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Page No 13.23:
Question 1:
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x − y = 0
(iii) −x + y = 6
(iv) y = 2x
(v) 3x + 5y = 15
(vi) $\frac{x}{2}-\frac{y}{3}=2$
(vii) $\frac{x-2}{3}=y-3$
(viii) 2y = −x + 1
Answer:
(i) We are given,
x + y = 4
We get,
y = 4 – x,
Now, substituting x = 0 in y = 4 – x, we get
y = 4
Substituting x = 4 in y = 4 – x, we get
y = 0
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given
x |
0 |
4 |
y |
4 |
0 |
(ii) We are given,
We get,
Now, substituting x = 0 in y = x – 2, we get
Substituting x = 2 in y = x – 2, we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
2 |
y |
–2 |
0 |
(iii) We are given,
We get,
Now, substituting in ,we get
Substituting x = –6 in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
−6 |
y |
6 |
0 |
(iv) We are given,
Now, substituting x = 1 in ,we get
Substituting x = 3 in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
1 |
3 |
y |
2 |
6 |
(v) We are given,
We get,
Now, substituting x = 0 in ,we get
Substituting x = 5 in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
5 |
y |
3 |
0 |
(vi) We are given,
We get,
Now, substituting x = 0 in ,we get
Substituting x = 4 in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
4 |
y |
–6 |
0 |
(vii) We are given,
We get,
Now, substituting x = 5 in ,we get
Substituting x = 8 in ,we get
y = 5
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
5 |
8 |
y |
4 |
5 |
(viii) We are given,
We get,
Now, substituting x = 1 in ,we get
Substituting x = 5 in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
1 |
5 |
y |
–1 |
–2 |
Page No 13.23:
Question 2:
Give the equations of two lines passing through (3, 12). How many more such lines are there, and why?
Answer:
We observe that x = 3 and y = 12 is the solution of the following equations
So, we get the equations of two lines passing through (3, 12) are, 4x – y = 0 and 3x – y + 3 = 0.
We know that passing through the given point infinitely many lines can be drawn. So, there are infinitely many lines passing through (3,12)
Page No 13.23:
Question 3:
A three-wheeler scooter charges Rs 15 for first kilometer and Rs 8 each for every subsequent kilometer. For a distance of x km, an amount of Rs y is paid. Write the linear equation representing the above information.
Answer:
Total fare of Rs y for covering the distance of x km is given by
y = 15 + 8(x − 1)
y = 15 + 8x − 8
y = 8x + 7
Where, Rs y is the total fare (x – 1) is taken as the cost of first kilometer is already given Rs 15 and 1 has to subtracted from the total distance travelled to deduct the cost of first kilometer.
Page No 13.23:
Question 4:
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Aarushi paid Rs 27 for a book kept for seven days. If fixed charges are Rs x and per day charges are Rs y. Write the linear equation representing the above information.
Answer:
Total charges of Rs 27 of which Rs x for first three days and Rs y per day for 4 more days is given by
Here, is taken as the charges for the first three days are already given at Rs x and we have to find the charges for the remaining four days as the book is kept for the total of 7 days.
Page No 13.23:
Question 5:
A number is 27 more than the number obtained by reversing its digits. If its unit's and ten's digit are x and y respectively, write the linear equation representing the above statement.
Answer:
The number given to us is in the form of yx,
where y represents the ten’s place of the number
And x represents the unit’s place of the number
Now, the given number is
number obtained by reversing the digits of the number is
It is given to us that the original number is 27 more than the number obtained by reversing its digits
So,
Page No 13.23:
Question 6:
The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten's digit of the number are x and y respectively, then write the linear equation representing the above statement.
Answer:
The number given to us is in the form of yx.
where y represents the ten^{’}s place of the number
And x represents the units place of the number
Now, the given number is
number obtained by reversing the digits of the number is
It is given to us that the sum of these two numbers is 121
So,
Page No 13.23:
Question 7:
Plot the points (3,5) and (−1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).
Answer:
The required graph is below:-
By plotting the given points (3, 5) and (–1, 3) on a graph paper, we get the line BC.
We have already plotted the point A (1, 4) on the given plane by the intersecting lines.
Therefore, it is proved that the straight line passing through (3, 5) and (–1, 3) also passes through A (1, 4).
Page No 13.23:
Question 8:
From the choices given below, choose the equation whose graph is given in the figure.
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
Answer:
We are given co-ordinates (1, –1) and (–1, 1) as the solution of one of the following equations.
We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.
(i) We are given,
Substituting ,we get
Substituting ,we get
Therefore, the given equationdoes not represent the graph in the figure.
(ii) We are given,
Substituting ,we get
Substituting ,we get
Therefore, the given solutions satisfy this equation. Thus, it is the equation whose graph is given.
Page No 13.24:
Question 9:
From the choices given below, choose the equation whose graph is given in the figure.
(i) y= x + 2
(ii) y = x −2
(iii) y = x + 2
(iv) x + 2y = 6
Answer:
We are given co-ordinates (–1, 3) and (2, 0) as the solution of one of the following equations.
We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.
(i) We are given,
Substituting ,we get
Substituting ,we get
Therefore, the given solutions does not satisfy this equation.
(ii) We are given,
Substituting ,we get
Substituting ,we get
Therefore, the given solutions does not completely satisfy this equation.
(iii) We are given,
Substituting ,we get
Substituting ,we get
Therefore, the given solutions satisfy this equation. Thus, it is the equation whose graph is given.
Page No 13.24:
Question 10:
If the point (2, −2) lies on the graph of the linear equation 5x + ky = 4, find the value of k.
Answer:
It is given that the point lies on the given equation,
Clearly, the given point is the solution of the given equation.
Now,
Substituting in the given equation, we get
Page No 13.25:
Question 11:
Draw the graph of the equation 2x + 3y = 12. From the graph, find the coordinates of the point:
(i) whose y-coordinates is 3.
(ii) whose x-coordinate is −3.
Answer:
We are given,
We get,
Substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
6 |
y |
4 |
0 |
By plotting the given equation on the graph, we get the point B (0, 4) and C (6,0).
(i) Co-ordinates of the point whose y axis is 3 are A
(ii) Co-ordinates of the point whose x -coordinate is –3 are D
Page No 13.25:
Question 12:
Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
(i) 6x − 3y = 12
(ii) −x + 4y = 8
(iii) 2x + y = 6
(iv) 3x + 2y + 6 = 0
Answer:
(i) We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
2 |
y |
–4 |
0 |
Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and
at x axis.
(ii) We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
–8 |
y |
2 |
0 |
Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and
at x axis.
(iii) We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
3 |
y |
6 |
0 |
Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and
at x axis.
(iv) We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
–2 |
y |
–3 |
0 |
Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and
at x axis.
Page No 13.25:
Question 13:
Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.
Answer:
We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
3 |
y |
6 |
0 |
The region bounded by the graph is ABC which forms a triangle.
AC at y axis is the base of triangle having AC = 6 units on y axis.
BC at x axis is the height of triangle having BC = 3 units on x axis.
Therefore,
Area of triangle ABC, say A is given by
Page No 13.25:
Question 14:
Draw the graph of the equation $\frac{x}{3}+\frac{y}{4}=1$. Also, find the area of the triangle formed by the line and the coordinates axes.
Answer:
We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
3 |
y |
4 |
0 |
The region bounded by the graph is ABC which form a traingle.
AC at y axis is the base of traingle having AC = 4 units on y axis.
BC at x axis is the height of traingle having BC = 3 units on x axis.
Therefore,
Area of traingle ABC,say A is given by
Page No 13.25:
Question 15:
Draw the graph of y = | x |.
Answer:
We are given,
Substituting, we get
Substituting, we get
Substituting, we get
Substituting, we get
For every value of x, whether positive or negative, we get y as a positive number.
Page No 13.25:
Question 16:
Draw the graph of y = | x | + 2.
Answer:
We are given,
Substituting, we get
Substituting, we get
Substituting, we get
Substituting, we get
Substituting, we get
For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units.
Page No 13.25:
Question 17:
Draw the graphs of the following linear equations on the same graph paper:
2x + 3y = 12, x − y = 1
Find the coordinates of the vertices of the triangle formed by the two straight lines and the area bounded by these lines and x-axis.
Answer:
We are given,
We get,
Now, substituting in , we get
Substituting in , we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
6 |
y |
4 |
0 |
Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation .
We are given,
We get,
$y=x-1$
Now, substituting in $y=x-1$,we get
$y=-1$
Substituting in $y=x-1$,we get
$y=-2$
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
–1 |
y |
-1 |
-2 |
Plotting D(0,1) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation .
By the intersection of lines formed by and on the graph, triangle ABC is formed on y axis.
Therefore,
AC at y axis is the base of triangle ABC having AC = 5 units on y axis.
Draw FE perpendicular from F on y axis.
FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.
Therefore,
Area of triangle ABC, say A is given by
$A=\frac{1}{2}\left(\mathrm{Base}\times \mathrm{Height}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\mathrm{AC}\times \mathrm{FE}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(5\times 3\right)\phantom{\rule{0ex}{0ex}}=\frac{15}{2}\mathrm{sq}.\mathrm{units}$
Page No 13.25:
Question 18:
Draw the graphs of the linear equations 4x − 3y + 4 = 0 and 4x + 3y −20 = 0. Find the area bounded by these lines and x-axis.
Answer:
We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
–1 |
y |
0 |
Plotting E(0, ) and A(-1,0) on the graph and by joining the points , we obtain the graph of equation .
We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x | 0 | 5 |
y | 0 |
Plotting D(0, ) and B(5,0) on the graph and by joining the points , we obtain the graph of equation .
By the intersection of lines formed by and on the graph, triangle ABC is formed on x axis.
Therefore,
AB at x axis is the base of triangle ABC having AB = 6 units on x axis.
Draw CF perpendicular from C on x axis.
CF parallel to y axis is the height of triangle ABC having CF = 4 units on y axis.
Therefore,
Area of triangle ABC, say A is given by
Page No 13.25:
Question 19:
The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.
Answer:
We are given the path of train A,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
4 |
y |
3 |
0 |
Plotting A(4,0) and E(0,3) on the graph and by joining the points , we obtain the graph of equation .
We are given the path of train B,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
8 |
y |
6 |
0 |
Plotting C(0,6) and D(8,0) on the graph and by joining the points , we obtain the graph of equation
Page No 13.25:
Question 20:
Ravish tells his daughter Aarushi, ''Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be''. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.
Answer:
We are given the present age of Ravish as y years and Aarushi as x years.
Age of Ravish seven years ago
Age of Aarushi seven years ago
It has already been said by Ravish that seven years ago he was seven times old then Aarushi was then
So,
Age of Ravish three years from now
Age of Aarushi three years from now
It has already been said by Ravish that three years from now he will be three times old then Aarushi will be then
So,
(1) and (2) are the algebraic representation of the given statement.
We are given,
We get,
Now, substituting in ,we get
Substituting in, we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
6 |
y |
–42 |
0 |
We are given,
We get,
Now, substituting in ,we get
Substituting in ,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
–2 |
y |
6 |
0 |
The red -line represents the equation.
The blue-line represents the equation.
Page No 13.25:
Question 21:
Aarushi was driving a car with uniform speed of 60km/h. Draw distance-time graph. From the graph, find the distance travelled by Aarushi in
(i) $2\frac{1}{2}$ Hours
(ii) $\frac{1}{2}$ Hour
Answer:
Aarushi is driving the car with the uniform speed of 60 km/h.
We represent time on X-axis and distance on Y-axis
Now, graphically
We are given that the car is travelling with a uniform speed 60 km/hr. This means car travels 60 km distance each hour. Thus the graph we get is of a straight line.
Also, we know when the car is at rest, the distance travelled is 0 km, speed is 0 km/hr and the time is also 0 hr.
Thus, the given straight line will pass through O(0,0) and M(1,60).
Join the points O and M and extend the line in both directions.
Now, we draw a dotted line parallel to y-axis from x = $\frac{1}{2}$ that meets the straight line graph at L from which we draw a line parallel to x-axis that crosses y-axis at 30. Thus, in $\frac{1}{2}$hr, distance travelled by the car is 30 km.
Now, we draw a dotted line parallel to y-axis from x = $2\frac{1}{2}$ that meets the straight line graph at N from which we draw a line parallel to x-axis that crosses y-axis at 150. Thus, in $2\frac{1}{2}$hr, distance travelled by the car is 150 km.
(i)
Distance travelled in hours is given by
(ii)
Distance travelled in hours is given by
Page No 13.3:
Question 1:
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) −2x + 3y = 12
(ii) $x-\frac{y}{2}-5=0$
(iii) 2x + 3y = 9.35
(iv) 3x = −7y
(v) 2x + 3 = 0
(vi) y − 5 = 0
(vii) 4 = 3x
(viii) $y=\frac{x}{2}$
Answer:
(i) We are given
$-2x+3y-12=0$
Comparing the given equation with ,we get
$\overline{)a=-2;b=3;c=-12}$
(ii) We are given
Comparing the given equation with ,we get
(iii) We are given
Comparing the given equation with ,we get
(iv) We are given
Comparing the given equation with ,we get
(v) We are given
Comparing the given equation with ,we get
(vi) We are given
Comparing the given equation with ,we get
(vii) We are given
Comparing the given equation with ,we get
(viii) We are given
Comparing the given equation with ,we get
Page No 13.3:
Question 2:
Write each of the following as an equation in two variables:
(i) 2x =−3
(ii) y = 3
(iii) 5x = $\frac{7}{2}$
(iv) y = $\frac{3}{2}x$
Answer:
(i) We are given,
Now, in two variable forms the given equation will be
(ii) We are given,
Now, in two variable forms the given equation will be
(iii) We are given,
Now, in two variable forms the given equation will be
(iv) We are given,
Now, in two variable forms the given equation will be
Page No 13.3:
Question 3:
The cost of ball pen is Rs 5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.
Answer:
Let the cost of fountain pen be x and cost of ball pen be y.
According to the given equation, we have
Here x is the cost of one fountain pen and y is that of one ball pen.
Page No 13.32:
Question 1:
Give the geometric representations of the following equations
(a) on the number line
(b) on the Cartesian plane:
(i) x = 2
(ii) y + 3 = 0
(iii) y = 3
(iv) 2x + 9 = 0
(v) 3x − 5 = 0
Answer:
(i) We are given,
x = 2
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point (2, 0) is shown below
(ii) We are given,
We get,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane, it is a line parallel to x axis passing through the point A(0, –3) is shown below
(iii) We are given,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane, it is a line parallel to x axis passing through the point (0, 3) is shown below
(iv) We are given,
We get,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane,it is a line parallel to y axis passing through the point is shown below
(v) We are given,
We get,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below
Page No 13.32:
Question 2:
Give the geometrical representation of 2x + 13 = 0 as an equation in
(i) on variable
(ii) two variable
Answer:
We are given,
We get,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below
Page No 13.32:
Question 3:
Solve the equation 3x + 2 = x − 8, and represent the solution on (i) the number line
(ii) the Cartesian plane.
Answer:
We are given,
we get,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point (–5, 0) is shown below
Page No 13.32:
Question 4:
Write the equation of the line that is parallel to x-axis and passing through the point.
(i) (0,3)
(ii) (0, −4)
(iii) (2, −5)
(iv) (3, 4)
Answer:
(i) We are given the co-ordinates of the Cartesian plane at (0,3).
For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to 3.
We get the equation as
(ii) We are given the co-ordinates of the Cartesian plane at (0,-4).
For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to -4.
We get the equation as
(iii) We are given the co-ordinates of the Cartesian plane at (2,-5).
For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to -5.
We get the equation as
(iv) We are given the co-ordinates of the Cartesian plane at (3,4).
For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to 4.
We get the equation as
Page No 13.32:
Question 5:
Write the equation of the line that is parallel to y-axis and passing through the point
(i) (4,0)
(ii) (−2,0)
(iii) (3, 5)
(iv) (−4, −3)
Answer:
(i) We are given the co-ordinates of the Cartesian plane at (4,0).
For the equation of the line parallel to y axis ,we assume the equation as a one variable equation independent of y containing x equal to 4.
We get the equation as
(ii) We are given the co-ordinates of the Cartesian plane at (–2,0).
For the equation of the line parallel to y axis ,we assume the equation as a one variable equation independent of y containing x equal to –2.
We get the equation as
(iii) We are given the co-ordinates of the Cartesian plane at (3,5).
For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to 3.
We get the equation as
(iv) We are given the co-ordinates of the Cartesian plane at (−4,−3).
For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to −4.
We get the equation as
Page No 13.32:
Question 1:
Write the equation representing x-axis.
Answer:
The equation of line representing x axis is given by
Page No 13.32:
Question 2:
Write the equation representing y-axis.
Answer:
The equation of line representing y axis is given by
Page No 13.32:
Question 3:
Write the equation of a line passing through the point (0, 4) and parallel to x-axis.
Answer:
We are given the co-ordinates of the Cartesian plane at (0,4).
For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 4.
We get the equation as
Page No 13.32:
Question 4:
Write the equation of a line passing through the point (3, 5) and parallel to x-axis.
Answer:
We are given the co-ordinates of the Cartesian plane at (3,5).
For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 5.
We get the equation as
Page No 13.32:
Question 5:
Write the equation of a line parallel to y-axis and passing through the point (−3, −7).
Answer:
We are given the co-ordinates of the Cartesian plane at (–3,–7).
For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to –3.
We get the equation as
Page No 13.32:
Question 6:
A line passes through the point (−4, 6) and is parallel to x-axis. Find its equation.
Answer:
We are given the co-ordinates of the Cartesian plane at (–4,6).
For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 6.
We get the equation as
Page No 13.32:
Question 7:
Solve the equation 3x − 2 = 2x + 3 and represent the solution on the number line.
Answer:
We are given,
we get,
The representation of the solution on the number line, when given equation is treated as an equation in one variable.
Page No 13.32:
Question 8:
Solve the equation 2y − 1 = y + 1 and represent it graphically on the coordinate plane.
Answer:
We are given,
we get,
The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below
Page No 13.32:
Question 9:
If the point (a, 2) lies on the graph of the linear equation 2x − 3y + 8 = 0, find the value of a.
Answer:
We are given lies on the graph of linear equation.
So, the given co-ordinates are the solution of the equation.
Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation.
Substituting in equation, we get
Page No 13.32:
Question 10:
Find the value of k for which the point (1, −2) lies on the graph of the linear equation x − 2y + k = 0.
Answer:
We are given lies on the graph of linear equation.
So, the given co-ordinates are the solution of the equation.
Therefore, we can calculate the value of k by substituting the value of given co-ordinates in equation.
Substituting in equation, we get
Page No 13.33:
Question 1:
Mark the correct alternative in each of the following:
If (4, 19) is a solution of the equation y = ax + 3, then a=
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
We are given (4, 19)as the solution of equation
Substituting x = 4 and y = 19, we get
Therefore, the correct answer is (b).
Page No 13.33:
Question 2:
If (a, 4) lies on the graph of 3x + y = 10, then the value of a is
(a) 3
(b) 1
(c) 3
(d) 4
Answer:
We are given (a, 4) lies on the graph of linear equation 3x + y = 10.
So, the given co-ordinates are the solution of the equation 3x + y = 10.
Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation 3x + y = 10.
Substituting x = a and y = 4 in equation 3x + y = 10, we get
No option is correct.
Page No 13.33:
Question 3:
The graph of the linear equation 2x − y = 4 cuts x- axis at
(a) (2, 0)
(b) (−2, 0)
(c) (0, −4)
(d) (0, 4)
Answer:
We are given,
we get,
We will substitute in to get the co-ordinates for the graph of at x axis
Co-ordinates for the graph of are .
Therefore, the correct answer is (a).
Page No 13.33:
Question 4:
How many linear equations are satisfied by x = 2 and y = −3?
(a) Only one
(b) Two
(c) Three
(d) Infinitely many
Answer:
There are infinite numbers of linear equations that are satisfied by as
(i) Every solution of the linear equation represent a point on the line.
(ii) Every point on the line is the solution of the linear equation.
Therefore, the correct answer is (d).
Page No 13.33:
Question 5:
The equation x − 2 = 0 on number line is represented by
(a) a line
(b) a point
(c) infinitely many lines
(d) two lines
Answer:
The equation is represented by a point on the number line.
Therefore, the correct answer is (b).
Page No 13.33:
Question 6:
x = 2, y = −1 is a solution of the linear equation
(a) x + 2y = 0
(b) x + 2y = 4
(c) 2x + y = 0
(d) 2x + y = 5
Answer:
We are given as the solution of linear equation, which we have to find?
The equation is which can be proved by
Substituting in the equation, we get
Therefore, the correct answer is (a).
Page No 13.33:
Question 7:
If (2k − 1, k) is a solution of the equation 10x − 9y = 12, then k =
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
We are given as the solution of equation
Substituting, we get
Therefore, the correct answer is (b).
Page No 13.33:
Question 8:
The distance between the graph of the equations x = −3 and x = 2 is
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
Distance between the graph of equations, say D
D = Distance of co-ordinate on negative side of x axis + Distance of co-ordinate on positive side of x axis
Distance of co-ordinate on negative side of x axis = x = 3 units
Distance of co-ordinate on positive side of x axis = x = 2 units
Therefore, the correct answer is (d).
Page No 13.33:
Question 9:
The distance between the graphs of the equations y = −1 and y = 3 is
(a) 2
(b) 4
(c) 3
(d) 1
Answer:
Distance between the graph of equations, say D
D = Distance of co-ordinate on negative side of y axis + Distance of co-ordinate on positive side of y axis
Distance of co-ordinate on negative side of y axis = y = 1 units
Distance of co-ordinate on positive side of y axis = y = 3 units
Therefore, the correct answer is (b).
Page No 13.33:
Question 10:
If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length
(a) 4 units
(b) 3 units
(c) 5 units
(d) none of these
Answer:
(i) We are given,
We get,
Now, substituting in, we get
Substituting in,we get
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
x |
0 |
3 |
y |
4 |
0 |
We are given that the graph of equation cuts the co-ordinate axes and and forms the right angle triangle AOB.
Length of AB in right angled triangle AOB
The length of hypotenuse AB of triangle AOB is 5 units.
Therefore, the correct answer is (c).
Page No 13.6:
Question 1:
Write two solutions for each of the following equations:
(i) 3x + 4y = 7
(ii) x = 6y
(iii) x + xy = 4
(iv) $\frac{2}{3}x-y=4$
Answer:
(i) We are given,
Substituting x = 1 in the given equation, we get
Substituting x = 2 in the given equation, we get
(ii) We are given,
Substituting in the given equation, we get
Substituting in the given equation, we get
(iii) We are given,
Substituting x = 0 in the given equation, we get
Substituting x = 4 in the given equation, we get
(iv) We are given,
Substituting x = 0 in the given equation, we get
Substituting x = 3 in the given equation, we get
Page No 13.6:
Question 2:
Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations:
(i) 5x − 2y = 10
(ii) −4x + 3y = 12
(iii) 2x + 3y =24
Answer:
(i) We are given,
Substituting x = 0 in the given equation, we get
Substituting y = 0 in the given equation, we get
(ii) We are given,
Substituting in the given equation, we get
$-4\times 0+3y=12\phantom{\rule{0ex}{0ex}}3y=12\phantom{\rule{0ex}{0ex}}y=4$
Thus x = 0 and y = 4 is the solution of the −4x + 3y = 12
Substituting y = 0 in the given equation, we get
(iii) We are given,
Substituting x = 0 in the given equation, we get
Substituting y = 0 in the given equation, we get
Page No 13.6:
Question 3:
Check which of the following are solutions of the equations 2x − y = 6 and which are not:
(i) (3,0)
(ii) (0,6)
(iii) (2, −2)
(iv) $\left(\sqrt{3},0\right)$
(v) $\left(\frac{1}{2},-5\right)$
Answer:
We are given,
2x – y = 6
(i) In the equation 2x – y = 6,we have
Substituting x = 3 and y = 0 in 2x – y, we get
is the solution of 2x – y = 6.
(ii) In the equation 2x – y = 6, we have
Substituting x = 0 and y = 6 in 2x – y,we get
is not the solution of 2x – y = 6.
(iii) In the equation 2x – y = 6,we have
Substituting x = 2 and y = –2 in 2x – y, we get
is the solution of 2x – y = 6.
(iv) In the equation 2x – y = 6, we have
Substituting and y = 0 in 2x – y, we get
is not the solution of 2x – y = 6.
(v) In the equation 2x – y = 6, we have
Substituting and y = –5 in 2x – y, we get
is the solution of 2x – y = 6.
Page No 13.6:
Question 4:
If x = −1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.
Answer:
We are given,
is the solution of equation .
Substituting and in ,we get
Page No 13.6:
Question 5:
Find the value of λ, if x = −λ and y = $\frac{5}{2}$ is a solution of the equation x + 4y − 7 = 0.
Answer:
We are given,
is the solution of equation .
Substituting and in ,we get
Page No 13.7:
Question 6:
If x = 2α + 1 and y = α − 1 is a solution of the equation 2x − 3y + 5 = 0, find the value of α.
Answer:
We are given,
is the solution of equation .
Substituting and in ,we get
$2\times \left(2a+1\right)-3\times \left(a-1\right)+5=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 4a+2-3a+3+5=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow a+10=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow a=-10\left(\mathrm{answer}\right)$
Page No 13.7:
Question 7:
If x = 1 and y = 6 is a solution of the equation 8x − ay + a^{2} = 0, find the value of a.
Answer:
We are given,
is the solution of equation .
Substituting and in ,we get
Using quadratic factorization
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