Mix Examples - Linear Equations in Two Variables Questions in English

(A) Given the linear equation $3x - y = x - 1$.
Rearranging the terms to bring variables to one side: $3x - x - y = -1$.
This simplifies to $2x - y = -1$,or $y = 2x + 1$.
This is a linear equation in two variables ($x$ and $y$).
$A$ linear equation in two variables of the form $ax + by + c = 0$ always has infinitely many solutions because for every value of $x$,there exists a corresponding value of $y$ that satisfies the equation.
Therefore,the equation has infinitely many solutions.

(A) linear equation in two variables is defined as an equation that can be written in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
This means that at least one of the coefficients of the variables $x$ or $y$ must be non-zero.
If $a = 0$ and $b = 0$,the equation becomes $c = 0$,which is not a linear equation in two variables.
Therefore,the condition for a linear equation in two variables is $a^2 + b^2 \neq 0$,which implies that $a$ and $b$ cannot both be zero simultaneously.

(C) In a Cartesian coordinate system,the position of any point is represented by an ordered pair $(x, y)$.
For any point lying on the $y$-axis,the horizontal distance from the origin (the $x$-coordinate) is always $0$.
Therefore,any point on the $y$-axis is of the form $(0, y)$,where $y$ represents the vertical distance from the origin.

(D) The given equation $2x - 5y = 7$ is a linear equation in two variables ($x$ and $y$).
For any linear equation in two variables of the form $ax + by = c$,there exist infinitely many pairs of $(x, y)$ that satisfy the equation.
By assigning any arbitrary real value to $x$,we can find a corresponding value for $y$,and vice versa.
Therefore,the equation has infinitely many solutions.

(A) The linear equation $2x + 5y = 7$ represents a line in the Cartesian plane,which has infinitely many solutions in the set of real numbers.
However,if we restrict $x$ and $y$ to natural numbers (i.e.,$x, y \in \{1, 2, 3, ...\}$),we can test for solutions:
If $x = 1$,then $2(1) + 5y = 7 \implies 5y = 5 \implies y = 1$.
If $x = 2$,then $2(2) + 5y = 7 \implies 5y = 3 \implies y = 3/5$ (not a natural number).
If $x \ge 2$,$y$ will not be a natural number.
Thus,$(1, 1)$ is the only solution in natural numbers.

(B) For the linear equation $2x + 5y = 7$,we look for solutions where $x$ and $y$ are natural numbers (i.e.,$x, y \in \{1, 2, 3, ...\}$).
If $x = 1$,then $2(1) + 5y = 7$,which gives $5y = 5$,so $y = 1$.
If $x = 2$,then $2(2) + 5y = 7$,which gives $4 + 5y = 7$,so $5y = 3$,or $y = 3/5$,which is not a natural number.
If $x \ge 2$,$5y$ would be $7 - 2x \le 3$,meaning $y$ would not be a natural number.
Thus,the only pair of natural numbers that satisfies the equation is $(1, 1)$,making it a unique solution.

(C) To find the value of $k,$ we substitute the given solution $(x, y) = (2, 0)$ into the linear equation $2x + 3y = k.$
Substituting $x = 2$ and $y = 0$ in the equation:
$2(2) + 3(0) = k$
$4 + 0 = k$
$k = 4$
Therefore,the value of $k$ is $4.$

(D) The given linear equation is $2x + 0y + 9 = 0$.
This simplifies to $2x = -9$.
Dividing by $2$,we get $x = -9/2$.
Since the coefficient of $y$ is $0$,the value of $y$ can be any real number $m$.
Therefore,any solution of the equation is of the form $(-9/2, m)$,where $m$ is any real number.

(A) The graph of the linear equation $2x + 3y = 6$ cuts the $y$-axis at the point where the $x$-coordinate is $0$.
By substituting $x = 0$ into the equation $2x + 3y = 6$,we get:
$2(0) + 3y = 6$
$0 + 3y = 6$
$3y = 6$
$y = 6 / 3 = 2$
Therefore,the point where the graph cuts the $y$-axis is $(0, 2)$.

(B) linear equation in two variables is of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation $x = 7$,we can rewrite it as $x - 7 = 0$.
To express this in two variables $x$ and $y$,we include the variable $y$ with a coefficient of $0$.
Thus,the equation becomes $1 \cdot x + 0 \cdot y = 7$.

(C) In the Cartesian coordinate system,the $x$-axis is the horizontal line where the vertical distance from the origin is zero.
Therefore,for any point lying on the $x$-axis,the $y$-coordinate (ordinate) must be $0$.
Thus,any point on the $x$-axis is represented in the form $(x, 0)$,where $x$ is any real number.

(D) For any point on the line $y=x$,the $x$-coordinate and the $y$-coordinate must be equal.
If we let the $x$-coordinate be $a$,then the $y$-coordinate must also be $a$.
Therefore,any point on the line $y=x$ is of the form $(a, a)$.

(A) In the Cartesian coordinate system,any point on the $x$-axis has a $y$-coordinate equal to $0$. Therefore,the equation representing all points on the $x$-axis is $y = 0$.

(B) The given equation $y=6$ does not contain the variable $x$.
This implies that for any value of $x$,the value of $y$ remains constant at $6$.
Therefore,the graph of $y=6$ is a horizontal line that is parallel to the $x$-axis.
Since the $y$-coordinate is always $6$,the line is at a constant distance of $6$ units from the origin along the $y$-axis.

(C) To check if $(x=5, y=2)$ is a solution,substitute the values into each equation:
For option $A$: $x + 2y = 5 + 2(2) = 5 + 4 = 9 \neq 7$.
For option $B$: $5x + 2y = 5(5) + 2(2) = 25 + 4 = 29 \neq 7$.
For option $C$: $x + y = 5 + 2 = 7$. This matches the equation.
For option $D$: $5x + y = 5(5) + 2 = 25 + 2 = 27 \neq 7$.
Therefore,the correct equation is $x + y = 7$.

(D) The graph of the linear equation $2x + 3y = 6$ is a line that meets the $x$-axis at the point where $y = 0$.
Substituting $y = 0$ into the equation $2x + 3y = 6$,we get:
$2x + 3(0) = 6$
$2x = 6$
$x = 6 / 2 = 3$
Therefore,the line meets the $x$-axis at the point $(3, 0)$.

(A) The given linear equation is $y = x$.
For a point $(x, y)$ to lie on the graph of this equation,the $x$-coordinate must be equal to the $y$-coordinate.
Checking the options:
$A$: $(1, 1)$ where $x = 1$ and $y = 1$. Since $1 = 1$,the point $(1, 1)$ satisfies the equation $y = x$.
$B$: $(0, 3/2)$ where $x = 0$ and $y = 1.5$. Since $0 \neq 1.5$,it does not lie on the line.
$C$: $(3/2, -3/2)$ where $x = 1.5$ and $y = -1.5$. Since $1.5 \neq -1.5$,it does not lie on the line.
$D$: $(-1/2, 1/2)$ where $x = -0.5$ and $y = 0.5$. Since $-0.5 \neq 0.5$,it does not lie on the line.
Therefore,the graph passes through the point $(1, 1)$.

(B) If we multiply or divide both sides of a linear equation with a non-zero number,the equality holds true. Since the relationship between the variables remains balanced,the solution of the linear equation remains the same.

(C) There are infinitely many linear equations that can be satisfied by the values $x = 1$ and $y = 2$.
$A$ linear equation in two variables is of the form $ax + by + c = 0$. Since we have a specific point $(1, 2)$,we can construct infinitely many lines passing through this point by varying the coefficients $a, b,$ and $c$.
For example:
$1$. $x + y = 3$ (since $1 + 2 = 3$)
$2$. $y = 2x$ (since $2 = 2(1)$)
$3$. $y - x = 1$ (since $2 - 1 = 1$)
$4$. $2y - x = 3$ (since $2(2) - 1 = 3$)
Since we can choose infinitely many combinations of $a, b,$ and $c$ such that $a(1) + b(2) + c = 0$,there are infinitely many such linear equations.

(D) The points of the form $(a, a)$ have the same $x$ and $y$ coordinates.
Since the $x$-coordinate is equal to the $y$-coordinate,the point satisfies the equation $y=x$.
Therefore,the point of the form $(a, a)$ always lies on the line $y=x$.

(A) point of the form $(a, -a)$ has an $x$-coordinate equal to $a$ and a $y$-coordinate equal to $-a$.
To find the line on which this point lies,we can observe the relationship between the coordinates:
$x = a$
$y = -a$
Adding these two equations gives:
$x + y = a + (-a)$
$x + y = 0$
Therefore,the point $(a, -a)$ always satisfies the equation $x + y = 0$.

(A) $(i)$ False,because $ax + by + c = 0$ is a linear equation in two variables only if both $a$ and $b$ are non-zero (i.e.,$a^2 + b^2 \neq 0$).
$(ii)$ False,because a linear equation in two variables has infinitely many solutions.
$(iii)$ False,the points $(2, 0)$ and $(-3, 0)$ lie on the $x$-axis because their $y$-coordinates are $0$. The point $(4, 2)$ lies in the first quadrant,and the point $(0, 5)$ lies on the $y$-axis.

(A) $(i)$ True. $A$ line parallel to the $y$-axis at a distance $a$ units to the left of the $y$-axis is represented by the equation $x = -a$. Since the distance is $4$ units,the equation is $x = -4$.
$(ii)$ False. For a graph to pass through the origin $(0, 0)$,the coordinates must satisfy the equation. Substituting $x = 0$ and $y = 0$ into $y = mx + c$ gives $0 = m(0) + c$,which implies $c = 0$. Since $c$ is not necessarily $0$ for all lines of the form $y = mx + c$,the statement is false.

(A) The statement is False.
To verify if the points $(x, y)$ satisfy the equation $2x + 2 = y$,we substitute the $x$-values from the table into the equation and check if the resulting $y$-value matches the table.
For $x = 0$: $2(0) + 2 = 2$. (Matches $y = 2$)
For $x = 1$: $2(1) + 2 = 4$. (Matches $y = 4$)
For $x = 2$: $2(2) + 2 = 6$. (Matches $y = 6$)
For $x = 3$: $2(3) + 2 = 8$. (Matches $y = 8$)
For $x = 4$: $2(4) + 2 = 10$. (Matches $y = 10$)
Wait,checking the equation $2x + 2 = y$ against the table values:
If $x=0, y=2(0)+2=2$.
If $x=1, y=2(1)+2=4$.
If $x=2, y=2(2)+2=6$.
If $x=3, y=2(3)+2=8$.
If $x=4, y=2(4)+2=10$.
All points satisfy the equation $2x + 2 = y$. Therefore,the statement is True.

(TRUE) To check if the point $(0, 3)$ lies on the graph of the linear equation $3x + 4y = 12$,we substitute $x = 0$ and $y = 3$ into the equation.
Substituting the values:
$3(0) + 4(3) = 0 + 12 = 12$.
Since the left-hand side equals the right-hand side $(12 = 12)$,the point $(0, 3)$ satisfies the equation.
Therefore,the given statement is True.

(FALSE) To check if the graph of the linear equation $x + 2y = 7$ passes through the point $(0, 7)$,we substitute $x = 0$ and $y = 7$ into the equation.
Substituting the values:
$0 + 2(7) = 7$
$0 + 14 = 7$
$14 = 7$
Since $14 \neq 7$,the point $(0, 7)$ does not satisfy the equation.
Therefore,the statement is False.

(A) The given equation is $x + y = 0$,which can be rewritten as $y = -x$.
Any point on the graph of $y = -x$ must have $x$ and $y$ coordinates that are additive inverses of each other (i.e.,they have opposite signs and the same absolute value).
Looking at the graph,we can identify two points: $(-1, 1)$ and $(-3, 3)$.
For the point $(-1, 1)$,$x = -1$ and $y = 1$. Substituting these into the equation $x + y = 0$,we get $-1 + 1 = 0$,which is true.
For the point $(-3, 3)$,$x = -3$ and $y = 3$. Substituting these into the equation $x + y = 0$,we get $-3 + 3 = 0$,which is true.
Since both points satisfy the equation and two points determine a unique line,the given statement is True.

(TRUE) We know that the graph of the equation $x = a$ is a line parallel to the $y$-axis and at a distance of $|a|$ units from it.
If $a > 0$,the line lies to the right of the $y$-axis.
In the given graph,the line is parallel to the $y$-axis and passes through the point $(3, 0)$ on the $x$-axis,which is at a distance of $3$ units to the right of the $y$-axis.
Therefore,the given statement is True.

(B) To determine if the points are solutions,we substitute the values of $x$ and $y$ into the equation $x-y+2=0$.
For $(0, 2)$: $0 - 2 + 2 = 0$. (True)
For $(1, 3)$: $1 - 3 + 2 = 0$. (True)
For $(2, 4)$: $2 - 4 + 2 = 0$. (True)
For $(3, -5)$: $3 - (-5) + 2 = 3 + 5 + 2 = 10 \neq 0$. (False)
For $(4, 6)$: $4 - 6 + 2 = 0$. (True)
Since the point $(3, -5)$ does not satisfy the equation,the statement is False.

(B) The given statement is $False$.
By definition,the graph of a linear equation in two variables is the collection of all points $(x, y)$ that satisfy the equation.
Therefore,every point lying on the line represented by the linear equation is a solution to that equation.

(FALSE) The statement is False.
By definition,a linear equation in two variables is of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero. The geometric representation of such an equation in a Cartesian plane is always a straight line. Therefore,it is incorrect to say that the graph need not be a line.

(N/A) The graph of the linear equation $3x + 4y = 12$ cuts the $x$-axis at the point where $y = 0$.
Putting $y = 0$ in the equation,we get $3x + 4(0) = 12$,which simplifies to $3x = 12$,so $x = 4$.
Thus,the point on the $x$-axis is $(4, 0)$.
The graph of the linear equation $3x + 4y = 12$ cuts the $y$-axis at the point where $x = 0$.
Putting $x = 0$ in the equation,we get $3(0) + 4y = 12$,which simplifies to $4y = 12$,so $y = 3$.
Thus,the point on the $y$-axis is $(0, 3)$.

(A) The line parallel to the $y$-axis at a distance of $2$ units from the origin in the positive direction of the $x$-axis is given by the equation $x=2$.
To find the point of intersection,we substitute $x=2$ into the given linear equation $x+y=5$.
$2+y=5$
$y=5-2$
$y=3$
Therefore,the point of intersection is $(2, 3)$.

(C) Let the point be $(x, y)$.
According to the problem,the $x$-coordinate is $\frac{5}{2}$ times its ordinate $(y)$,so $x = \frac{5}{2}y$.
Substitute $x = \frac{5}{2}y$ into the equation $2x + 5y = 20$:
$2(\frac{5}{2}y) + 5y = 20$
$5y + 5y = 20$
$10y = 20$
$y = 2$
Now,find the $x$-coordinate using $x = \frac{5}{2}y$:
$x = \frac{5}{2}(2) = 5$
Thus,the required point is $(5, 2)$.

(N/A) Any straight line parallel to the $x$-axis is given by the equation $y=k$,where $k$ is the distance of the line from the $x$-axis.
Here,the line is $4$ units above the $x$-axis,so $k=4$.
Therefore,the equation of the line is $y=4$.
To draw the graph of this equation,plot the points $(1,4)$,$(2,4)$,and $(3,4)$ on the Cartesian plane and join them with a straight line. This line represents all points where the $y$-coordinate is $4$.

(N/A) Any point on the graph of $y = x$ will have $x$ and $y$ coordinates that are identical. The line passes through the points $(0,0), (1,1),$ and $(-1,-1).$
Similarly,any point on the graph of $y = -x$ will have $x$ and $y$ coordinates of opposite signs. The line passes through the points $(1,-1)$ and $(-1,1).$
Also,the point $(0,0)$ satisfies the equation $y = -x.$
The graphs of the linear equations $y = x$ and $y = -x$ on the same Cartesian plane are shown in the figure below.
We observe that the graphs of both these equations intersect at the origin $(0,0).$

(A) Let the abscissa of the point be $x$ and the ordinate be $y$.
According to the problem,the ordinate is $1 \frac{1}{2}$ times the abscissa,so $y = 1 \frac{1}{2} x = \frac{3}{2} x$.
Substitute $y = \frac{3}{2} x$ into the given linear equation $2x + 5y = 19$:
$2x + 5(\frac{3}{2} x) = 19$
$2x + \frac{15}{2} x = 19$
Multiply the entire equation by $2$ to clear the fraction:
$4x + 15x = 38$
$19x = 38$
$x = 2$
Now,find the ordinate $y$ using $y = \frac{3}{2} x$:
$y = \frac{3}{2} (2) = 3$
Thus,the point is $(2, 3)$.

(N/A) The equation of a line parallel to the $x$-axis at a distance $a$ units below it is given by $y = -a$.
Here,the distance is $3$ units below the $x$-axis,so $a = 3$.
Therefore,the equation of the line is $y = -3$.
This line is parallel to the $x$-axis and passes through every point where the $y$-coordinate is $-3$,such as $(0, -3)$,$(1, -3)$,$(-1, -3)$,etc.
The graph of the equation $y = -3$ is a horizontal line passing through the point $(0, -3)$ on the $y$-axis,as shown in the figure.

(N/A) linear equation whose solutions are represented by the points having the sum of coordinates as $10$ units is $x+y=10$.
When $x=0$,$y=10$ and when $x=10$,$y=0$.
Now,plot these two points $(0, 10)$ and $(10, 0)$ on a graph paper and join them to obtain a straight line.
The graph of $x+y=10$ is a straight line as shown in the figure.

(C) In a Cartesian coordinate system,any point is represented as $(x, y)$,where $x$ is the abscissa and $y$ is the ordinate.
According to the problem,the ordinate is $3$ times the abscissa.
Therefore,the relationship is $y = 3x$.
This is the required linear equation.

(A) The point $(3,4)$ lies on the graph of $3y = ax + 7$.
Substituting $x = 3$ and $y = 4$ in the given equation $3y = ax + 7$,we get:
$3 \times 4 = a \times 3 + 7$
$12 = 3a + 7$
$3a = 12 - 7$
$3a = 5$
$a = \frac{5}{3}$

(N/A) $(i)$ The number of solution$(s)$ of the equation $2x + 1 = x - 3$ on the number line is $1$.
Solving the equation: $2x + 1 = x - 3 \Rightarrow 2x - x = -3 - 1 \Rightarrow x = -4$.
Thus,there is exactly one unique solution,$x = -4$,on the number line.
$(ii)$ The number of solution$(s)$ of the equation $2x + 1 = x - 3$ on the Cartesian plane is infinitely many.
In the Cartesian plane,the equation $2x + 1 = x - 3$ can be written as $x + 4 = 0$,which represents a vertical line parallel to the $y$-axis. Every point on this line $( -4, y )$ for any real value of $y$ satisfies the equation,hence there are infinitely many solutions.

(N/A) We know that any point lying on the $x$-axis has its ordinate (y-coordinate) equal to $0$.
Putting $y = 0$ in the equation $x + 2y = 8$,we get:
$x + 2(0) = 8 \Rightarrow x = 8$.
Thus,the point on the $x$-axis is $(8, 0)$.
We also know that any point lying on the $y$-axis has its abscissa (x-coordinate) equal to $0$.
Putting $x = 0$ in the equation $x + 2y = 8$,we get:
$0 + 2y = 8 \Rightarrow 2y = 8 \Rightarrow y = 4$.
Thus,the point on the $y$-axis is $(0, 4)$.

(B) Given the linear equation: $2x + cy = 8$.
We are looking for the value of $c$ such that $x = y$.
Substitute $y = x$ into the given equation:
$2x + c(x) = 8$
$x(2 + c) = 8$
$x = \frac{8}{2 + c}$
Since the question implies a specific value for $c$ that satisfies the condition for any solution where $x=y$,we look at the structure of the equation. If we assume a specific point like $(2, 2)$ which satisfies $x=y$:
$2(2) + c(2) = 8$
$4 + 2c = 8$
$2c = 4$
$c = 2$
Thus,for $c = 2$,the equation becomes $2x + 2y = 8$,or $x + y = 4$,which clearly allows for the solution $x = y = 2$.

(A) $y$ varies directly as $x$.
$\Rightarrow y \propto x$
$\therefore y = kx$
Substituting $y = 12$ when $x = 4$,we get:
$12 = k \times 4 \Rightarrow k = 12 \div 4 = 3$
Hence,the required linear equation is $y = 3x$.
The value of $y$ when $x = 5$ is $y = 3 \times 5 = 15$.

(N/A) The given equation is $2x + 3y = 12$. To draw the graph of this equation,we need at least two points lying on the graph.
From the equation,we have $y = \frac{12 - 2x}{3}$.
For $x = 0$,$y = \frac{12 - 0}{3} = 4$. Therefore,$(0, 4)$ lies on the graph.
For $y = 0$,$2x = 12$,so $x = 6$. Therefore,$(6, 0)$ lies on the graph.
Now,plot the points $A(0, 4)$ and $B(6, 0)$ and join them to get the line $AB$.
Line $AB$ is the required graph. The graph (line $AB$) cuts the $x$-axis at the point $(6, 0)$ and the $y$-axis at the point $(0, 4)$.

(N/A) From the table,we get two points $A(1, 1)$ and $B(2, 3)$ which lie on the graph of the linear equation. Obviously,the graph will be a straight line. So,we first plot the points $A$ and $B$ and join them as shown in the figure.
To find the equation of the line passing through $(1, 1)$ and $(2, 3)$:
Slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{2 - 1} = 2$.
Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - 1 = 2(x - 1) \implies y - 1 = 2x - 2 \implies y = 2x - 1$.
$(i)$ To find where it cuts the $x$-axis,set $y = 0$:
$0 = 2x - 1 \implies 2x = 1 \implies x = \frac{1}{2}$.
So,it cuts the $x$-axis at the point $\left(\frac{1}{2}, 0\right)$.
$(ii)$ To find where it cuts the $y$-axis,set $x = 0$:
$y = 2(0) - 1 \implies y = -1$.
So,it cuts the $y$-axis at the point $(0, -1)$.

(N/A) Let the total distance covered be $x \text{ km}$ and the total fare charged be $Rs. y$.
For the first $1 \text{ km}$,the fare is $Rs. 10$.
For the remaining distance $(x - 1) \text{ km}$,the fare is $Rs. 4(x - 1)$.
Therefore,the total fare $y$ is given by:
$y = 10 + 4(x - 1)$
$y = 10 + 4x - 4$
$y = 4x + 6$
Thus,the required linear equation is $4x - y + 6 = 0$.
To draw the graph,we find two points on the line:
If $x = 0$,then $y = 4(0) + 6 = 6$. Point: $(0, 6)$.
If $x = -1$,then $y = 4(-1) + 6 = 2$. Point: $(-1, 2)$.
Plotting these points on a Cartesian plane and joining them gives the required straight line graph.

(N/A) Work done $= (\text{constant force}) \times (\text{distance})$
$= 3 \times (\text{distance})$
i.e., $y = 3x$, where $y$ (units) is the work done and $x$ (units) is the distance travelled.
Since $x = 2$ units (given), therefore, work done $= 3 \times 2 = 6$ units.
To plot the graph of the linear equation $y = 3x$, we need at least two solutions of the equation.
We see that $x = 0, y = 0$ satisfies the given equation, and $x = 1, y = 3$ also satisfies the equation.
Now we plot the points $A(0, 0)$ and $B(1, 3)$ and join them to form a line. The graph of the equation is a straight line. [We have not shown the whole line because work done cannot be negative].
To verify from the graph, draw a perpendicular to the $x$-axis at the point $(2, 0)$ meeting the graph at the point $C$. Clearly, the coordinates of $C$ are $(2, 6)$. It means that the work done is $6$ units.

(N/A) To show that a point $(x, y)$ lies on the graph of the linear equation $y = 9x - 7$,we must substitute the coordinates of the point into the equation and verify if the left-hand side equals the right-hand side.
For point $A(1, 2)$:
Substitute $x = 1$ and $y = 2$ into the equation:
$2 = 9(1) - 7$
$2 = 9 - 7$
$2 = 2$
Since the left-hand side equals the right-hand side,point $A$ lies on the graph.
For point $B(-1, -16)$:
Substitute $x = -1$ and $y = -16$ into the equation:
$-16 = 9(-1) - 7$
$-16 = -9 - 7$
$-16 = -16$
Since the left-hand side equals the right-hand side,point $B$ lies on the graph.
For point $C(0, -7)$:
Substitute $x = 0$ and $y = -7$ into the equation:
$-7 = 9(0) - 7$
$-7 = 0 - 7$
$-7 = -7$
Since the left-hand side equals the right-hand side,point $C$ lies on the graph.
Thus,all three points $A$,$B$,and $C$ satisfy the equation $y = 9x - 7$ and therefore lie on its graph.