5 Marks Questions

Question 15 Marks

Draw the graphs of linear equations $y = x$ and $y = -x$ on the same Cartesian plane. What do you observe$?$

Answer

The given equation is $y = x$. To draw the graph of this equations, we need atleast two points lying on the given line.

For $x = 1, y = 1,$ therefore $(1, 1)$ satisfies the linear equation $y = x.$ For $x = 4, y = 4,$
therefore $(4, 4)$ satisfies the linear equation $y = x.$
By plotting the points $(1, 1)$ and $(4, 4)$ on the graph paper and joining them by a line,
we obtain the graph of $y = x.$
The given equation is $y = – x.$ To draw the graph of this equation,
we need atleast two points lying on the given line.
For $x = 3, y = -3,$ therefore, $(3, -3)$ satisfies the linear equation $y = -x.$ For $x = -4, y = 4,$
therefore, $(-4, 4)$ satisfies the linear equation $y = -x.$
By plotting the points $(3, –3)$ and $(-4, 4)$ on the graph paper and joining them by a line,
we obtain the graph of $y = -x.$ We observe that,
the line $y = x$ and $y = -x$ intersect at the point $O(0, 0).$

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Question 25 Marks

The following observed values of $x$ and $y$ are thought to satisfy a linear equation. Write the linear equation

$x$	$6$	$-6$
$y$	$-2$	$6$

Draw the graph, using the values of $x, y$ as given in the above table. At what points the graph of the linear equation
$i.$ Cuts the $X-$axis$?$
$ii.$ Cuts the $Y- $axis$?$

Answer

Given, points are $(6, -2)$ and $(-6, 6) $
Let the linear eqution $y = mx + c$ is satisfied by the points $(6, -2)$ and $(-6, 6)$ then at point $(6, -2)$
$-2=6\text{m}+\text{c}\ ....(\text{i})$
$6=-6\text{m}+\text{c}\ ....(\text{ii})$ and at point $(-6, 6).$
On subtracting Eq. $(ii)$ from Eq. $(i),$ we get
$12\text{m}=-8$
$\Rightarrow\text{m}=\frac{-8}{12}$
$\Rightarrow\text{m}=-\frac{2}{3}$
On putting the value of min Eq. $(i),$
we get $-2=6\Big(\frac{-2}{3}\Big)+\text{c}$
$\Rightarrow-2=-4+\text{c}$
$\Rightarrow\text{c}=-2+4$
$\Rightarrow\text{c}=2$
Om putting $\text{m}=-\frac{2}{3}$ and $c = 2$ in linear eqution $y = mx + c$ we get

$\Rightarrow\text{y}=-\frac{2}{3}\text{x}+2$
$\Rightarrow\text{y}=\frac{-2\text{x}+6}{3}$
$\Rightarrow3\text{y}=-2\text{x}+6$
$\Rightarrow3\text{y}+2\text{x}=6$ When the graph of the linear equation
$i.$ Cuts the $X-$axis
Then, put $y = 0$ in eqution $2x + 3y = 6,$ we get
$2x + 3.0 = 6$
$\Rightarrow 2x = 6$
$\therefore x = 3$
When the graph of the linear equation
$ii.$ Cuts the $Y-$axis
Then, put $x = 0$ in eqution $2x + 3y = 6,$ we get
$2.0 + 3y = 6$
$\Rightarrow 3y = 6$
$\therefore y = 2$
Therefore, the graph the linear eqution cuts the $X-$axis at the point $(3, 0)$ and the $Y-$axis at the point $(0, 2).$

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