Fill in the blanks.

Question 11 Mark

If the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2), then values of m and c are ___________ and ___________ respectively.

Answer

$m=\frac{1}{2}$ and $c=\frac{5}{2}$, because
Let the perpendicular $O M$ is drawn from the origin to $A B . M$ is the foot of the perpendicular.
$\therefore $ Slope of $O M=\frac{2-0}{-1-0}=-2$
Thus, $$ Slope of $A B=-\left(\frac{1}{-2}\right)=\frac{1}{2} [\because O M \perp A B]$
Since, $M(-1,2)$ lies on $A B$ whose equation is
$y=m x+c$
⇒ $y=\frac{1}{2} x+c$
⇒ $2=-\frac{1}{2}+c$
⇒ $c =\frac{5}{2}$
Thus, $m=\frac{1}{2}$ and $c=\frac{5}{2}$.

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Question 21 Mark

If 'p' is the length of the perpendicular from the origin to the line whose intercepts on axes are 'a' and 'b', then relation between a, b and p is ___________ .

Answer

$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$, because
Equation of the line which makes intercepts ' $a$ ' and $' b$ ' on the axes is $\frac{x}{a}+\frac{y}{b}=1$.
$\therefore$ The perpendicular distance 'p' from the origin is
$p=\left|\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right|$
$\Rightarrow \frac{1}{p}=\left|\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}\right|$
$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

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Question 31 Mark

The distance between parallel lines l(x + y) + p = 0 and l(x + y) - r = 0 is ___________ .

Answer

$\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right|$, because
Given lines are:
l(x + y) + p = 0
⇒ ly = - lx - p
⇒ $y=-x-\frac{p}{l}$
and l(x + y) - r = 0
⇒ ly = - lx + r
⇒ $y=-x+\frac{r}{l}$
$\therefore m=-1, c_1=\frac{-p}{l}$ and $c_2=\frac{r}{l}$
$\therefore$ Distance between the lines
$\begin{aligned} d & =\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}} \\ & =\frac{\left|-\frac{p}{l}-\frac{r}{l}\right|}{\sqrt{1+(-1)^2}}=\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right| .\end{aligned}$

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Question 41 Mark

The distance of (2, 3) from the line x + 4y = 5 is ___________ .

Answer

$\frac{9 \sqrt{17}}{17}$ units, because
Given line is x + 4y - 5 = 0
Here, A = 1, B = 4, C = - 5 and the given point is (x₁, y₁) = (2, 3)
$\therefore$ Distance, $d=\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right|$
$\begin{array}{l}=\left|\frac{1.2+4.3-5}{\sqrt{1^2+4^2}}\right| \\ =\frac{9}{\sqrt{17}}=\frac{9 \sqrt{17}}{17} \text { units. }\end{array}$

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Question 51 Mark

The equation of the line which is parallel to $Y$-axis and passes through $(-4,3)$ is ___________.

Answer

$x=-4$, because
Equation of a line parallel to $Y$-axis is $x=a$
Since, the line passes through the point $(-4,3)$.
Therefore, $-4=a$
$\Rightarrow a=-4$
$\therefore$ The required equation of line is $x=-4$ or $x+4=0$.

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Question 61 Mark

The equation of the line passing through the points $(2,2)$ and $(4,-6)$ is ___________.

Answer

$4 x+y-10=0$, because
Equation of a line in two-point form is
$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$
Given, $x_1=2, x_2=4, y_1=2$ and $y_2=-6$
$\therefore ~y-2=\frac{-6-2}{4-2}(x-2)$
$\begin{array}{lr}\Rightarrow 2(y-2)=-8(x-2) \\ \Rightarrow 2 y-4=-8 x+16\end{array}$
$\Rightarrow y-2=-4 x+8$
$\Rightarrow 4 x+y-10=0$

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Question 71 Mark

The equation of line with slope '5' and $\gamma$-intercept '3' above the axis is ___________.

Answer

$y=5 x+3$, because
Equation of line in slope intercept form is
$y=m x+c$
Here,$\quad m=5 \text { and } c=3$
$\therefore y=5 x+3$

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Question 81 Mark

The slope of the line $6 x+3 y-5=0$ is ___________ .

Answer

-2 , because
Given, line is $6 x+3 y-5=0$
$\text{or}\quad 3 y =-6 x+5$
$\text{or}\quad y =\frac{-6}{3} x+\frac{5}{3}$
$=-2 x+\frac{5}{3}$
$\therefore $ Slope $=-2$
(On comparing with $y=m x+c$)

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Question 91 Mark

If the origin is shifted to (2, 3), then the new co-ordinate of point (-1, 2) is ___________.

Answer

(-3,-1) because
O(0, 0) is changed to O'(2, 3).
$\therefore$ The new coordinate of (-1, 2) are
(- 1 - 2, 2 - 3) = (- 3, - 1).

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Question 101 Mark

The value of y, so that the line through (3, y) and (2, 7) is parallel to the line through (-1, 4) and (0, 6) is ___________.

Answer

y = 9 because
Slope of the line passing through (3, y) and (2, 7)
$m_1=\frac{7-y}{2-3}=y-7$
and slope of the line passing through (-1, 4) and (0, 6),
$m_2=\frac{6-4}{0+1}=2$
Since, the lines are parallel,
So,$\quad$m₁= m₂
⇒ y - 7 = 2
⇒ y = 9.

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Question 111 Mark

The angle between the lines whose slopes are $(2-\sqrt{3})$ and $(2+\sqrt{3})$ is ___________ .

Answer

$60^{\circ}$, because
Let the slopes of the two lines are $m_1=2-\sqrt{3}$ and $m_2=(2+\sqrt{3})$.
Then, acute angle '$\theta$' between the lines is given by
$\theta=\tan ^{-1} \frac{m_2-m_1}{1+m_1 m_2}$
$=\tan ^{-1} \frac{(2+\sqrt{3})-(2-\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}$
$\begin{array}{l}=\tan ^{-1} \frac{2 \sqrt{3}}{1+4-3} \\ =\tan ^{-1}(\sqrt{3}) \\ =\tan ^{-1}\left(\tan 60^{\circ}\right) \\ =60^{\circ} .\end{array}$

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Question 121 Mark

The slope of a line perpendicular to the line passing through (-3, 8) and (2,-2) is ___________ .

Answer

$\frac{1}{2}$, because
Slope of the line passing through $(-3,8)$ and $(2,-2)$,
$m_1=\frac{8+2}{-3-2}=-2$
If m₂ be the slope of the perpendicular line,
then $\quad m_{1}m_{2}=-1$
$m_2=\frac{1}{2}.$

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