Graphical Solution — Test Yourself

For the line 2x - y = 7 and for x = 3 the value of y is :

1. 1

2. -1

3. 5

4. -5

Answer

Given, equation of lines 2x - y = 7 and x = 3

Substituting x = 3 in 2x - y = 7, we get :

⇒ 2 x 3 - y = 7

⇒ 6 - y = 7

⇒ y = 6 - 7

⇒ y = -1.

Hence, option 2 is the correct option.

The line x + 3y + 2 = 0 passes through the point (4, k); then the value of k is;

1. 2

2. 1

3. -1

4. -2

Answer

If the line x + 3y + 2 = 0 passes through the point (4, k), then substituting x = 4 and y = k into the equation of the line,

⇒ 4 + 3 x k + 2 = 0

⇒ 6 + 3k = 0

⇒ 3k = -6

⇒ k = $-\dfrac{6}{3}$

⇒ k = -2.

Hence, option 4 is the correct option.

Line x - 5 = 0 and y + 3 = 0 intersect each other at point:

1. (5, 3)

2. (5, -3)

3. (-5, 3)

4. (-5, -3)

Answer

⇒ x - 5 = 0

⇒ x = 5

This line represents a vertical line parallel to y-axis.

⇒ y + 3 = 0

⇒ y = ​-3

This represents a horizontal line parallel to x-axis.

Point of intersection of lines x = 5 and y = -3 is (5, -3).

Hence, option 2 is the correct option.

For the line 4x - 7y + 6 = 0 if x = 2; the value of y is :

1. 2

2. -2

3. 1

4. -1

Answer

Given,
equation of line 4x - 7y + 6 = 0

Substitute the given value of x (which is 2) into the equation:

⇒ 4 x 2 - 7y + 6 = 0

⇒ 8 - 7y + 6 = 0

⇒ 14 - 7y = 0

⇒ 7y = 14

⇒ y = $\dfrac{14}{7}$

⇒ y = 2.

Hence, option 1 is the correct option.

Statement 1: The graph of 2x - 5 = 0 is a line parallel to y-axis.

Statement 2: When the equations of line are of the form x = ± k (a constant), the lines are parallel to y-axis.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

⇒ 2x - 5 = 0

⇒ 2x = 5

⇒ x = $\dfrac{5}{2}$

This represents a vertical line (parallel to the y-axis) at a distance of $\dfrac{5}{2}$ units.

This line is at a distance of $\dfrac{5}{2}$ units to the right of the y-axis because the x-coordinate of every point on the line is $\dfrac{5}{2}$.

∴ Statement 1 is true.

For equations of the form x = ± k.

These represent vertical lines (parallel to the y-axis) because the value of x remains constant at k or -k for all values of y.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

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.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

A line passes through the origin (0, 0) if the coordinates of the origin satisfy the equation of the line.

Substituting x = 0 and y = 0 in L.H.S. of the equation ax ± by = 0, we get :

⇒ a(0) ± b(0)

⇒ 0, which is equal to R.H.S.

Since, (0, 0) satisfies the equation ax ± by = 0, thus lines of the form ax ± by = 0 always pass through the origin.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): y + 5 = 0 is the equation of line parallel to x-axis and at the distance of 5 unit in the negative direction from it.

Reason (R): For all the points on the y = a (a constant), the value of abscissa is a.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

⇒ y + 5 = 0

⇒ y = -5

This represents a horizontal line (parallel to the x-axis) because the y-coordinate remains constant at -5 for all values of x. The line is 5 units below the x-axis since the y-coordinate of every point is −5, which is 5 units in the negative direction from the x-axis.

∴ Assertion (R) is true.

For all points on the line y = a, the value of ordinate (y-coordinate) = a.

∴ Reason (R) is false.

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Assertion (A): For the line 3x + 4y = 7, the abscissa is $-\dfrac{3}{4}$.

Reason (R): For abscissa of a point, y = 0 and 3x + 4y = 7

⇒ 3x = 7 and x = $\dfrac{7}{3}$.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, equation of line : 3x + 4y = 7

When, y = 0

⇒ 3x + 4 x 0 = 7

⇒ 3x = 7

⇒ x = $\dfrac{7}{3}$

∴ Reason (R) is true.

So, the abscissa is $\dfrac{7}{3} \text{ and not } -\dfrac{3}{4}$.

∴ Assertion (A) is false.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Find the distance of point (8, -4) from y-axis.

Answer

Plot the point (8, -4) on the graph paper.

Draw a perpendicular line from the point (8, -4) to the y-axis.

From graph it is clear that the distance between the point (8,−4) and the y-axis is 8 units.

Hence, the distance of point (8, -4) from y-axis is 8 units.

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

Plot the points A(-5, -1), B(3, -1) and C(1, -6) on the graph paper. Join point A with B and B with C.

From the graph, it is clear that the horizontal distance between the points A (-5, -1) and B (3, -1) is 8 units and the vertical distance between the points B (3, -1) and C (1, -6) is 5 units. Therefore, the vertical distance between the points A (-5, -1) and D must be 5 units and the horizontal distance between the points C (1, -6) and D must be 8 units.

Now, complete the parallelogram ABCD and read the coordinates of point D. As shown on the graph, D = (-7, -6).

Hence, the co-ordinates of fourth vertex D are (-7, -6).

In the given figure, ABC is an equilateral triangle. Find the co-ordinates of A.

Answer

Given:

The co-ordinates of B = (2, 0)

The co-ordinates of C = (6, 0)

So, the length of BC = 6 - 2 = 4 units

Since ABC is an equilateral triangle,

The height of the triangle = $\dfrac{\sqrt3}{2}a$

= $\dfrac{\sqrt3}{2} \times 4$

= 2 $\sqrt3$ units

The mid-point of BC = $\Big(\dfrac{2+6}{2}, \dfrac{0+0}{2}\Big)$

= (4, 0)

Since Δ ABC is equilateral, and BC lies on the x-axis:

• The abscissa (x-coordinate) of A is the same as the midpoint of BC, i.e., x=4.

• The ordinate (y-coordinate) of A is the height of the triangle, $\sqrt{3}$.

Hence, the co-ordinates of A are (4, 2 $\sqrt{3}$).

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:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 + 2y = 6 ⇒ y = 3

Let x = 2, then 3 $\times$ 2 + 2y = 6 ⇒ y = 0

Let x = 4, then 3 $\times$ 4 + 2y = 6 ⇒ y = -3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

The area of the triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x OA x OB

= $\dfrac{1}{2}$ x 2 x 3

= $\dfrac{6}{2}$ sq. units

= 3 sq. units

Hence, the area of triangle formed by the line drawn and the co-ordinate axes is 3 sq. units.

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

(i) 2x + 3y = 1

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 2 $\times$ 0 + 3y = 1 ⇒ y = 0.3

Let x = 2, then 2 $\times$ 2 + 3y = 1 ⇒ y = -1

Let x = 4, then 2 $\times$ 4 + 3y = 1 ⇒ y = -2.3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Since, point (k, -3) lies on the straight line drawn, through y = -3, draw a horizontal line which meets the straight line at point, say Q. Through point Q, draw a vertical line which meets the x-axis at 5.

Hence, the value of k = 5.

(ii) x - 2y + 1 = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then -1 - 2y + 1 = 0 ⇒ y = 0

Let x = 0, then 0 - 2y + 1 = 0 ⇒ y = 0.5

Let x = 1, then 1 - 2y + 1 = 0 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Since, point (5, k - 2) lies on the straight line drawn, through x = 5, draw a vertical line which meets the graph at a point, say Q. Through point Q, draw a horizontal line which meets the y-axis at point 3.

k - 2 = 3

⇒ k = 3 + 2

⇒ k = 5

Hence, the value of k = 5.

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

First equation: 2y - x = 8

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then 2y - (-4) = 8 ⇒ y = 2

Let x = -2, then 2y - (-2) = 8 ⇒ y = 3

Let x = 0, then 2y - 0 = 8 ⇒ y = 4

Let x = 2, then 2y - 2 = 8 ⇒ y = 5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 5y - x = 14

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then 5y - (-4) = 14 ⇒ y = 2

Let x = 0, then 5y - 0 = 14 ⇒ y = 2.8

Let x = 2, then 5y - 2 = 14 ⇒ y = 3.2

Let x = 4, then 5y - 4 = 14 ⇒ y = 3.6

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Third equation: y - 2x = 1

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then y - 2 $\times$ 0 = 1 ⇒ y = 1

Let x = 2, then y - 2 $\times$ 2 = 1 ⇒ y = 5

Let x = 4, then y - 2 $\times$ 4 = 1 ⇒ y = 9

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

The vertices of the triangle are:

The coordinate of A = (2, 5)

The coordinate of B = (-4, 2)

The coordinate of C = (1, 3)

Hence, the coordinates of the vertices of triangle = (1, 3), (-4, 2) and (2, 5).

Using the same axes of co-ordinates and the same unit, solve graphically :

x + y = 0 and 3x - 2y = 10.

(Take at least 3 points for each line drawn).

Answer

First equation: x + y = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 0 + y = 0 ⇒ y = 0

Let x = 1, then 1 + y = 0 ⇒ y = -1

Let x = 2, then 2 + y = 0 ⇒ y = -2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 3x - 2y = 10

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 - 2y = 10 ⇒ y = -5

Let x = 1, then 3 $\times$ 1 - 2y = 10 ⇒ y = -3.5

Let x = 2, then 3 $\times$ 2 - 2y = 10 ⇒ y = -2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Both the straight line drawn meet the point P. As it is clear from the graph, co-ordinates of the common point P are (2, -2).

Solution of the given equation x = 2 and y = -2.

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

First equation: x + 2y = 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 0 + 2y = 4 ⇒ y = 2

Let x = 2, then 2 + 2y = 4 ⇒ y = 1

Let x = 4, then 4 + 2y = 4 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 3x - 2y = 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 - 2y = 4 ⇒ y = -2

Let x = 2, then 3 $\times$ 2 - 2y = 4 ⇒ y = 1

Let x = 4, then 3 $\times$ 4 - 2y = 4 ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Both the straight line drawn meet the point A. As it is clear from the graph, co-ordinates of the common point A are (2, 1).

Solution of the given equation x = 2 and y = 1.

The area of the triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x BC x AD

= $\dfrac{1}{2}$ x 2.6 x 1

= 1.3 sq. units

Hence, the area of the triangle formed between the lines and the x-axis = 1.3 sq. units.

Use the graphical method to find the value of 'x' for which the expressions $\dfrac{3x + 2}{2}$ and $\dfrac{3}{4}x - 2$ are equal.

Answer

First equation: y = $\dfrac{3x + 2}{2}$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then y = $\dfrac{3 \times (-4) + 2}{2}$ ⇒ y = -5

Let x = -2, then y = $\dfrac{3 \times (-2) + 2}{2}$ ⇒ y = -2

Let x = 2, then y = $\dfrac{3 \times 2 + 2}{2}$ ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: y = $\dfrac{3}{4}x - 2$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then y = $\dfrac{3}{4} \times (-4) - 2$ ⇒ y = -5

Let x = 4, then y = $\dfrac{3}{4} \times 4 - 2$ ⇒ y = 1

Let x = 8, then y = $\dfrac{3}{4} \times 8 - 2$ ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

After plotting both lines on the graph, observe where the two lines intersect. The intersection point represents the value of x where both equations are equal.

From the graph, we see that the lines intersect at x = -4.

Hence, the value of x for which the two equations are equal is -4.

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.

Answer

Enemy equation: 2x + 3y = 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then 2 $\times$ (-4) + 3y = 4 ⇒ y = 4

Let x = -1, then 2 $\times$ (-1) + 3y = 4 ⇒ y = 2

Let x = 5, then 2 $\times$ 5 + 3y = 4 ⇒ y = -2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Destroyer equation: x - y = 7

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 5, then 5 - y = 7 ⇒ y = -2

Let x = 9, then 9 - y = 7 ⇒ y = 2

Let x = 11, then 11 - y = 7 ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Both the straight lines intersect at point P. As it is clear from the graph, co-ordinates of point P are (-2, -1).

Hence, (5, -2) is the point at which the paths of the submarine and the destroyer intersect.