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 - k²)y + k² - 7k + 6 = 0 is passing through the origin.

Answer

The given equation of the line is
(k–3) x – (4 – k²) 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-k²)(0) + k² -7k +6 = 0
$\Rightarrow$ k² -7k +6 = 0
$\Rightarrow$ k² -6k-k +6 = 0
$\Rightarrow$ (k-6)(k-1)=0
$\Rightarrow$ k = 1or 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 – k² ) y + k² –7k + 6 = 0 is parallel to the y-axis.

Answer

The given equation of the line is
(k–3) x – (4 – k²) 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 k² = 4
$\Rightarrow$ k² = 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 – k² ) y + k² –7k + 6 = 0 is parallel to the x-axis.

Answer

The given equation of the line is
(k–3) x – (4 – k²) y + $k^2-7k+6=0$
The given line can be written as
(4 – k²) = (k–3) x + k² –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 (x₁, y₁) is given by
y - y₁ = m(x - x₁)
$\therefore$ 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 (x₀, y₀), if and only if, its coordinates satisfy the equation y - y₀ = m (x - x₀)
$\therefore$ 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 x₁ = 1, y₁ = – 1, x₂ = 3 and y₂ = 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 a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where b₁, b₂ $\ne$ 0 are: Perpendicular if a₁b₁+ b₁b₂= 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 m₁.m₂ = –1,
$\Rightarrow \frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}}=-1 $ or a₁a₂ + b₁b₂ = 0

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Question 121 Mark

Show that two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where b₁, b₂ $\ne$ 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 m₁ = 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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