[3 marks sum]

Question 13 Marks

Draw a graph of each of the following equations: $y=\frac{3}{5}, x-1$

Answer

$y=\frac{3}{5}, x-1$
When $x=5, y=\frac{3}{5}(5)-1=2$
When $x=-5, y=\frac{3}{5}(-5)-1=-4$
When $x=10, y=\frac{3}{5}(10)-1=5$

$x$	$5$	$-5$	$10$
$y$	$2$	$-4$	$5$

Plotting the points $(5, 2), (-5, -4)$ and $(10, 5)$, we get a line $AB$ as shown in the figure.

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Question 23 Marks

Find the inclination and slope of a line which is equidistant from the $x-$axis.

Answer

Equidistant from the $x-$axis.
Indination $= 0^\circ $
Slope
$= \tan\theta $
$= \tan 0^\circ $
Slope $= 0$

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Question 33 Marks

Plot the points $A (3, 4)$ and $C (-3, -2)$ on a graph. Find the coordinates of the point $B$ and $D$ such $\text{ABCD}$ is a square. Also find the area of the square.

Answer

$B=(-3,4)$
$D=(3,-2)$
Area of the square
$=$ sid$e^2$
$= 6^2$
$= 36sq.$ units
$∴$ The area of the square is $36\ sq.$ units.

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Question 43 Marks

A rectangle $\text{PQRS}$ is drawn on the coordinate axes such that $P$ is the origin, $PQ = 6$ units and $PS = 5$ units. Find the coordinates of the vertices $P, Q R$ and $S.$ Also, find the area of the rectangle.

Answer

$P=(0,0)$
$Q=(6,0)$
$R=(6,5)$
$S=(0,5)$
Area of a rectangle
$=$ length $\times$ breadth
$= 6 \times 5$
$= 30sq.$ units
The area of rectangle is $30sq.$ units.

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Question 53 Marks

In each of the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation.$(7-x)+7 x =(x+5) ; \frac{2+3 y}{2}=2 y -6$

Answer

For abscissa,
$(7-x)+7 x=(x+5) \ldots ($given$)$
$ 7-x+7 x=x+5$
$ 6 x-x=5-7$
$ 5 x=-2$
$ \therefore x=-\frac{2}{5}$
$($given$)$
$7- x +7 x = x +5$
$6 x - x =5-7$
$5 x=-2$
$\therefore x =-\frac{2}{5}$
For ordinate,
$\frac{2+3 y}{2}=2 y-6$
$ 2+3 y=2(2 y-6)$
$ 2+3 y=4 y-12$
$ 3 y-4 y=-12-2$
$ -y=-14$
$ \therefore y=14$
$($given$)$
$\therefore$ The coordinates of the point are $\left(-\frac{2}{5}, 14\right)$.

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Question 63 Marks

In each of the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation.$5+2 x=9: 3 \frac{1}{2} y+1=12-3 y$

Answer

For abscissa,
$ 5+2 x=9 \ldots ($given$)$
$ 2 x=9-5$
$ x=\frac{4}{2}$
$ \therefore x=2$
For ordinate,
$3 \frac{1}{2} y+1=12-3 y \ldots ($given$)$
$ \frac{7}{2} y+3 y=12-1$
$ \frac{(7+6) y}{2}=11$
$ 13 y=22$
$ \therefore y=\frac{22}{13}$
$\therefore$ The coordinates of the point are $\left(2, \frac{22}{13}\right)$.

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Question 73 Marks

Find the value of $'a\ '$ and $'b\ '$ if $a. (a + 2,5 + b) = (1, 6),b. (2a + b, a - 2b) = (7, 6)$

Answer

$a.$ Given, two ordered pairs are equal.
$\therefore a + 2 = 1$ and $5 + b = 6$
$\therefore a = -1$ and $b = 1$
$b.$ Given, two ordered pairs are equal.
$\therefore 2a + b = 7 .....(1)$
$a - 2b = 6 .....(2)$
On multiplying equation $(1)$ with $2,$ we get:
$4a + 2b = 14 .....(3)$
Add equation $(2)$ and $(3),$
$a - 2b = 6$
$(+) 4a + 2b = 6$
$5a = 20$
$\therefore a = 4$
Substituting $a = 4$ in equation $(1),$ we get:
$2(4) + b = 7$
$\therefore b = -1$
$\therefore a = 4$ and $b = -1.$

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MATHEMATICS [3 marks sum]

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