + 1 = 10 - 2.5y. We get 4.5y = 9, so y = 2 (ordinate). Co-ordinates: (-2, 2).
Step 3: For (ii), solve 2a/3 - a/2 = 1. Common denominator 6 gives (4a - 3a)/6 = 1, so a = 6 (abscissa).
Step 4: Solve (15 - 4b)/7 = (2b - 1)/3. Cross multiply: 3(15 - 4b) = 7(2b - 1). 45 - 12b = 14b - 7. 26b = 52, so b = 2 (ordinate). Co-ordinates: (6, 2).
Step 5: For (iii), solve 5x - 5 + x = 1.5 - 0.5x. 6x - 5 = 1.5 - 0.5x. 6.5x = 6.5, so x = 1 (abscissa).
Step 6: Solve 4 - 3y = (4 + y)/3. Multiply by 3: 12 - 9y = 4 + y. 10y = 8, so y = 8/10 = 4/5 (ordinate). Co-ordinates: (1, 4/5).
8. In each of the following, the co-ordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, in each case, the co-ordinates of the fourth vertex :
(i) A (2, 0), B (8, 0) and C (8, 4).
(ii) A (4, 2), B (-2, 2) and D (4, -2).
(iii) A (-4, -6), C (6, 0) and D (-4, 0).
(iv) B (10, 4), C (0, 4) and D (0, -2).
Step 2: For (i), D must share the x-coordinate of A (2) and the y-coordinate of C (4). So, D is (2, 4).
Step 3: For (ii), C must share the x-coordinate of B (-2) and the y-coordinate of D (-2). So, C is (-2, -2).
Step 4: For (iii), B must share the x-coordinate of C (6) and the y-coordinate of A (-6). So, B is (6, -6).
Step 5: For (iv), A must share the x-coordinate of B (10) and the y-coordinate of D (-2). So, A is (10, -2).
9. A (- 2, 2), B (8, 2) and C (4, - 4) are the vertices of a parallelogram ABCD. By plotting the given points on a graph paper; find the co-ordinates of the fourth vertex D.
Also, from the same graph, state the co-ordinates of the mid-points of the sides AB and CD.
Step 2: From B(8, 2) to A(-2, 2), x decreases by 10 and y stays the same.
Step 3: Apply this shift to C(4, -4): decrease x by 10. D = (4 - 10, -4) = (-6, -4).
Step 4: Mid-point of AB = ((-2 + 8)/2, (2 + 2)/2) = (6/2, 4/2) = (3, 2).
Step 5: Mid-point of CD = ((4 + (-6))/2, (-4 + (-4))/2) = (-2/2, -8/2) = (-1, -4).
10. A (- 2, 4), C (4, 10) and D (- 2, 10) are the vertices of a square ABCD. Use the graphical method to find the co-ordinates of the fourth vertex B. Also, find :
(i) the co-ordinates of the mid-point of BC;
(ii) the co-ordinates of the mid-point of CD and
(iii) the co-ordinates of the point of intersection of the diagonals of the square ABCD.
Step 2: C and D have the same y-coordinate (10), meaning CD is horizontal.
Step 3: B must share the x-coordinate of C (4) and the y-coordinate of A (4). So, B is (4, 4).
Step 4: (i) Mid-point of BC: B(4, 4) and C(4, 10). Mid-point = ((4+4)/2, (4+10)/2) = (4, 7).
Step 5: (ii) Mid-point of CD: C(4, 10) and D(-2, 10). Mid-point = ((4-2)/2, (10+10)/2) = (1, 10).
Step 6: (iii) Intersection of diagonals is the mid-point of AC: ((-2+4)/2, (4+10)/2) = (1, 7).
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) Line y + 7 = 0 is :
(i) parallel to x-axis (ii) parallel to y-axis (iii) not parallel to x-axis (iv) not parallel to y-axis
Step 2: Any equation of the form y = a represents a horizontal line.
Step 3: A horizontal line is parallel to the x-axis.
Step 4: Therefore, the correct option is (i) parallel to x-axis.
(b) A line is parallel to y-axis and at a distance of 5 units on the positive side of the x-axis. The equation of the line is :
(i) y = 5 (ii) y + 5 = 0 (iii) x = 5 (iv) x + 5 = 0
Step 2: Because it is 5 units on the positive side, a = 5.
Step 3: The equation is x = 5.
Step 4: Therefore, the correct option is (iii) x = 5.
(c) 6x - 5y = 7 is the equation of a line. If x = 2 then the value of y will be :
(i) 1 (ii) -1 (iii) 5 (iv) -5
Step 2: 12 - 5y = 7.
Step 3: -5y = 7 - 12, which gives -5y = -5.
Step 4: Solving for y gives y = 1.
Step 5: Therefore, the correct option is (i) 1.
(d) For equation x/3 - y/2 = 1, the value of y for x = 9 is :
(i) -4 (ii) 6 (iii) 4 (iv) -6
Step 2: 3 - y/2 = 1.
Step 3: -y/2 = 1 - 3, so -y/2 = -2.
Step 4: Multiply by -2 to get y = 4.
Step 5: Therefore, the correct option is (iii) 4.
(e) Lines x - 4 = 0 and 3y = 1 intersect each other at point P. The co-ordinates of point P are :
(i) (4, -1/3) (ii) (4, 1/3) (iii) (-4, 1/3) (iv) (-4, -1/3)
Step 2: Solve the second equation for y: 3y = 1 means y = 1/3.
Step 3: The coordinates of intersection are (4, 1/3).
Step 4: Therefore, the correct option is (ii) (4, 1/3).
2. Draw the graph for each linear equation given below :
(i) x = 3 (ii) x + 3 = 0 (iii) x - 5 = 0 (iv) 2x - 7 = 0 (v) y = 4 (vi) y + 6 = 0 (vii) y - 2 = 0 (viii) 3y + 5 = 0 (ix) 2y - 5 = 0 (x) y = 0
Step 2: (ii) x = -3 is a vertical line crossing the x-axis at -3.
Step 3: (iii) x = 5 is a vertical line crossing the x-axis at 5.
Step 4: (iv) x = 3.5 is a vertical line crossing the x-axis at 3.5.
Step 5: (v) y = 4 is a horizontal line crossing the y-axis at 4.
Step 6: (vi) y = -6 is a horizontal line crossing the y-axis at -6.
Step 7: (vii) y = 2 is a horizontal line crossing the y-axis at 2.
Step 8: (viii) y = -5/3 is a horizontal line crossing the y-axis at approximately -1.67.
Step 9: (ix) y = 2.5 is a horizontal line crossing the y-axis at 2.5.
Step 10: (x) y = 0 represents the entire x-axis.
3. Draw the graph for each linear equation given below :
(i) y = 3x (ii) y = -x (iii) y = -2x (iv) y = x (v) 5x + y = 0 (vi) x + 2y = 0 (vii) 4x - y = 0 (viii) 3x + 2y = 0 (ix) x = -2y
Step 2: Plot the origin as the first point for each.
Step 3: Find a second point by picking an x-value and solving for y.
Step 4: Connect the origin and the second point with a straight line.
4. Draw the graph for each linear equation given below :
(i) y = 2x + 3 (ii) y = 2/3 x - 1 (iii) y = -x + 4 (iv) y = 4x - 5/2 (v) y = 3/2 x + 2/3 (vi) 2x - 3y = 4 (vii) (x - 1)/3 - (y + 2)/2 = 0 (viii) x - 3 = 2/5 (y + 1) (ix) x + 5y + 2 = 0
Step 2: Substitute these x-values into the equation to calculate the corresponding y-values.
Step 3: Plot these three ordered pairs on the graph paper.
Step 4: Draw a straight line passing through all three points to complete the graph.
5. Draw the graph for each equation given below :
(i) 3x + 2y = 6 (ii) 2x - 5y = 10 (iii) x/2 + 2y/3 = 5 (iv) (2x - 1)/3 - (y - 2)/5 = 0
In each case, find the co-ordinates of the points where the graph (line) drawn meets the co-ordinate axes.
Step 2: (i) If y=0, 3x=6 => x=2. If x=0, 2y=6 => y=3. Meets axes at (2, 0) and (0, 3).
Step 3: (ii) If y=0, 2x=10 => x=5. If x=0, -5y=10 => y=-2. Meets axes at (5, 0) and (0, -2).
Step 4: (iii) 3x + 4y = 30. If y=0, 3x=30 => x=10. If x=0, 4y=30 => y=7.5. Meets axes at (10, 0) and (0, 7.5).
Step 5: (iv) simplifies to 10x - 3y + 1 = 0. If y=0, x=-0.1. If x=0, y=1/3. Meets axes at (-0.1, 0) and (0, 1/3).
6. For each linear equation, given above, draw the graph and then use the graph drawn (in each case) to find the area of a triangle enclosed by the graph and the co-ordinate axes :
(i) 3x - (5 - y) = 7
(ii) 7 - 3 (1 - y) = - 5 + 2x.
Step 2: Find intercepts: x-intercept is 4 (base = 4), y-intercept is 12 (height = 12).
Step 3: Area of triangle = 1/2 * base * height = 1/2 * 4 * 12 = 24 square units.
Step 4: (ii) Simplify: 7 - 3 + 3y = -5 + 2x, which gives 2x - 3y = 9.
Step 5: Find intercepts: x-intercept is 4.5 (base = 4.5), y-intercept is -3 (height = |-3| = 3).
Step 6: Area of triangle = 1/2 * 4.5 * 3 = 6.75 square units.
7. For each pair of linear equations given below, draw graphs on the same graph paper and then state, whether the lines drawn are parallel or perpendicular to each other.
(i) y = 3x - 1 ; y = 3x + 2
(ii) y = x - 3 ; y = -x + 5
(iii) 2x - 3y = 6 ; x/2 + y/3 = 1
(iv) 3x + 4y = 24 ; x/4 + y/3 = 1
Step 2: (ii) Slopes are 1 and -1. Since 1 * (-1) = -1, they are perpendicular.
Step 3: (iii) Simplify second equation: 3x + 2y = 6. First equation slope is 2/3. Second equation slope is -3/2. Their product is -1, so they are perpendicular.
Step 4: (iv) Simplify second equation: 3x + 4y = 12. Both have slope -3/4. They are parallel.
8. On the same graph paper, plot the graph of y = x - 2, y = 2x + 1 and y = 4 from x = - 4 to 3.
Step 2: Plot the respective points for y = x - 2 and connect them.
Step 3: Plot the respective points for y = 2x + 1 and connect them.
Step 4: Draw a horizontal line at y = 4 across the specified x range.
9. On the same graph paper, plot the graphs of y = 2x - 1, y = 2x and y = 2x + 1 from x = - 2 to x = 4. Are the graphs (lines) drawn parallel to each other ?
Step 2: Plot the points and draw the straight lines on the graph.
Step 3: Notice that all three equations have the same coefficient for x (m = 2).
Step 4: Because they share the exact same slope, yes, the graphs drawn are parallel to each other.
10. The graph of 3x + 2y = 6 meets the x-axis at point P and the y-axis at point Q. Use the graphical method to find the co-ordinates of points P and Q. (HOTS)
Step 2: Observe where the drawn line physically crosses the x-axis. It intersects at (2, 0), so P is (2, 0).
Step 3: Observe where the drawn line physically crosses the y-axis. It intersects at (0, 3), so Q is (0, 3).
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The inclination of a line is 60°. The slope of the line is :
(i) 1/√3 (ii) -1/√3 (iii) √3 (iv) -√3°
Step 2: m = tan(60°).
Step 3: The value of tan(60°) is √3.
Step 4: Therefore, the correct option is (iii) √3.
(b) For the equation 2x - 5y = 8; slope is :
(i) 5 (ii) -2/5 (iii) 8 (iv) 2/5
Step 2: 5y = 2x - 8.
Step 3: Divide by 5: y = (2/5)x - 8/5.
Step 4: The coefficient of x is the slope, so m = 2/5.
Step 5: Therefore, the correct option is (iv) 2/5.
(c) For the equation 5x - 6y = 9, the y-intercept is :
(i) 3/2 (ii) 5/6 (iii) 6/5 (iv) -3/2
Step 2: 6y = 5x - 9.
Step 3: Divide by 6: y = (5/6)x - 9/6, which simplifies to y = (5/6)x - 3/2.
Step 4: The constant term is the y-intercept, which is -3/2.
Step 5: Therefore, the correct option is (iv) -3/2.
(d) If the slope of a line is -2 and its y-intercept is -7, the equation of the line is :
(i) 2x + y + 7 = 0 (ii) 2x - y + 7 = 0 (iii) 2x - y - 7 = 0 (iv) 2x + y - 7 = 0
Step 2: Substitute m = -2 and c = -7 to get y = -2x - 7.
Step 3: Rearrange the terms to one side: 2x + y + 7 = 0.
Step 4: Therefore, the correct option is (i) 2x + y + 7 = 0.
(e) For the equation x - y + 1 = 0; the values of slope (m) and y-intercept (c) are :
(i) m = 1, c = 1 (ii) m = -1, c = 1 (iii) m = 1, c = -1 (iv) m = -1, c = -1
Step 2: y = x + 1.
Step 3: Comparing with y = mx + c, the coefficient of x is 1 and constant is 1.
Step 4: Therefore, slope m = 1 and y-intercept c = 1.
Step 5: The correct option is (i) m = 1, c = 1.
2. In each of the following, find the inclination of line AB :
[Note: Questions are based on images of 3 graphs]
Step 2: (ii) The line AB is perfectly horizontal. The angle it makes with the x-axis is 0°.
Step 3: (iii) The line makes an angle 2x with the positive x-axis and an adjacent angle x. As they form a straight line, 2x + x = 180° => 3x = 180° => x = 60°. The inclination is the angle with the positive x-axis, which is 2x = 120°.
3. Write the inclination of a line which is :
(i) parallel to x-axis.
(ii) perpendicular to x-axis.
(iii) parallel to y-axis.
(iv) perpendicular to y-axis.
Step 2: (ii) A line perpendicular to the x-axis crosses it at a right angle, so its inclination is 90°.
Step 3: (iii) A line parallel to the y-axis is perfectly vertical, so its inclination is 90°.
Step 4: (iv) A line perpendicular to the y-axis is perfectly horizontal, so its inclination is 0°.
4. Write the slope of the line whose inclination is :
(i) 0° (ii) 30° (iii) 45° (iv) 60°
Step 2: (i) tan(0°) = 0.
Step 3: (ii) tan(30°) = 1/√3.
Step 4: (iii) tan(45°) = 1.
Step 5: (iv) tan(60°) = √3.
5. Find the inclination of the line whose slope is :
(i) 0 (ii) 1 (iii) √3 (iv) 1/√3
Step 2: (i) tan θ = 0 implies inclination θ = 0°.
Step 3: (ii) tan θ = 1 implies inclination θ = 45°.
Step 4: (iii) tan θ = √3 implies inclination θ = 60°.
Step 5: (iv) tan θ = 1/√3 implies inclination θ = 30°.
6. Write the slope of the line which is :
(i) parallel to x-axis.
(ii) perpendicular to x-axis.
(iii) parallel to y-axis.
(iv) perpendicular to y-axis.
Step 2: (ii) Perpendicular to x-axis means inclination is 90°. Slope = tan(90°) = Infinity (not defined).
Step 3: (iii) Parallel to y-axis means inclination is 90°. Slope = tan(90°) = Infinity (not defined).
Step 4: (iv) Perpendicular to y-axis means inclination is 0°. Slope = tan(0°) = 0.
7. For each of the equations given below, find the slope and the y-intercept :
(i) x + 3y + 5 = 0 (ii) 3x - y - 8 = 0 (iii) 5x = 4y + 7 (iv) x = 5y - 4 (v) y = 7x - 2 (vi) 3y = 7 (vii) 4y + 9 = 0
Step 2: (i) 3y = -x - 5 => y = (-1/3)x - 5/3. Slope = -1/3, y-intercept = -5/3.
Step 3: (ii) y = 3x - 8. Slope = 3, y-intercept = -8.
Step 4: (iii) 4y = 5x - 7 => y = (5/4)x - 7/4. Slope = 5/4, y-intercept = -7/4.
Step 5: (iv) 5y = x + 4 => y = (1/5)x + 4/5. Slope = 1/5, y-intercept = 4/5.
Step 6: (v) y = 7x - 2 is already in form. Slope = 7, y-intercept = -2.
Step 7: (vi) y = 7/3. Slope = 0, y-intercept = 7/3.
Step 8: (vii) y = -9/4. Slope = 0, y-intercept = -9/4.
8. Find the equation of the line, whose :
(i) slope = 2 and y-intercept = 3
(ii) slope = 5 and y-intercept = -8
(iii) slope = -4 and y-intercept = 2
(iv) slope = -3 and y-intercept = -1
(v) slope = 0 and y-intercept = -5
(vi) slope = 0 and y-intercept = 0
Step 2: (i) y = 2x + 3.
Step 3: (ii) y = 5x - 8.
Step 4: (iii) y = -4x + 2.
Step 5: (iv) y = -3x - 1.
Step 6: (v) y = 0x - 5, which simplifies to y = -5.
Step 7: (vi) y = 0x + 0, which simplifies to y = 0.
9. Draw the line 3x + 4y = 12 on a graph paper. From the graph paper, read the y-intercept of the line.
Step 2: Draw the line passing through (4,0) and (0,3).
Step 3: Observe where the line crosses the y-axis.
Step 4: It crosses at y = 3, so the y-intercept is 3.
10. Draw the line 2x - 3y - 18 = 0 on a graph paper. From the graph paper, read the y-intercept of the line.
Step 2: Draw the line passing through (9,0) and (0,-6).
Step 3: Observe where the line crosses the y-axis.
Step 4: It crosses at y = -6, so the y-intercept is -6.
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) For point P(-5, 4) :
(i) abscissa = 4, ordinate = -5 (ii) ordinate = 4, abscissa = -5 (iii) abscissa = ordinate = -5 (iv) abscissa = ordinate = 4
Step 2: For P(-5, 4), the abscissa is -5 and the ordinate is 4.
Step 3: Therefore, the correct option is (ii) ordinate = 4, abscissa = -5.
(b) If the perpendicular distance of a point P from the x-axis is 5 unit then the point P has :
(i) x-co-ordinate = -5 (ii) y-co-ordinate = 5 only (iii) y-co-ordinate = -5 only (iv) x-co-ordinate = 5 or -5
Step 2: A distance of 5 means the y-coordinate must be either 5 or -5.
Step 3: The provided options seem incorrectly formulated in the book as none strictly capture 'y = 5 or -5', but logically the y-coordinate holds the distance value.
(c) A point lies on y-axis at a distance of 2 unit from x-axis. Its co-ordinates are :
(i) (2, 0) only (ii) (0, 2) only (iii) (2, 2) (iv) (0, 2) or (0, -2)
Step 2: A distance of 2 units from the x-axis means y can be 2 (above) or -2 (below).
Step 3: Therefore, the coordinates can be (0, 2) or (0, -2).
Step 4: The correct option is (iv) (0, 2) or (0, -2).
(d) Three vertices of a square ABCD are A(2, 0), B(-3, 0) and C(-3, -5). Its fourth vertex D is :
(i) (2, 5) (ii) (2, -5) (iii) (-2, 5) (iv) (-2, -5)
Step 2: D takes the x-coordinate of A, which is 2.
Step 3: D takes the y-coordinate of C, which is -5.
Step 4: The coordinates of D are (2, -5).
Step 5: The correct option is (ii) (2, -5). (Wait, the first option in the original text might have been a typo in my reading, but it's clearly x=2, y=-5).
(e) Statement (1) : In the given diagram, OAB is an equilateral triangle.
Statement (2) : B = (4, 4√3)
Step 2: If the triangle is equilateral, all sides are 8. The midpoint of the base OA is at x = 4.
Step 3: The height (y-coordinate of B) can be found using Pythagoras: h² + 4² = 8² => h² = 64 - 16 = 48 => h = √48 = 4√3.
Step 4: Thus, B is indeed at (4, 4√3). Both statements are true and consistent.
Step 5: The correct option is (i) Both the statements are true.
(f) Statement (1) : The vertex B of square OABC with each side 4 units lies in fourth quadrant and its sides are along the co-ordinate axes. The co-ordinates of vertex B are (4, -4).
Statement (2) : B = (4, 4)
Step 2: Since sides are 4 units, B must be (4, -4). Statement 1 is correct.
Step 3: Statement 2 says B = (4, 4), which would put it in the first quadrant. Statement 2 is false.
Step 4: The correct option is (iii) Statement 1 is true, and statement 2 is false.
(g) Assertion (A) : PQR is an equilateral triangle. The co-ordinates of point Q are (0, 2√2).
Reason (R) : In Δ OPQ, OQ² = PQ² - OP² = 4² - 2² = 12.
Step 2: A point at (0, 2√2) would be on the y-axis, not the x-axis, meaning the Assertion is completely false regarding Q's coordinates.
Step 3: In the Reason, it assumes PQ = 4. However, PR = √(2²+2²) = 2√2. If equilateral, PQ must equal 2√2, not 4. Thus the Reason's calculation is also completely false.
Step 4: Both A and R are false.
(h) Assertion (A) : (2x - 3y, 8) = (2, x + 2y) => x = 1 and y = -2
Reason (R) : 2x - 3y = 2 and 8 = x + 2y which on solving give x = 4 and y = 2
Step 2: Solve the system: x = 8 - 2y. Substitute into first: 2(8 - 2y) - 3y = 2 => 16 - 4y - 3y = 2 => 7y = 14 => y = 2.
Step 3: Substitute y = 2 back: x = 8 - 4 = 4. The actual solution is x=4, y=2.
Step 4: Assertion A claims x=1, y=-2, which is false.
Step 5: Reason R correctly states the equations and their correct solution (x=4, y=2), so R is true.
Step 6: The correct option is (ii) A is false, R is true.
2. By plotting the following points on the same graph paper, check whether they are collinear or not :
(i) (3, 5), (1, 1) and (0, -1)
(ii) (- 2, - 1), (- 1, - 4) and (- 4, 1)
Step 2: (i) is collinear. (Can also be checked by slope: (1-5)/(1-3) = 2 and (-1-1)/(0-1) = 2).
Step 3: Plot the points for (ii) on a graph. They do not fall on a single straight line.
Step 4: (ii) is not collinear.
3. Plot the point A (5, - 7). From point A, draw AM perpendicular to x-axis and AN perpendicular to y-axis. Write the co-ordinates of points M and N.
Step 2: Dropping a perpendicular to the x-axis means the x-coordinate stays 5, but the y-coordinate becomes 0.
Step 3: Therefore, M is (5, 0).
Step 4: Dropping a perpendicular to the y-axis means the y-coordinate stays -7, but the x-coordinate becomes 0.
Step 5: Therefore, N is (0, -7).
4. In square ABCD; A = (3, 4), B = (- 2, 4) and C = (- 2, - 1). By plotting these points on a graph paper, find the co-ordinates of vertex D. Also, find the area of the square.
Step 2: Therefore, D is (3, -1).
Step 3: Calculate the side length by finding the distance between A(3, 4) and B(-2, 4), which is 3 - (-2) = 5 units.
Step 4: Area of a square is side × side = 5 × 5 = 25 square units.
5. In rectangle OABC; point O is the origin, OA = 10 units along x-axis and AB = 8 units. Find the co-ordinates of vertices A, B and C.
Step 2: Since it is a rectangle, AB is perpendicular to OA. It goes 8 units up parallel to the y-axis.
Step 3: Therefore, B has the same x as A, and y is 8, making B (10, 8).
Step 4: C must align horizontally with B and vertically with O, so C is (0, 8).
6. Draw the graph of equation x + 2y - 3 = 0. From the graph, find :
(i) x₁, the value of x, when y = 3
(ii) x₂, the value of x, when y = - 2.
Step 2: Read the graph where y = 3. Or solve algebraically: x + 2(3) - 3 = 0 => x + 3 = 0 => x = -3. So x₁ = -3.
Step 3: Read the graph where y = -2. Or solve algebraically: x + 2(-2) - 3 = 0 => x - 7 = 0 => x = 7. So x₂ = 7.
7. Draw the graph of equation 3x - 4y = 12. Use the graph drawn to find :
(i) y₁, the value of y, when x = 4
(ii) y₂, the value of y, when x = 0.
Step 2: Where x = 4, read the y-value from graph. 3(4) - 4y = 12 => 12 - 4y = 12 => y = 0. So y₁ = 0.
Step 3: Where x = 0, read the y-value from graph. 3(0) - 4y = 12 => -4y = 12 => y = -3. So y₂ = -3.
8. Draw the graph of equation x/4 + y/5 = 1. Use the graph drawn to find :
(i) x₁, the value of x, when y = 10
(ii) y₁, the value of y, when x = 8.
Step 2: Find x₁ when y = 10. Graphically it extends, algebraically: x/4 + 10/5 = 1 => x/4 + 2 = 1 => x/4 = -1 => x = -4. So x₁ = -4.
Step 3: Find y₁ when x = 8. Algebraically: 8/4 + y/5 = 1 => 2 + y/5 = 1 => y/5 = -1 => y = -5. So y₁ = -5.
9. Use the graphical method to show that the straight lines given by the equations x + y = 2, x - 2y = 5 and x/3 + y = 0 pass through the same point.
Step 2: You will observe they all intersect at exactly one single point on the graph.
Step 3: To verify algebraically, the intersection of the first two is x = 3, y = -1.
Step 4: Check if (3, -1) satisfies the third equation: 3/3 + (-1) = 1 - 1 = 0. It does. Thus they pass through the same point.
10. Draw the graph of line x + y = 5. Use the graph paper drawn to find the inclination and the y-intercept of the line.
Step 2: Notice it crosses the y-axis at y = 5, so the y-intercept is 5.
Step 3: Rewrite equation as y = -x + 5. The slope is -1.
Step 4: The angle whose tangent is -1 is 135°. The inclination is 135°.
11. Draw the graph of line 2x + y = 5.
Step 2: Plot (0, 5), (2, 1), and (3, -1) on a graph.
Step 3: Draw a straight line connecting these points.
12. Draw the graph of line 4x - y = 5. Use this graph to find :
(i) x₁, the value of x when y = 3.
(ii) y₁, the value of y when x = 3.
Step 2: From the graph, find where the line reaches y = 3. Or algebraically: 4x - 3 = 5 => 4x = 8 => x = 2. So x₁ = 2.
Step 3: From the graph, find where the line reaches x = 3. Or algebraically: 4(3) - y = 5 => 12 - y = 5 => y = 7. So y₁ = 7.