GRAPHICAL SOLUTION - Questions & Answers
EXERCISE 24(A)
Choose the correct answer from the options given below.
(a) Point (k, 2) lies on the line x - 4y = 2; the value of k is :
(i) 10 (ii) -6 (iii) -10 (iv) 6
Answer:
Step 1: Substitute x = k and y = 2 in the equation x - 4y = 2.
Step 2: We get k - 4(2) = 2.
Step 3: k - 8 = 2, which gives k = 10.
Step 4: The correct option is (i) 10.
(b) The line y = mx - 8 passes through the point (5, 2); the value of m is :
(i) 1 (ii) 2 (iii) -2 (iv) -1
Answer:
Step 1: Substitute x = 5 and y = 2 in the equation y = mx - 8.
Step 2: We get 2 = m(5) - 8.
Step 3: 2 + 8 = 5m, so 10 = 5m, which gives m = 2.
Step 4: The correct option is (ii) 2.
(c) The line 5x - 2y - 10 = 0 intersects x-axis at point P. The co-ordinates of point P are:
(i) (0, 2) (ii) (0, -2) (iii) (2, 0) (iv) (-2, 0)
Answer:
Step 1: Any point on the x-axis has y-coordinate 0. So, substitute y = 0.
Step 2: 5x - 2(0) - 10 = 0.
Step 3: 5x = 10, which gives x = 2.
Step 4: The coordinates of point P are (2, 0). The correct option is (iii) (2, 0).
(d) The line 5x - 4y - 20 = 0 intersects y-axis at point A. The co-ordinates of point A are :
(i) (-5, 0) (ii) (5, 0) (iii) (0, 5) (iv) (0, -5)
Answer:
Step 1: Any point on the y-axis has x-coordinate 0. So, substitute x = 0.
Step 2: 5(0) - 4y - 20 = 0.
Step 3: -4y = 20, which gives y = -5.
Step 4: The coordinates of point A are (0, -5). The correct option is (iv) (0, -5).
(e) The point (0, 0) lies on :
(i) x-axis (ii) y-axis (iii) x-axis or y-axis (iv) x-axis and y-axis both
Answer:
Step 1: The point (0, 0) is the origin.
Step 2: The origin is the intersection of both the x-axis and the y-axis.
Step 3: The correct option is (iv) x-axis and y-axis both.
2. Draw the graph for each equation, given below :
(i) x = 5 (ii) x + 5 = 0 (iii) y = 7 (iv) y + 7 = 0
(v) 2x + 3y = 0 (vi) 3x + 2y = 6 (vii) x - 5y + 4 = 0 (viii) 5x + y + 5 = 0
Answer:
Step 1: For (i) x = 5, draw a line parallel to the y-axis, passing through (5, 0).
Step 2: For (ii) x + 5 = 0 (x = -5), draw a line parallel to the y-axis, passing through (-5, 0).
Step 3: For (iii) y = 7, draw a line parallel to the x-axis, passing through (0, 7).
Step 4: For (iv) y + 7 = 0 (y = -7), draw a line parallel to the x-axis, passing through (0, -7).
Step 5: For (v) 2x + 3y = 0, y = -2x/3. Plot points (0, 0), (3, -2), and (-3, 2), then draw a straight line.
Step 6: For (vi) 3x + 2y = 6, y = (6 - 3x)/2. Plot points (0, 3), (2, 0), and (4, -3), then draw a straight line.
Step 7: For (vii) x - 5y + 4 = 0, x = 5y - 4. Plot points (-4, 0), (1, 1), and (6, 2), then draw a straight line.
Step 8: For (viii) 5x + y + 5 = 0, y = -5x - 5. Plot points (0, -5), (-1, 0), and (-2, 5), then draw a straight line.
3. Draw the graph for each equation given below; hence find the co-ordinates of the points where the graph drawn meets the co-ordinate axes :
(i) 1/3 x + 1/5 y = 1 (ii) (2x + 15) / 3 = y - 1
Answer:
Step 1: For (i) Multiply the equation by 15 to get 5x + 3y = 15.
Step 2: Substitute y = 0 to get the x-intercept: 5x = 15 => x = 3. Point is (3, 0).
Step 3: Substitute x = 0 to get the y-intercept: 3y = 15 => y = 5. Point is (0, 5).
Step 4: Draw a line through (3, 0) and (0, 5). The graph meets the axes at (3, 0) and (0, 5).
Step 5: For (ii) Cross multiply: 2x + 15 = 3y - 3 => 2x - 3y + 18 = 0.
Step 6: Substitute y = 0 to get the x-intercept: 2x + 18 = 0 => x = -9. Point is (-9, 0).
Step 7: Substitute x = 0 to get the y-intercept: -3y + 18 = 0 => y = 6. Point is (0, 6).
Step 8: Draw a line through (-9, 0) and (0, 6). The graph meets the axes at (-9, 0) and (0, 6).
4. Draw the graph of the straight line given by the equation 4x - 3y + 36 = 0. Calculate the area of the triangle formed by the line drawn and the co-ordinate axes.
Answer:
Step 1: Find the x-intercept by putting y = 0: 4x + 36 = 0 => x = -9. Point is (-9, 0).
Step 2: Find the y-intercept by putting x = 0: -3y + 36 = 0 => y = 12. Point is (0, 12).
Step 3: Draw a line passing through (-9, 0) and (0, 12).
Step 4: The base of the triangle on the x-axis is 9 units.
Step 5: The height of the triangle on the y-axis is 12 units.
Step 6: Area = 1/2 × base × height = 1/2 × 9 × 12 = 54 sq. units.
5. Draw the graph of the equation 2x - 3y - 5 = 0. From the graph, find :
(i) x1, the value of x, when y = 7
(ii) x2, the value of x, when y = -5
Answer:
Step 1: Write the equation as 2x = 3y + 5. Plot the graph using suitable points like (1, -1), (4, 1), and (-2, -3).
Step 2: For (i) when y = 7, substitute into the equation: 2x - 3(7) - 5 = 0.
Step 3: 2x - 21 - 5 = 0 => 2x = 26 => x1 = 13.
Step 4: For (ii) when y = -5, substitute into the equation: 2x - 3(-5) - 5 = 0.
Step 5: 2x + 15 - 5 = 0 => 2x = -10 => x2 = -5.
6. Draw the graph of the equation 4x + 3y + 6 = 0. From the graph, find :
(i) y1, the value of y, when x = 12
(ii) y2, the value of y, when x = -6
Answer:
Step 1: Write the equation as 3y = -4x - 6. Plot the graph using suitable points like (0, -2), (3, -6), and (-3, 2).
Step 2: For (i) when x = 12, substitute into the equation: 4(12) + 3y + 6 = 0.
Step 3: 48 + 3y + 6 = 0 => 3y = -54 => y1 = -18.
Step 4: For (ii) when x = -6, substitute into the equation: 4(-6) + 3y + 6 = 0.
Step 5: -24 + 3y + 6 = 0 => 3y = 18 => y2 = 6.
7. Use the table given below to draw the graph.
x = -5, -1, 3, b, 13
y = -2, a, 2, 5, 7
From your graph, find the values of 'a' and 'b'. State a linear relation between the variables x and y.
Answer:
Step 1: Plot the given complete pairs (-5, -2), (3, 2), and (13, 7) on a graph paper and draw a straight line through them.
Step 2: Find the slope (m) using (-5, -2) and (3, 2): m = (2 - (-2)) / (3 - (-5)) = 4 / 8 = 1/2.
Step 3: Use the point-slope form: y - 2 = 1/2 (x - 3) => 2y - 4 = x - 3 => x - 2y + 1 = 0. This is the linear relation.
Step 4: To find 'a' (when x = -1): substitute in relation => -1 - 2a + 1 = 0 => -2a = 0 => a = 0.
Step 5: To find 'b' (when y = 5): substitute in relation => b - 2(5) + 1 = 0 => b - 9 = 0 => b = 9.
8. Draw the graph obtained from the table below :
x = a, 3, -5, 5, c, -1
y = -1, 2, b, 3, 4, 0
Use the graph to find the values of a, b and c. State a linear relation between the variables x and y.
Answer:
Step 1: Plot the given complete pairs (3, 2), (5, 3), and (-1, 0) on a graph paper and draw a straight line.
Step 2: Find the slope (m) using (3, 2) and (-1, 0): m = (2 - 0) / (3 - (-1)) = 2 / 4 = 1/2.
Step 3: Use point-slope form: y - 0 = 1/2 (x - (-1)) => 2y = x + 1 => x - 2y + 1 = 0. This is the linear relation.
Step 4: To find 'a' (when y = -1): substitute => a - 2(-1) + 1 = 0 => a + 3 = 0 => a = -3.
Step 5: To find 'b' (when x = -5): substitute => -5 - 2b + 1 = 0 => -2b - 4 = 0 => b = -2.
Step 6: To find 'c' (when y = 4): substitute => c - 2(4) + 1 = 0 => c - 7 = 0 => c = 7.
9. A straight line passes through the points (2, 4) and (5, -2). Taking 1 cm = 1 unit; mark these points on a graph paper and draw the straight line through these points. If points (m, -4) and (3, n) lie on the line drawn; find the values of m and n.
Answer:
Step 1: Find the slope using (2, 4) and (5, -2): slope = (-2 - 4) / (5 - 2) = -6 / 3 = -2.
Step 2: The equation is y - 4 = -2(x - 2) => y - 4 = -2x + 4 => 2x + y = 8.
Step 3: For point (m, -4) to lie on the line, substitute x = m, y = -4: 2m + (-4) = 8 => 2m = 12 => m = 6.
Step 4: For point (3, n) to lie on the line, substitute x = 3, y = n: 2(3) + n = 8 => 6 + n = 8 => n = 2.
10. Draw the graph (straight line) given by equation x - 3y = 18. If the straight line drawn passes through the points (m, -5) and (6, n); find the values of m and n. (HOTS)
Answer:
Step 1: Draw the graph for x - 3y = 18 by plotting points like (0, -6), (6, -4), and (18, 0).
Step 2: Since point (m, -5) lies on the line, substitute x = m, y = -5: m - 3(-5) = 18.
Step 3: m + 15 = 18 => m = 3.
Step 4: Since point (6, n) lies on the line, substitute x = 6, y = n: 6 - 3n = 18.
Step 5: -3n = 12 => n = -4.
EXERCISE 24(B)
1. Multiple Choice Type :Choose the correct answer from the options given below.
(a) The point of intersection of lines x = 8 and y - 8 = 0 is :
(i) (8, 8) (ii) (8, -8) (iii) (-8, 8) (iv) (-8, -8)
Answer:
Step 1: The first line gives x = 8.
Step 2: The second line gives y = 8.
Step 3: Therefore, the intersection point is (8, 8). The correct option is (i).
(b) For line y = 20 + 2x, if x = 20, the value of y is :
(i) 40 (ii) 60 (iii) 80 (iv) 0
Answer:
Step 1: Substitute x = 20 into the equation y = 20 + 2x.
Step 2: y = 20 + 2(20) = 20 + 40 = 60.
Step 3: The correct option is (ii).
(c) A point that lies on the line 2x - 3y = 4 can be taken as :
(i) (-5, -2) (ii) (-5, 2) (iii) (5, 2) (iv) (5, -2)
Answer:
Step 1: Test option (iii) where x = 5, y = 2.
Step 2: 2(5) - 3(2) = 10 - 6 = 4. This matches the equation.
Step 3: The correct option is (iii).
(d) The line x + 5 = 0 :
(i) is parallel to x-axis (ii) is parallel to y-axis (iii) passes through (0, 0) (iv) passes through (5, 0)
Answer:
Step 1: The equation is x = -5.
Step 2: Equations of the form x = constant are always parallel to the y-axis.
Step 3: The correct option is (ii).
(e) The lines y - 2 = 0 and 2x + 3y = 12 cut each other at point :
(i) (-3, 2) (ii) (3, -2) (iii) (-3, -2) (iv) (3, 2)
Answer:
Step 1: From the first line, we get y = 2.
Step 2: Substitute y = 2 into the second line: 2x + 3(2) = 12 => 2x + 6 = 12 => 2x = 6 => x = 3.
Step 3: The point is (3, 2). The correct option is (iv).
2. Solve, graphically, the following pairs of equations :
(i) x - 5 = 0; y + 4 = 0
Answer:
Step 1: Graph x = 5 (vertical line) and y = -4 (horizontal line).
Step 2: The intersection point is (5, -4).
(ii) 2x + y = 23; 4x - y = 19
Answer:
Step 1: Add the equations: (2x + y) + (4x - y) = 23 + 19 => 6x = 42 => x = 7.
Step 2: Substitute x = 7: 2(7) + y = 23 => 14 + y = 23 => y = 9.
Step 3: The solution point is (7, 9).
(iii) 3x + 7y = 27; 8 - y = 5/2 x
Answer:
Step 1: Rewrite second equation: 16 - 2y = 5x => 5x + 2y = 16.
Step 2: Multiply first equation by 2: 6x + 14y = 54. Multiply second by 7: 35x + 14y = 112.
Step 3: Subtract: 29x = 58 => x = 2.
Step 4: Substitute x = 2 in first eq: 3(2) + 7y = 27 => 7y = 21 => y = 3.
Step 5: The solution point is (2, 3).
(iv) (x+1)/4 = 2/3 (1-2y); (2+5y)/3 = x/7 - 2
Answer:
Step 1: Simplify first eq: 3x + 3 = 8 - 16y => 3x + 16y = 5.
Step 2: Simplify second eq: 14 + 35y = 3x - 42 => 3x - 35y = 56.
Step 3: Subtract: (3x + 16y) - (3x - 35y) = 5 - 56 => 51y = -51 => y = -1.
Step 4: Substitute y = -1 in 3x + 16y = 5 => 3x - 16 = 5 => 3x = 21 => x = 7.
Step 5: The solution point is (7, -1).
3. Solve graphically the simultaneous equations given below. Take the scale as 2 cm = 1 unit on both the axes.
x - 2y - 4 = 0; 2x + y = 3
Answer:
Step 1: Multiply the second equation by 2: 4x + 2y = 6.
Step 2: Add it to the first equation: (x - 2y - 4) + (4x + 2y - 6) = -10 => 5x - 10 = 0 => x = 2.
Step 3: Substitute x = 2 into second eq: 2(2) + y = 3 => y = -1.
Step 4: The solution point is (2, -1).
4. Use graph paper for this question. Draw the graph of 2x - y - 1 = 0 and 2x + y = 9 on the same axes. Use 2 cm = 1 unit on both axes and plot only 3 points per line. Write down the co-ordinates of the point of intersection of the two lines.
Answer:
Step 1: Add the two equations: (2x - y) + (2x + y) = 1 + 9 => 4x = 10 => x = 2.5.
Step 2: Substitute x = 2.5 in first eq: 2(2.5) - y = 1 => 5 - y = 1 => y = 4.
Step 3: The coordinates of the point of intersection are (2.5, 4).
5. Use graph paper for this question. Take 2 cm = 2 units on x-axis and 2 cm = 1 unit on y-axis. Solve graphically the following equations :
3x + 5y = 12; 3x - 5y + 18 = 0
Answer:
Step 1: Add the equations: (3x + 5y) + (3x - 5y) = 12 + (-18) => 6x = -6 => x = -1.
Step 2: Substitute x = -1 into the first eq: 3(-1) + 5y = 12 => -3 + 5y = 12 => 5y = 15 => y = 3.
Step 3: The graphical solution is x = -1, y = 3, which is the point (-1, 3).
6. Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot only three points per line.
(ii) Write down the co-ordinates of the point of intersection of the lines.
(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.
Answer:
Step 1: From the first eq, x + y = -3. Multiply by 2: 2x + 2y = -6.
Step 2: Add to the second eq: (2x + 2y) + (3x - 2y) = -6 + (-4) => 5x = -10 => x = -2.
Step 3: Substitute x = -2 in the first eq: -2 + y = -3 => y = -1.
Step 4: For (ii), the co-ordinates of intersection are (-2, -1).
Step 5: For (iii), distance from origin = √( (-2)² + (-1)² ) = √(4 + 1) = √5 ≈ 2.24 cm.
7. The sides of a triangle are given by the equations y - 2 = 0; y + 1 = 3(x - 2) and x + 2y = 0. (HOTS)
Find, graphically :
(i) the area of triangle;
(ii) the co-ordinates of the vertices of the triangle.
Answer:
Step 1: Simplify the second equation: y = 3x - 7.
Step 2: Intersection of y = 2 and y = 3x - 7 => 2 = 3x - 7 => 3x = 9 => x = 3. First vertex is (3, 2).
Step 3: Intersection of y = 2 and x + 2y = 0 => x + 4 = 0 => x = -4. Second vertex is (-4, 2).
Step 4: Intersection of y = 3x - 7 and x + 2y = 0 => x + 2(3x - 7) = 0 => x + 6x - 14 = 0 => 7x = 14 => x = 2, y = -1. Third vertex is (2, -1).
Step 5: For (ii), vertices are (-4, 2), (3, 2), and (2, -1).
Step 6: For (i), the base lies on the line y = 2. Length of base = 3 - (-4) = 7 units.
Step 7: The height of the third vertex (2, -1) from y = 2 is 2 - (-1) = 3 units.
Step 8: Area = 1/2 × base × height = 1/2 × 7 × 3 = 10.5 sq. units.
8. By drawing a graph for each of the equations 3x + y + 5 = 0; 3y - x = 5 and 2x + 5y = 1 on the same graph paper; show that the lines given by these equations are concurrent (i.e. they pass through the same point). Take 2 cm = 1 unit on both the axes.
Answer:
Step 1: Solve the first two equations: y = -3x - 5, substitute into second: 3(-3x - 5) - x = 5.
Step 2: -9x - 15 - x = 5 => -10x = 20 => x = -2.
Step 3: y = -3(-2) - 5 = 6 - 5 = 1. Intersection point is (-2, 1).
Step 4: Check if (-2, 1) satisfies the third equation 2x + 5y = 1.
Step 5: 2(-2) + 5(1) = -4 + 5 = 1. It satisfies the equation.
Step 6: Therefore, all three lines meet at (-2, 1) and are concurrent.
9. Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations : 6y = 5x + 10, y = 5x - 15.
From the graph find :
(i) the co-ordinates of the point where the two lines intersect;
(ii) the area of the triangle between the lines and the x-axis.
Answer:
Step 1: Solve 6y = 5x + 10 and y = 5x - 15. Substitute y: 6(5x - 15) = 5x + 10.
Step 2: 30x - 90 = 5x + 10 => 25x = 100 => x = 4. Then y = 5(4) - 15 = 5. For (i), Intersection is (4, 5).
Step 3: Find x-intercepts by setting y = 0. First line: 0 = 5x + 10 => x = -2.
Step 4: Second line x-intercept: 0 = 5x - 15 => x = 3.
Step 5: Base on x-axis = 3 - (-2) = 5 units. Height of triangle from x-axis = 5 units.
Step 6: For (ii), Area = 1/2 × 5 × 5 = 12.5 sq. units.
10. The cost of manufacturing x articles is Rs (50 + 3x). The selling price of x articles is Rs 4x. (LS)
On a graph sheet, with the same axes, and taking suitable scales draw two graphs, first for the cost of manufacturing against no. of articles and the second for the selling price against number of articles.
Use your graph to determine :
(i) No. of articles to be manufactured and sold to break even (no profit and no loss),
(ii) The profit or loss made when (a) 30 (b) 60 articles are manufactured and sold.
Answer:
Step 1: Set Cost (C) = 50 + 3x and Selling Price (S) = 4x.
Step 2: For (i) breakeven occurs when C = S, so 50 + 3x = 4x => x = 50. Break even is 50 articles.
Step 3: For (ii)(a) at 30 articles, C = 50 + 3(30) = 140, S = 4(30) = 120. Loss = 140 - 120 = Rs 20.
Step 4: For (ii)(b) at 60 articles, C = 50 + 3(60) = 230, S = 4(60) = 240. Profit = 240 - 230 = Rs 10.
TEST YOURSELF
1. Multiple Choice Type :Choose the correct answer from the options given below.
(a) For the line 2x - y = 7 and for x = 3 the value of y is :
(i) 1 (ii) -1 (iii) 5 (iv) -5
Answer:
Step 1: Substitute x = 3 into the equation 2(3) - y = 7.
Step 2: 6 - y = 7 => y = -1.
Step 3: The correct option is (ii).
(b) The line x + 3y + 2 = 0 passes through the point (4, k); then the value of k is :
(i) 2 (ii) 1 (iii) -1 (iv) -2
Answer:
Step 1: Substitute x = 4, y = k into the equation 4 + 3k + 2 = 0.
Step 2: 3k + 6 = 0 => 3k = -6 => k = -2.
Step 3: The correct option is (iv).
(c) Lines x - 5 = 0 and y + 3 = 0 intersect each other at point :
(i) (5, 3) (ii) (5, -3) (iii) (-5, 3) (iv) (-5, -3)
Answer:
Step 1: The lines are x = 5 and y = -3.
Step 2: Their intersection point is (5, -3).
Step 3: The correct option is (ii).
(d) For the line 4x - 7y + 6 = 0 if x = 2; the value of y is :
(i) 2 (ii) -2 (iii) 1 (iv) -1
Answer:
Step 1: Substitute x = 2 into the equation 4(2) - 7y + 6 = 0.
Step 2: 8 - 7y + 6 = 0 => 14 = 7y => y = 2.
Step 3: The correct option is (i).
(e) Statement (1) : The graph of 2x - 5 = 0 is a line parallel to y-axis.
Statement (2) : When the equations of lines are of the form x = ±k (a constant), the lines are parallel to y-axis.
(i) Both the statements are true. (ii) Both the statements are false. (iii) Statement 1 is true, and statement 2 is false. (iv) Statement 1 is false, and statement 2 is true.
Answer:
Step 1: 2x - 5 = 0 means x = 2.5, which is parallel to y-axis. So Statement 1 is true.
Step 2: Statement 2 correctly explains the rule for such lines. So Statement 2 is true.
Step 3: The correct option is (i).
(f) Statement (1) : The lines of the form ax ± by = 0 always pass through the origin.
Statement (2) : On substituting x = 0 and y = 0; we get a x 0 ± b x 0 = 0.
(i) Both the statements are true. (ii) Both the statements are false. (iii) Statement 1 is true, and statement 2 is false. (iv) Statement 1 is false, and statement 2 is true.
Answer:
Step 1: If there is no constant term (c=0), the line passes through (0,0). Statement 1 is true.
Step 2: Substituting x = 0 and y = 0 verifies this. Statement 2 is true.
Step 3: The correct option is (i).
(g) Assertion (A) : y + 5 = 0 is the equation of the line parallel to x-axis and at a distance of 5 unit in the negative direction from it.
Reason (R) : For all the points on the line y = a (a constant), the value of abscissa is a.
(i) A is true, R is false. (ii) A is false, R is true. (iii) Both A and R are true and R is the correct reason for A. (iv) Both A and R are true and R is the incorrect reason for A.
Answer:
Step 1: y + 5 = 0 => y = -5, which is parallel to x-axis at distance 5 downwards. Assertion (A) is true.
Step 2: For line y = a, the value of ordinate is a, not abscissa. Reason (R) is false.
Step 3: The correct option is (i).
(h) Assertion (A) : For the line 3x + 4y = 7, the abscissa is -3/4.
Reason (R) : For abscissa of a point, y = 0 and 3x + 4y = 7 => 3x = 7 and x = 7/3.
(i) A is true, R is false. (ii) A is false, R is true. (iii) Both A and R are true and R is the correct reason for A. (iv) Both A and R are true and R is the incorrect reason for A.
Answer:
Step 1: The value -3/4 is the slope of the line, not the abscissa. Assertion (A) is false.
Step 2: To find the abscissa of the x-intercept, we put y = 0, which yields x = 7/3. Reason (R) is true.
Step 3: The correct option is (ii).
2. Find the distance of point (8, -4) from y-axis.
Answer:
Step 1: The distance of any point (x, y) from the y-axis is the absolute value of its x-coordinate.
Step 2: Here, the x-coordinate is 8.
Step 3: Therefore, the distance is 8 units.
3. Three vertices of parallelogram ABCD are A(-5, -1), B(3, -1) and C(1, -6). Use graphical method to find the co-ordinates of fourth vertex D.
Answer:
Step 1: Observe that the y-coordinates of A and B are both -1, meaning side AB is horizontal.
Step 2: The length of AB is 3 - (-5) = 8 units.
Step 3: In a parallelogram, opposite sides are equal and parallel. So side CD is also horizontal and 8 units long.
Step 4: Since C is at (1, -6), and the order is ABCD, D must be 8 units to the left of C.
Step 5: x-coordinate of D = 1 - 8 = -7. The y-coordinate is the same as C, which is -6.
Step 6: The co-ordinates of D are (-7, -6).
4. In the given figure, ABC is an equilateral triangle. Find the co-ordinates of A.
Answer:
Step 1: From the graph, B is at (2, 0) and C is at (6, 0) on the x-axis.
Step 2: The length of base BC = 6 - 2 = 4 units.
Step 3: Since it is an equilateral triangle, all sides are 4 units long.
Step 4: The x-coordinate of A is exactly in the middle of B and C, so x = (2 + 6)/2 = 4.
Step 5: The height of the triangle is √(4² - 2²) = √(16 - 4) = √12 = 2√3.
Step 6: Therefore, the co-ordinates of A are (4, 2√3).
5. Draw the graph of 3x + 2y = 6. Use the graph drawn to find the area of triangle formed by the line drawn and the co-ordinate axes.
Answer:
Step 1: Find the x-intercept by putting y = 0 => 3x = 6 => x = 2. Point is (2, 0).
Step 2: Find the y-intercept by putting x = 0 => 2y = 6 => y = 3. Point is (0, 3).
Step 3: The base of the triangle is 2 units and the height is 3 units.
Step 4: Area = 1/2 × base × height = 1/2 × 2 × 3 = 3 sq. units.
6. Use the graphical method to find the value of k, if :
(i) (k, -3) lies on the straight line 2x + 3y = 1
(ii) (5, k - 2) lies on the straight line x - 2y + 1 = 0
Answer:
Step 1: For (i), substitute x = k, y = -3 into 2x + 3y = 1.
Step 2: 2k + 3(-3) = 1 => 2k - 9 = 1 => 2k = 10 => k = 5.
Step 3: For (ii), substitute x = 5, y = k - 2 into x - 2y + 1 = 0.
Step 4: 5 - 2(k - 2) + 1 = 0 => 6 - 2k + 4 = 0 => 10 - 2k = 0 => 2k = 10 => k = 5.
7. Find graphically, the vertices of the triangle whose sides have the equations 2y - x = 8; 5y - x = 14 and y - 2x = 1 respectively.
Take 1 cm = 1 unit on both the axes.
Answer:
Step 1: Solve 2y - x = 8 and 5y - x = 14 by subtracting: (5y - x) - (2y - x) = 14 - 8 => 3y = 6 => y = 2. Then x = -4. Vertex 1 is (-4, 2).
Step 2: Solve 2y - x = 8 and y - 2x = 1 (y = 2x + 1). Substitute y: 2(2x + 1) - x = 8 => 4x + 2 - x = 8 => 3x = 6 => x = 2, y = 5. Vertex 2 is (2, 5).
Step 3: Solve 5y - x = 14 and y - 2x = 1. Substitute y: 5(2x + 1) - x = 14 => 10x + 5 - x = 14 => 9x = 9 => x = 1, y = 3. Vertex 3 is (1, 3).
Step 4: The vertices are (-4, 2), (2, 5), and (1, 3).
8. Using the same axes of co-ordinates and the same unit, solve graphically :
x + y = 0 and 3x - 2y = 10.
Answer:
Step 1: From the first equation, y = -x.
Step 2: Substitute y = -x into the second equation: 3x - 2(-x) = 10.
Step 3: 3x + 2x = 10 => 5x = 10 => x = 2.
Step 4: y = -(2) = -2.
Step 5: The graphical solution intersection point is (2, -2).
9. Solve graphically, the following equations.
x + 2y = 4; 3x - 2y = 4.
Take 2 cm = 1 unit on each axis.
Also, find the area of the triangle formed by the lines and the x-axis.
Answer:
Step 1: Add the equations: (x + 2y) + (3x - 2y) = 4 + 4 => 4x = 8 => x = 2.
Step 2: Substitute x = 2 into first eq: 2 + 2y = 4 => 2y = 2 => y = 1. Intersection is (2, 1).
Step 3: Find x-intercepts by setting y = 0. First line: x = 4.
Step 4: Second line: 3x = 4 => x = 4/3.
Step 5: Base on x-axis = 4 - 4/3 = 8/3 units. Height is y-coordinate of intersection = 1 unit.
Step 6: Area = 1/2 × base × height = 1/2 × 8/3 × 1 = 4/3 sq. units (or 1.33 sq. units).
10. Use the graphical method to find the value of 'x' for which the expressions (3x + 2) / 2 and (3/4)x - 2 are equal.
Answer:
Step 1: Set y = (3x + 2) / 2 => 2y = 3x + 2.
Step 2: Set y = 3x/4 - 2 => 4y = 3x - 8.
Step 3: Equating them: (3x + 2) / 2 = (3x - 8) / 4.
Step 4: Multiply by 4: 2(3x + 2) = 3x - 8 => 6x + 4 = 3x - 8.
Step 5: 3x = -12 => x = -4.
11. The course of an enemy submarine, as plotted on rectangular co-ordinate axes, gives the equation 2x + 3y = 4. On the same axes, a destroyer's course is indicated by the graph x - y = 7. Use the graphical method to find the point at which the paths of the submarine and the destroyer intersect. (LS)
Answer:
Step 1: From the destroyer's equation, x = y + 7.
Step 2: Substitute into submarine's equation: 2(y + 7) + 3y = 4.
Step 3: 2y + 14 + 3y = 4 => 5y = -10 => y = -2.
Step 4: x = (-2) + 7 = 5.
Step 5: The paths intersect at point (5, -2).
Case-Study Based Question
An equation which can be put in the form ax + by + c = 0 is called a linear equation, where :(i) x and y are variables,
(ii) a, b and c are real numbers and
(iii) a and b are both not zero.
On drawing a graph for the two linear equations on the same plane, it is seen that only one of the following three possibilities can happen :
(Table with properties omitted here for brevity)
1. On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
(i) 7x - 5y + 10 = 0 and 6x + 2y - 15 = 0
Answer:
Step 1: Here a1/a2 = 7/6 and b1/b2 = -5/2.
Step 2: Since a1/a2 ≠ b1/b2, the lines intersect at a point.
(ii) 5x + 2y + 8 = 0 and 15x + 6y + 24 = 0
Answer:
Step 1: Here a1/a2 = 5/15 = 1/3, b1/b2 = 2/6 = 1/3, and c1/c2 = 8/24 = 1/3.
Step 2: Since a1/a2 = b1/b2 = c1/c2, the lines are coincident.
(iii) 4x - 8y + 9 = 0 and 2x - 4y + 7 = 0
Answer:
Step 1: Here a1/a2 = 4/2 = 2, b1/b2 = -8/-4 = 2, and c1/c2 = 9/7.
Step 2: Since a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel.
(iv) x - 2y = 0 and 3x - 4y - 20 = 0
Answer:
Step 1: Here a1/a2 = 1/3 and b1/b2 = -2/-4 = 1/2.
Step 2: Since a1/a2 ≠ b1/b2, the lines intersect at a point.
(v) 2x + 3y - 9 = 0 and 4x + 6y - 18 = 0
Answer:
Step 1: Here a1/a2 = 2/4 = 1/2, b1/b2 = 3/6 = 1/2, and c1/c2 = -9/-18 = 1/2.
Step 2: Since a1/a2 = b1/b2 = c1/c2, the lines are coincident.
(vi) x + 2y - 4 = 0 and 2x + 4y - 12 = 0
Answer:
Step 1: Here a1/a2 = 1/2, b1/b2 = 2/4 = 1/2, and c1/c2 = -4/-12 = 1/3.
Step 2: Since a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel.
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What is the general form of a linear equation in two variables?
Answer
$ax + by + c = 0$
Question
In the linear equation $ax + by + c = 0$, what do $x$ and $y$ represent?
Answer
Variables
Question
In the equation $ax + by + c = 0$, what do $a$, $b$, and $c$ represent?
Answer
Constants
Question
What is the recommended first step to draw the graph of a linear equation $ax + by + c = 0$?
Answer
Make either $x$ or $y$ the subject of the equation.
Question
When plotting a linear equation, what is the minimum number of suitable values required for the variable on the right-hand side?
Answer
At least three
Question
What geometric shape is formed by the graph of a linear equation in two variables?
Answer
A straight line
Question
What is the equation of the $x$-axis?
Answer
$y = 0$
Question
What is the equation of the $y$-axis?
Answer
$x = 0$
Question
Describe the graph of $x = a$, where $a$ is a constant.
Answer
A straight line parallel to the $y$-axis at a distance of $a$ units.
Question
Describe the graph of $y = b$, where $b$ is a constant.
Answer
A straight line parallel to the $x$-axis at a distance of $b$ units.
Question
How is the solution of two simultaneous linear equations determined graphically?
Answer
By finding the coordinates of the point of intersection of the two lines.
Question
If two lines drawn on a graph are parallel to each other, how many solutions do the equations have?
Answer
No solution
Question
If two lines drawn on a graph coincide, how many solutions do the equations have?
Answer
Infinite number of solutions
Question
Condition for two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have a unique solution.
Answer
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
Question
Condition for two linear equations to have no solution (parallel lines).
Answer
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Question
Condition for two linear equations to have infinite solutions (coincident lines).
Answer
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
Question
To find the $x$-intercept of a linear graph, what value must be assigned to $y$?
Answer
$0$
Question
To find the $y$-intercept of a linear graph, what value must be assigned to $x$?
Answer
$0$
Question
In the linear relation $y = mx + c$, what does the constant $c$ represent?
Answer
The $y$-intercept
Question
What formula is used to calculate the area of a triangle formed by a line and the coordinate axes?
Answer
$\frac{1}{2} \times \text{base} \times \text{height}$
Question
How do you graphically determine the value of $y$ for a specific value of $x$?
Answer
Draw a vertical line from $x$ to the graph, then a horizontal line to the $y$-axis.
Question
How do you graphically determine the value of $x$ for a specific value of $y$?
Answer
Draw a horizontal line from $y$ to the graph, then a vertical line to the $x$-axis.
Question
In a factory cost graph, what is the 'breakeven point'?
Answer
The point where the cost price equals the selling price (no profit and no loss).
Question
How is profit or loss calculated graphically from Cost Price ($C.P.$) and Selling Price ($S.P.$) lines at a specific quantity?
Answer
The vertical distance between the $S.P.$ and $C.P.$ lines at that quantity.
Question
If a point $(k, 2)$ lies on the line $x - 4y = 2$, what is the value of $k$?
Answer
$10$
Question
If the line $y = mx - 8$ passes through the point $(5, 2)$, what is the value of $m$?
Answer
$2$
Question
What are the co-ordinates of the point where the line $5x - 2y - 10 = 0$ intersects the $x$-axis?
Answer
$(2, 0)$
Question
What are the co-ordinates of the point where the line $5x - 4y - 20 = 0$ intersects the $y$-axis?
Answer
$(0, -5)$
Question
On which axis or axes does the point $(0, 0)$ lie?
Answer
Both the $x$-axis and the $y$-axis.
Question
The distance of the point $(8, -4)$ from the $y$-axis is _____ units.
Answer
$8$
Question
For all points on the line $y = a$, what is the value of the ordinate?
Answer
$a$
Question
What is the name for the $x$-coordinate of a point?
Answer
Abscissa
Question
What is the name for the $y$-coordinate of a point?
Answer
Ordinate
Question
If the point $(m, -2)$ lies on the line $2x - 3y + 12 = 0$, what is the value of $m$?
Answer
$-9$
Question
If the point $(3, n)$ lies on the line $2x - 3y + 12 = 0$, what is the value of $n$?
Answer
$6$
Question
A line parallel to the $x$-axis at a distance of $5$ units in the negative direction is represented by the equation _____.
Answer
$y + 5 = 0$
Question
Why do all lines of the form $ax \pm by = 0$ always pass through the origin?
Answer
Substituting $x = 0$ and $y = 0$ results in $0 = 0$, satisfying the equation.
Question
What is the area of the triangle formed by the line $3x + 2y = 6$ and the co-ordinate axes?
Answer
$3$ square units
Question
Lines that pass through a common point are known as _____ lines.
Answer
Concurrent
Question
How can you verify graphically if three given lines are concurrent?
Answer
Plot the three lines and observe if they intersect at exactly one common point.
Question
If $2x + 3y = 12$, what is the value of $x$ when $y = 2$?
Answer
$3$
Question
If $2x + 3y = 12$, what is the value of $y$ when $x = 0$?
Answer
$4$
Question
State the linear relation between $x$ and $y$ if the line passes through $(-2, -3)$ and $(0, 1)$.
Answer
$y = 2x + 1$
Question
For the line $2x - 3y - 5 = 0$, find the value of $x_1$ when $y = 7$.
Answer
$13$
Question
If the straight line $x - 3y = 18$ passes through $(m, -5)$, find $m$.
Answer
$3$
Question
If the straight line $x - 3y = 18$ passes through $(6, n)$, find $n$.
Answer
$-4$
Question
In the comparison of ratios for the lines $x - 2y = 0$ and $3x - 4y - 20 = 0$, do the lines intersect at a point?
Answer
Yes, because $\frac{1}{3} \neq \frac{-2}{-4}$.
Question
For the lines $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$, are the lines coincident?
Answer
Yes, because $\frac{2}{4} = \frac{3}{6} = \frac{-9}{-18}$.
Question
For the lines $7x - 5y + 10 = 0$ and $6x + 2y - 15 = 0$, what is the solution type?
Answer
Unique solution
Question
What is the equation of a line parallel to the $y$-axis passing through the point $(8, -4)$?
Answer
$x = 8$
Question
What is the equation of a line parallel to the $x$-axis passing through the point $(8, -4)$?
Answer
$y = -4$
Question
How do you represent the situation where a destroyer's course and a submarine's course intersect?
Answer
By the coordinates of the point of intersection of their respective graphical paths.
Question
In the equation $y = mx + c$, if $m = 0$, describe the graph.
Answer
A horizontal line (parallel to the $x$-axis).
Question
In the equation $ax + by + c = 0$, if $c = 0$, where must the line pass?
Answer
Through the origin $(0, 0)$.
Question
If $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, what can be concluded about the two lines relative to each other?
Answer
They are either parallel or coincident.
Question
What is the distance between the lines $y = 3$ and $y = -2$?
Answer
$5$ units
Question
What is the distance between the lines $x = 4$ and $x = 10$?
Answer
$6$ units
Question
If the area of a triangle formed by a line and the axes is $0$, what does this imply about the line?
Answer
The line passes through the origin.
Question
Find the co-ordinates of the vertices of a triangle formed by the lines $x = 0$, $y = 0$, and $x + y = 5$.
Answer
$(0, 0), (5, 0), (0, 5)$
Question
What is the relationship between the abscissa and ordinate for all points on the line $y = x$?
Answer
They are equal.