1 Marks Question - Maths STD 11 Science Questions - Vidyadip

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark

Find the values of $k$ for which the line $(k - 3) x - (4 - k^2)y + k^2 - 7k + 6 = 0$ is passing through the origin.

Answer

The given equation of the line is
$(k–3) x – (4 – k^2) y + k^2-7k+6=0$
If the given line is passing through the origin, then point $(0,0)$ satisfies the given equation of line.
$\Rightarrow(k -3) (0) – (4-k^2)(0) + k^2 -7k +6 = 0$
$\Rightarrow k^2 -7k +6 = 0$
$\Rightarrow k^2 -6k-k +6 = 0$
$\Rightarrow (k-6)(k-1)=0$
$\Rightarrow k = 1$ or $6$
Thus, we get the value of $k$ is either $1$ or $6.$

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Question 21 Mark

Find the values of $k$ for which the line $(k–3) x – (4 – k^2 ) y + k^2 –7k + 6 = 0$ is parallel to the y-axis.

Answer

The given equation of the line is
$(k–3) x – (4 – k^2) y + k^2-7k+6=0$
The slope of the line is = $\frac{(k-3)}{\left(4-k^{2}\right)}$
Now, $\frac{(k-3)}{\left(4-k^{2}\right)}$ is undefined at $k^2 = 4$
$\Rightarrow k^2 = 4$
$\Rightarrow k = \pm 2$
Thus, if the given line is parallel to the $y –$ axis, then the value of k is $\pm 2.$

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Question 31 Mark

Find the values of $k$ for which the line $(k–3) x – (4 – k^2 ) y + k^2 –7k + 6 = 0$ is parallel to the x-axis.

Answer

The given equation of the line is
$(k–3) x – (4 – k^2) y + k^2-7k+6=0$
The given line can be written as
$(4 – k^2) = (k–3) x + k^2 –7k + 6 = 0$
$\Rightarrow y = \frac{(k-3)}{\left(4-k^{2}\right)} x+\frac{k^{2}-7 k+6}{\left(4-k^{2}\right)}$ which is the form $y = mx+c$
Therefore, Slope of the given line $=\frac{(k-3)}{\left(4-k^{2}\right)}$
Slope of $x-$ axis = 0
Then, \frac{(k-3)}{\left(4-k^{2}\right)}=0
$\Rightarrow k – 3 = 0$
$\Rightarrow k = 3$
Thus if the given line is parallel to the $x – $axis, then the value of $k$ is $3$

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Question 41 Mark

Find the equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through the point $(–2, 3).$

Answer

Given equation of the line is $3x – 4y + 2 = 0$
$\Rightarrow y=\frac{3 x}{4}+\frac{2}{4}$
$\Rightarrow y=\frac{3 x}{4}+\frac{1}{2}$
Which is of the form $y = mx + c,$ where $m$ is the slope.
$\therefore$ the slope of the given line is $\frac34$
We know that parallel lines have the same slope.
$\therefore$ the slope of another line $= m = \frac34$
Equation of line having slope m and passing through $(x_1, y_1)$ is given by
$y - y_1 = m(x - x_1)$
$\therefore$ Equation of line having slope $\frac34$ and passing through $(-2, 3)$ is
$y-3=\frac{3}{4}(x-(-2))$
$\Rightarrow 4y - 3 \times 4 = 3x + 3 \times 2$
$\Rightarrow 3x - 4y = 18$

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Question 51 Mark

Reduce the equation into slope-intercept form and find the slope and the y-intercept.
y = 0

Answer

Here y = 0
$\Rightarrow$ y = 0 . x + 0
Which is required slope intercept form,
Comparing it with y = mx + c, we have
m = 0 and c = 0

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Question 61 Mark

Reduce the equation into slope intercept form and find the slope and the y-intercepts.
6x + 3y - 5 = 0

Answer

Here 6x + 3y - 5 = 0
$\Rightarrow$ 3y = -6x + 5 $\Rightarrow y = \frac{{ - 6}}{3}x + \frac{5}{3} \Rightarrow y = - 2x + \frac{5}{3}$
Which is required slope intercept form,
Comparing it with y = mx + c, we have
m = -2 and $c = \frac{5}{3}$

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Question 71 Mark

Reduce the equation into slope-intercept form and find the slope and the y-intercept.
x + 7 y = 0

Answer

Here x + 7 y = 0
$\Rightarrow$ 7y = -x

$y = \frac{{ - 1}}{7}x \Rightarrow y = \frac{{ - 1}}{7}x + 0$ which is required slope intercept form.
Comparing it with y = mx +c, we have
$m = \frac{{ - 1}}{7}$ and c = 0

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Question 81 Mark

Find the equation of the line which satisfy the given condition:
Passing through $(0, 0)$ with slope m.

Answer

Given point $= (0, 0)$ and slope $= m$
We know that the point $(x, y)$ lies on the line with slope m through the fixed point $(x_0, y_0),$ if and only if, its coordinates satisfy the equation $y - y_0 = m (x - x_0)$
$\therefore y - 0 = m (x - 0)$
$\Rightarrow y = mx$
$\Rightarrow y - mx = 0$
Therefore, the required equation of the line is $y - mx = 0.$

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Question 91 Mark

Find the equation of the line which satisfy the given condition:

Write the equations for the x-and y-axes.

Answer

We know that the y-coordinate of every point on the x-axis is 0.
$\therefore$ Equation of x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
$\therefore$ Equation of y-axis is x = 0.

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Question 101 Mark

Write the equation of the line through the points $(1, –1)$ and $(3, 5)$

Answer

We know that,
$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$
Here $x_1 = 1, y_1 = – 1, x_2 = 3$ and $y_2 = 5.$
$y-(-1)=\frac{5-(-1)}{3-1}(x-1)$
or $ –3x + y + 4 = 0.$
which is the required equation.

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Question 111 Mark

Show that two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0,$ where $b_1, b_2 \ne 0$ are: Perpendicular if $a_1b_1+ b_1b_2= 0$

Answer

Given lines can be written as
$y=-\frac{a_{1}}{b_{1}} x-\frac{c_{1}}{b_{1}}$ ...(1)
and $y=-\frac{a_{2}}{b_{2}} x-\frac{c_{2}}{b_{2}}$ ...(2)
Slopes of the lines (1) and (2) are $m_{1}=-\frac{a_{1}}{b_{1}}$ and $m_{2}=-\frac{a_{2}}{b_{2}}$, respectively.
Now, lines are perpendicular, if $m_1.m_2 = –1,$
$\Rightarrow \frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}}=-1 $ or $a_1a_2 + b_1b_2 = 0$

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Question 121 Mark

Show that two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0,$ where $b_1, b_2 \ne 0$ are: Parallel if $\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}$

Answer

Given lines can be written as
$y=-\frac{a_{1}}{b_{1}} x-\frac{c_{1}}{b_{1}} ...(1)$
and $y=-\frac{a_{2}}{b_{2}} x-\frac{c_{2}}{b_{2}} ...(2)$
Slopes of the lines (1) and (2) are $m_{1}=-\frac{a_{1}}{b_{1}}$ and $m_{2}=-\frac{a_{2}}{b_{2}}$, respectively.
Now, lines are parallel, if $m_1 = m_{2}.$
$\Rightarrow -\frac{a_{1}}{b_{1}}=-\frac{a_{2}}{b_{2}} \text { or } \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}$

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Question 131 Mark

Find the slope of the lines. Making inclination of 60° with the positive direction of x-axis

Answer

Here inclination of the line α = 60°.
Thus, slope of the line is m = tan 60° = $\sqrt{3}$

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Question 141 Mark

Equation of a line is 3x – 4y + 10 = 0. Find its x - and y-intercepts.

Answer

Given equation 3x – 4y + 10 = 0 can be written as
$3 x-4 y=-10$ or $\frac{x}{-\frac{10}{3}}+\frac{y}{\frac{5}{2}}=1$ ... (2)
Comparing (2) with $\frac{x}{a}+\frac{y}{b}=1$ we have x-intercept as a $=-\frac{10}{3}$ and y-intercept as b =$\frac{5}{2}$.

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Question 151 Mark

Equation of a line is 3x – 4y + 10 = 0. Find its slope.

Answer

Here, equation 3x – 4y + 10 = 0 can be written as $y=\frac{3}{4} x+\frac{5}{2}$ ... (1)
Comparing (1) with y = mx + c, then we get hslope of the given line as $m=\frac{3}{4}$.

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Question 161 Mark

Find the slope of the line passing through the points (3, – 2) and (3, 4).

Answer

The slope of the line through the points (3, – 2) and (3, 4) is
$m=\frac{4-(-2)}{3-3}=\frac{6}{0}$ which is not defined.

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Question 171 Mark

Find the slope of the lines passing through the points (3, - 2) and (7, - 2).

Answer

Slope of the line through the points (3, - 2) and (7, - 2) is
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-(-2)}{7-3}=\frac{0}{4}=0$

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Question 181 Mark

Find the slope of the line passing through the points (3, -2) and (-1, 4).

Answer

Slope of the line through the points (3, -2) and (-1, 4) is
$m=\frac{4-(-2)}{-1-3}$ $\left[\because m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]$
$=\frac{6}{-4}=-\frac{3}{2}$

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