Question 14 Marks
The graph of $3x + 2y = 6$ meets the $x-$axis at point $P$ and the $y-$axis at point $Q.$ Use the graphical method to find the co$-$ordinates of points $P$ and $Q$.
AnswerTo draw the graph of $3x + 2y = 6$ follows the steps:
First, prepare a table as below:
| $X$ |
$- 2$ |
$0$ |
$2$ |
| $Y$ |
$6$ |
$3$ |
$0$ |
Now sketch the graph as shown:
From the graph it can verify that the line intersects the $x-$axis at $(2,0)$ and y at $(0,3)$, therefore the coordinates of $P(x-$axis$)$ and $Q(y-$axis$)$ are $(2,0)$ and $(0,3)$ respectively. View full question & answer→Question 24 Marks
On the same graph paper, plot the graphs of $y = 2x - 1, y = 2x$ and $y = 2x + 1$ from $x = - 2$ to $x = 4$. Are the graphs $($lines$)$ drawn parallel to each other?
AnswerFirst, prepare a table as follows:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y = 2x - 1$ |
$-3$ |
$-1$ |
$1$ |
| $Y = 2x$ |
$-2$ |
$0$ |
$2$ |
| $Y = 2x + 1$ |
$-1$ |
$1$ |
$3$ |
Now the graph can be drawn as follows:
The lines are parallel to each other. View full question & answer→Question 34 Marks
On the same graph paper, plot the graph of $y = x - 2, y = 2x + 1$ and $y = 4$ from $x= - 4$ to $3.$
AnswerFirst, prepare a table as follows:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y = x - 2$ |
$-3$ |
$-2$ |
$-1$ |
| $Y = 2x + 1$ |
$-1$ |
$1$ |
$3$ |
| $Y = 4$ |
$4$ |
$4$ |
$4$ |
Now the graph can be drawn as follows:
View full question & answer→Question 44 Marks
For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.$3x + 4y = 24,\frac{x}{4}+\frac{y}{3}=1$
AnswerTo draw the graph of $3x + 4y = 24$ and $\frac{x}{4}+\frac{y}{3}=1$ follows the steps:
First prepare a table as below
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y=-\frac{3}{4} \times+6$ |
$\frac{27}{4}$ |
$6$ |
$\frac{21}{4}$ |
| $Y=-\frac{3}{4} \times+3$ |
$\frac{15}{4}$ |
$3$ |
$\frac{9}{4}$ |
Now sketch the graph as shown:
From the graph it can verify that the lines are parallel. View full question & answer→Question 54 Marks
For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other$.y = x - 3,y = - x + 5$
AnswerTo draw the graph of $y = x - 3$ and $y = - x + 5$ follows the steps:
First, prepare a table as below:
| $X$ |
$- 1$ |
$0$ |
$1$ |
| $Y = x - 3$ |
$- 4$ |
$-3$ |
$- 2$ |
| $Y = - x + 5$ |
$6$ |
$5$ |
$4$ |
Now sketch the graph as shown:
From the graph it can verify that the lines are perpendicular. View full question & answer→Question 64 Marks
For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other$.y = 3x - 1,y = 3x + 2$
AnswerTo draw the graph of $y = 3x - 1$ and $y = 3x + 2$ follows the steps:
First, prepare a table as below:
| $X$ |
$- 1$ |
$0$ |
$1$ |
| $Y = 3x -1$ |
$- 4$ |
$- 1$ |
$2$ |
| $Y = 3x + 2$ |
$- 1$ |
$2$ |
$5$ |
Now sketch the graph as shown
:
From the graph it can verify that the lines are parallel. View full question & answer→Question 74 Marks
Draw the graph for the equation given below:$\frac{2 x-1}{3}-\frac{y-2}{5}=0$
AnswerTo draw the graph of $\frac{2 x-1}{3}-\frac{y-2}{5}=0$ follows the steps:
First prepare a table as below:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y$ |
$-3$ |
$\frac{1}{3}$ |
$\frac{11}{3}$ |
Now sketch the graph as shown:
From the graph it can verify that the line intersects the $x-$axis at $\left(-\frac{1}{10}, 0\right)$ and $y$ at $(0,4.5).$ View full question & answer→Question 84 Marks
Draw the graph for the equation given below:$\frac{1}{2} x+\frac{2}{3} y=5$
AnswerTo draw the graph of $\frac{x}{2}+\frac{2 y}{3}=5$ follows the steps:
First, prepare a table as below:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y$ |
$5.25$ |
$4.5$ |
$3.75$ |
Now sketch the graph as shown:
From the graph it can verify that the line intersect the $x-$axis at $(10,0)$ and $y$ at $(0,7.5).$ View full question & answer→Question 94 Marks
Draw the graph for the linear equation given below:$x-3=\frac{2}{5}(y+1)$
AnswerThe equation will become:
$5x - 2y = 17$
First prepare a table as follows:
| $x$ |
$- 1$ |
$0$ |
$1$ |
| $y$ |
$- 11$ |
$-\frac{17}{2}$ |
$- 6$ |
Thus the graph can be drawn as follows:
View full question & answer→Question 104 Marks
Draw the graph for the linear equation given below:$\frac{x-1}{3}-\frac{y+2}{2}=0$
AnswerThe equation will become:$2x - 3y = 8$
First prepare a table as follows:
| $x$ |
$-1$ |
$0$ |
$1$ |
| $y$ |
$-\frac{10}{3}$ |
$-\frac{8}{3}$ |
$-2$ |
Thus the graph can be drawn as follows:
View full question & answer→Question 114 Marks
Draw the graph ofequation $\frac{x}{4}+\frac{y}{5}=1$ Use the graph drawn to find:$(i) x_1$, the value of $x$, when $y = 10(ii) y_1$, the value of $y$, when $x = 8$.
AnswerFirst, prepare a table as follows:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y$ |
$\frac{25}{4}$ |
$5$ |
$\frac{15}{4}$ |
The graph of the equation can be drawn as follows:
From the graph, it can be verified that:
for $y = 10$, the value of $x = - 4.$
for $x = 8$ the value of $y = - 5.$ View full question & answer→Question 124 Marks
Draw the graph of the equation $3x - 4y = 12$.Use the graph drawn to find$:(i) y_1$, the value of y, when $x = 4.(ii) y_2$, the value of $y$, when $x = 0$.
AnswerFirst, prepare a table as follows:
| $X$ |
$- 1$ |
$0$ |
$1$ |
| $Y$ |
$-\frac{15}{4}$ |
$- 3$ |
$-\frac{9}{4}$ |
The graph of the equation can be drawn as follows:
From the graph, it can verify that
If $x = 4 $the value of $y = 0$
If $x = 0$ the value of y = - 3. View full question & answer→Question 134 Marks
Draw the graph of equation $x + 2y - 3 = 0$. From the graph, find:$(i) x_1$, the value of $x$, when $y = 3(ii) x_2$, the value of $x$, when $y = - 2$.
AnswerFirst, prepare a table as follows:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y$ |
$2$ |
$\frac{3}{2}$ |
$1$ |
Thus the graph can be drawn as shown:
$(i)$ For $y = 3$ we have $x = - 3$
$(ii)$ For $y = - 2$ we have $x = 7$ View full question & answer→Question 144 Marks
Draw the graph for the linear equation given below:$2x - 7 = 0$
AnswerThe equation can be written as:
$x = \frac{7}{2}$
First prepare a table as follows:
| $x$ |
$\frac{7}{2}$ |
$\frac{7}{2}$ |
$\frac{7}{2}$ |
| $y$ |
$-1$ |
$0$ |
$1$ |
Thus the graph can be drawn as follows:
View full question & answer→Question 154 Marks
Draw the graph for the linear equation given below$:x = 3$
AnswerSince $x = 3$, therefore the value of $y$ can be taken as any real no.
First prepare a table as follows:
| $x$ |
$3$ |
$3$ |
$3$ |
| $y$ |
$-1$ |
$0$ |
$1$ |
Thus the graph can be drawn as follows:
View full question & answer→Question 164 Marks
In the following, find the co$-$ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation:$5 x-(5-x)=\frac{1}{2}(3-x) ; 4-3 y=\frac{4+y}{3}$
Answer$5 x-(5-x)=\frac{1}{2}(3-x) ; 4-3 y=\frac{4+y}{3}$
Now
$5 x-(5-x)=\frac{1}{2}(3-x)$
$(5 x+x)-5=\frac{1}{2}(3-x)$
$12 x - 10 = 3 - x$
$12x + x = 3 +10$
$13x = 13$
$x = 1$
Again
$4-3 y=\frac{4+y}{3}$
$12-9 y=4+y$
$12-4=y+9 y$
$8=10 y$
$\frac{8}{10}=y$
$\frac{4}{5}=y$
$\therefore$ The co$-$ordinates of the point $\left(1, \frac{4}{5}\right)$
View full question & answer→Question 174 Marks
Find the values of $x$ and $y$ if:$(5x - 3y, y - 3x) = (4, -4)$
AnswerTwo ordered pairs are equal.
$\Rightarrow $ Therefore their first components are equal and their second components too are separately equal.
$(5x - 3y, y - 3x) = (4, -4)$
$5x - 3y = 4 .......(A)$ and
$ y - 3x = -4 ......(B)$
Now multiplying the equation $(B)$ by $3$, we get
$3y - 9x = -12......(C)$
Now adding both the equations $(A)$ and $(C),$ we get
$(5x - 3y) + (3y - 9x) = (4 + (-12))$
$-4x = -8$
$x = 2$
Putting the value of $x$ in the equation$(B)$, we get
$y - 3x = -4$
$\Rightarrow y = 3x -4$
$\Rightarrow y = 3(2) -4$
$\Rightarrow y = 2$
Therefore we get,
$x = 2, y = 2$
View full question & answer→Question 184 Marks
In rectangle $\text{OABC}$; point $O$ is the origin, $OA = 10$ units along $x-$axis and $AB = 8$ units. Find the co$-$ordinates of vertices $A, B$ and $C.$
AnswerGiven that in rectangle $\text{OABC};$
point $O$ is the origin and $OA = 10$ units along $x-$axis therefore we get $O(0,0)$ and $A(0,0)$.
Also it is given that $AB = 8$ units.
Therefore we get $B(10,8)$ and $C(0,8)$
After plotting the points $O(0,0), A(10,0), B(10,8)$ and $C(0,8)$ on a graph paper;
we get the above rectangle $\text{OABC}$ and the required co$-$ordinates of the vertices are $A(10,0), B(10,8)$ and $C(0,8)$ View full question & answer→Question 194 Marks
Draw the graph of the line $x + y = 5$. Use the graph paper drawn to find the inclination and the $y-$intercept of the line.
AnswerGiven line is
$x + y = 5$
The graph of the given line is shown below.
From the given line $x + y = 5,$ we get
$x + y = 5$
$y = - x + 5$
$y = (-1) . x + 5 \dots.......(A)$
Again we know that equation of any straight line in the form $y = mx + c,$
where $m$ is the gradient and $c$ is the intercept.
Again we have if slope of a line is $\tan \theta$ then inclination of the line is $\theta $
Now from the equation $(A),$ we have
$m = - 1$
$\tan \theta = - 1$
$\tan \theta = \tan 135^\circ $
$\theta = 135^\circ $
And $c = 5$
Therefore the required inclination is $\theta = 135^\circ $ and $y-$intercept is $c = 5$ View full question & answer→Question 204 Marks
For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.$2x - 3y = 6,\frac{x}{2}+\frac{y}{3}=1$
AnswerTo draw the graph of $2x - 3y = 6$ and $\frac{x}{2}+\frac{y}{3}=1$ follows the steps:
First prepare a table as below:
| $X$ |
$-1$ |
$0$ |
$1$ |
| $Y=\frac{2}{3} \times 2$ |
$-\frac{8}{3}$ |
$-2$ |
$-\frac{4}{3}$ |
| $Y=-\frac{3}{2} \times+3$ |
$\frac{9}{2}$ |
$3$ |
$\frac{3}{2}$ |
Now sketch the graph as shown:
From the graph it can verify that the lines are perpendicular. View full question & answer→Question 214 Marks
Use the graphical method to show that the straight lines given by the equations $x + y = 2, x - 2y = 5$ and $\frac{x}{3}+y=0$ pass through the same point.
AnswerThe equations can be written as follows:
$y = 2 - x$
$ y =\frac{1}{2}(x-5)$
$y =-\frac{x}{3}$
First prepare a table as follows:
| $X$ |
$Y = 2 - x$ |
$Y=\frac{1}{2}(x-5)$ |
$Y =-\frac{x}{3}$ |
| $- 1$ |
$3$ |
$- 3$ |
$\frac{1}{3}$ |
| $0$ |
$2$ |
$-\frac{5}{2}$ |
$0$ |
| $1$ |
$1$ |
$- 2$ |
$-\frac{1}{3}$ |
Thus the graph can be drawn as follows:
From the graph it is clear that the equation of lines are passes through the same point. View full question & answer→Question 224 Marks
$A (-2, 4),C(4, 10)$ and $D(-2, 10)$ are the vertices of a square $\text{ABCD}$. Use the graphical method to find the co$-$ordinates of the fourth vertex $B$. Also, find:$(i)$ The co$-$ordinates of the mid$-$point of $BC;(ii)$ The co$-$ordinates of the mipd$-$oint of $CD$ and $(iii)$ The co$-$ordinates of the point of intersection of the diagonals of the square $\text{ABCD}.$
AnswerGiven $A(-2, 4), C(4,10)$ and $D(-2,10)$ are the vertices of a square $ABCD$
After plotting the given points $A(- 2, 4), C(4,10)$ and $D(- 2,10)$ on a graph paper;
joining $D$ with $A$ and $D$ with $C$.
Now complete the square $\text{ABCD}$
As is clear from the graph $B(4, 4)$
Now from the graph we can find the midpoints of the sides $BC$ and $CD$ the co$-$4ordinates of the diagonals of the square.
Therefore the coordinates of the mid$-$point of $BC$ is $E(4, 7)$ and the coordinates of the mid$-$point of $CD$ is $F(1,10)$ and the coordinates of the diagonals of the square is $G(1, 7).$ View full question & answer→Question 234 Marks
$A (- 2, 2), B(8, 2)$ and $C(4, - 4)$ are the vertices of a parallelogram $\text{ABCD}$. By plotting the given points on a graph paper; find the co$-$ordinates of the fourth vertex $D$.Also, form the sazmegraph,state the co$-$ordinates of the mid$-$points of the sides $AB$ and $CD.$
AnswerPlot the points $A, B,$ and $C$ on the graph paper. Join the points to complete the parallelogram $\text{ABCD}$.
As the distance between points $A$ and $B$ is $10$ units, distance between the corners of the opposite side $C$ and $D$ will also be $10$ units.
Hence, the fourth point $D = (- 6, - 4)$
Mid point of $AB$ will lie at $5$ units distance from $A$ and $B$ both which is $= (3, 2)$
Mid point of $CD$ will lie at $5$ units distance from $C$ and $D$ both which is$ = (-1, -4)$ View full question & answer→Question 244 Marks
In the following, the coordinates of the three vertices of a rectangle $\text{ABCD}$ are given. By plotting the given points; find, in case, the coordinates of the fourth vertex:$B (10, 4),C(0, 4)$ and $D(0, -2).$
Answer$B (10, 4),C(0, 4)$ and $D(0, -2)$
After plotting the given points $B (10, 4), C(0, 4)$ and $D(0, - 2)$ on a graph paper;
joining $C$ with $B$ and $C$ with $D$
. From the graph it is clear that the vertical distance between the points $C(0, 4)$ and $D(0, - 2)$ is $6$ units and the horizontal distance between the points $C(0, 4)$ and $D(0, - 2)$ is $10$ units,
therefore the vertical distance between the points $B (10, 4)$ and $A$ must be $6$ units and the horizontal distance between the points $D(0, - 2)$ and $A$ must be $10$ units
Now complete the rectangle $\text{ABCD}$
As is clear from the graph $A(10, - 2).$ View full question & answer→Question 254 Marks
In the following, the coordinates of the three vertices of a rectangle $\text{ABCD}$ are given. By plotting the given points; find, in case, the coordinates of the fourth vertex:$A (- 4, - 6), C(6, 0)$ and $D(- 4, 0).$
Answer$A (- 4, - 6), C(6, 0)$ and $D(- 4, 0)$
After plotting the given points $A(- 4, - 6), C(6, 0)$ and $D(- 4, 0)$ on a graph paper;
joining $D$ with and 4 with $C.$
From the graph it is clear that the vertical distance between the points $D( - 4, 0)$ and $A(- 4, - 6)$ is $6$ units and the horizontal distance between the points $D(- 4, 0)$ and $C(6, 0)$ is $10$ units,
$\therefore$ the vertical distance between the points $C(6, 0)$ and $B$ must be 6 units and the horizontal distance between the points $A(- 4, - 6)$ and $B$ must be $10$ units.
Now complete the rectangle $\text{ABCD}$
As is clear from the graph $B(6, - 6)$ View full question & answer→Question 264 Marks
In the following, the coordinates of the three vertices of a rectangle $\text{ABCD}$ are given. By plotting the given points; find, in case, the coordinates of the fourth vertex:$A (4, 2),B(-2, 2)$ and $D(4, -2).$
Answer$A(4,2), B(-2,-2)$ and $D(4,-2)$
After plotting the given points $A(4,2), B(-2,2)$ and $D(4,-2)$ on a graph paper;
joining $A$ with $B$ and $A$ with $D.$
From the graph it is clear that the vertical distance between the points $A(4,2)$ and $D(4,-2)$ is $4$ units and the horizontal distance between the points $A(4,2)$ and $B(-2,2)$ is $6$ units,
$\therefore$ the vertical distance between the points $B(-2,2)$ and $C$ must be $4$ units and the horizontal distance between the points $B(-2,2)$ and $C$ must be 6 units.
Now complete the rectangle $\text{ABCD}$
As is clear from the graph $C(-2,2)$ View full question & answer→Question 274 Marks
In the following, the coordinates of the three vertices of a rectangle $\text{ABCD}$ are given. By plotting the given points; find, in case, the coordinates of the fourth vertex: $A(2, 0), B(8, 0)$ and $C(8, 4).$
Answer$A(2, 0), B(8, 0)$ and $C(8, 4).$
After plotting the given points $A(2,0), B(8,0)$ and $C(8,4)$ on a graph paper;
joining $A$ with Band Bwith $C.$
From the graph it is clear that the vertical distance between the points $B(8,0)$ and $C(8,4)$ is $4$ units,
$\therefore$ the vertical distance between the points $A(2,0)$ and $D$ must be $4$ units.
Now complete the rectangle $\text{ABCD}$
As is clear from the graph $D(2,4)$ View full question & answer→Question 284 Marks
Use the graph given alongside, to find the coordinates of the point $(s)$ satisfying the given condition:$(i)$ The abscissa is $2.(ii)$The ordinate is $0.(iii)$ The ordinate is $3.(iv)$ The ordinate is $-4.(v)$ The abscissa is $5.(vi)$ The abscissa is equal to the ordinate.$(vii)$ The ordinate is half of the abscissa.
Answer$(i)$ The abscissa is $2$
Now using the given graph the co$-$ordinate of the given point Ais given by $(2,2)$
$(ii)$ The ordinate is $0$
Now using the given graph the co$-$ordinate of the given point Bis given by $(5,0)$
$(iii)$ The ordinate is $3$
Now using the given graph the co$-$ordinate of the given point $C$ and $E$ is given by $(- 4, 3)\ \&\ (6, 3)$
$(iv)$ The ordinate is $-4$
Now using the given graph the co$-$ordinate of the given point $D$ is given by $( 2,- 4)$
$(v)$ The abscissa is $5$
Now using the given graph the co$-$ordinate of the given point $H, B$ and $G$ is given by $(5, 5),(5, 0)\ \&\ (5, -3)$
$(vi)$The abscissa is equal to the ordinate.
Now using the given graph the co-ordinate of the given point $I,$
$ A\ \&\ H$ is given by $(4, 4),(2, 2)\ \& \ (5, 5)$
$(vii)$The ordinate is half of the abscissa
Now using the given graph the co4ordinate of the given point $E$ is given by $(6, 3)$
View full question & answer→Question 294 Marks
Plot the following points on the same graph paper:$(i) (8, 7);(ii) (3, 6);(iii) (0, 4);(iv) (0, -4);(v) (3, -2);(vi) (-2, 5);(vii) (-3, 0)(viii) (5, 0);(ix) (-4, -3)$
AnswerLet us take the point as
$A(8,7), B(3,6), C(0,4), D(0,4), E(3,-2), F(-2,5), G(-3,0), H(5,0), I(-4,-3)$
On the graph paper, let us draw the co$-$ordinate axes $\text{XOX}\ '$ and $4$ intersecting at the origin $O$.
With proper scale, mark the numbers on the two co$-$ordinate axes.
Now for the point $A(8,7)$
Step $1$
Starting from origin $O,$ move $8$ units along the positive direction of $X-$axis, to the right of the origin $O$
Step $2$
Now from there, move $7$ units up and place a dot at the point reached.
Label this point as $A(8,7)$
Similarly plotting the other points
$B(3,6), C(0,4), D(0,-4), E(3,-2), F(-2,5), G(-3,0), H(5,0), I(-4,-3)$
View full question & answer→Question 304 Marks
In square $\text{ABCD}; A = (3, 4), B = (-2, 4)$ and $C = (-2, -1).$ By plotting these points on a graph paper, find the co$-$ordinates of vertex $D.$ Also, find the area of the square.
AnswerGiven that in square $\text{ABCD} ; A(3,4), B(-2,4)$ and $C(-2,-1)$
After plotting the given points $A(3,4), B(-2,4)$ and $C(-2,-1)$ on a graph paper;
joining $B$ with $C$ and $B$ with $A.$
From the graph it is clear that the vertical distance between the points $B(-2,4)$ and $C(-2,-1)$ is $5$ units and the horizontal distance between the points $B(-2,4)$ and $A(3,4)$ is $5$ units,
$\therefore$ the vertical distance between the points $A(3,4)$ and
$D$ must be $5$ units and the horizontal distance between the points $C(-2,-1)$ and $D$ must be $5$ units.
Now complete the square $\text{ABCD}$
As is clear from the graph $D(3,-1)$
Now the area of the square $\text{ABCD}$ is given by area of $\text{ABCD} = ($side$)^2 = (5)^2$
$A= 25$ units View full question & answer→Question 314 Marks
Plot point$ A(5, -7)$. From point $A$, draw $AM$ perpendicular to the $x-$axis and $AN$ perpendicular to the $y-$axis. Write the coordinates of points $M$ and $N.$
AnswerGiven $A(5, -7)$
After plotting the given point $A(5,-7)$ on a graph paper.
Now let us draw a perpendicular $AM$ from the point $A(5,-7)$ on the $x-$axis and a perpendicular $AN$ from the point $A(5,-7)$ on the $y-$axis.
As from the graph clearly we get the co$-$ordinates of the points $M$ and 4
Co$-$ordinate of the point $M$ is $(5,0)$
Co$-$ordinate of the point $N$ is $(0,-7)$ View full question & answer→