[1 Mark Question Answer]

Question 11 Mark

Calculate the distance between $A (7, 3)$ and $B$ on the x-axis, whose abscissa is $11.$

Answer

Here $B$ is $(11, 0)$
$A B=\sqrt{(11-7)^2}+(0-3)^2$
$=\sqrt{(4)^2+(-3)^2}$
$=\sqrt{16+9}$
$=\sqrt{25}$
$=5 \text { units. }$

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Question 21 Mark

Find the distance of the following points from origin.
$(a \cos \theta, a \sin \theta).$

Answer

Let $p(a \cos \theta, a \sin \theta)$ and $O(0 , 0)$
$\text { Then }|O P|=\sqrt{(a \cos \theta-0)^2+(a \sin \theta-0)^2}$
$=\sqrt{a^2 \cos ^2 \theta+a^2 \sin ^2 \theta}$
$=\sqrt{a^2\left(\sin ^2 \theta+\cos ^2 \theta\right)}$
$=a \sqrt{1}$
$=a \text { units. }$

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Question 31 Mark

Find the distance of the following points from origin.
$(a+b, a-b)$

Answer

Let $P(a + b, a - b)$ and $O(0 , 0)$
$\text { Then } OP =\sqrt{(a+b-0)^2+(a-b-0)^2}$
$=\sqrt{(a+b)^2+(a-b)^2}$
$=\sqrt{a^2+b^2+2 a b+a^2+b^2-2 a b} $.
$OP =\sqrt{2\left(a^2+b^2\right)} \text { units. } $

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Question 41 Mark

Find the distance of the following points from origin.
$(5, 6)$

Answer

Let $O(0, 0)$ be the origin.
Distance between
$O(0 , 0)\ \&\ P(5 , 6)$
$O P=\sqrt{(5-0)^2+(6-0)^2}$
$=\sqrt{25+36}$
$=\sqrt{61} \text { units. }$

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Question 51 Mark

Find the equation of a line with slope 1 and cutting off an intercept of 5 units on Y-axis.

Answer

We have
Slope of the line m = 1
and Y-intercept, c = 5 units
The equation of line is given by
y = mx + c
i.e., y = 1.x + 5
⇒ y = x + 5
or x - y + 5 = 0.

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Question 61 Mark

Find the equation of a line which is inclined to $x$ axis at an angle of $60^\circ$ and its $y –$ intercept $= 2.$

Answer

Hence, $m =\tan 60^\circ =\sqrt{3}$
and $c = 2$
The equation of line is given by
$y = mx + c$
$ y =\sqrt{3} \cdot x+2$
$y =\sqrt{3} x+2$
$\sqrt{3} x-y+2=0$

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Question 71 Mark

Find the value of x so that the line passing through $(3, 4)$ and $(x, 5)$ makes an angle $135^\circ$ with positive direction of X-axis.

Answer

Slope of the line which makes an angle $135^\circ$ with X-axis,
$m = \tan 135^\circ$
$= -1.$
Also slope m $=\frac{5-4}{x-3}=\frac{1}{x-3}$
$\text { Then, } \frac{1}{x-3}=-1$
$\Rightarrow x-3=-1$
$\therefore x=2 .$

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Question 81 Mark

Given that $(a, 2a)$ lies on line $\frac{y}{2}=3-6$.Find the value of a.

Answer

Point $(a, 2a)$ lies on the line
$\frac{y}{2}=3 x-6 $
$\therefore \frac{2 a}{2}=3 a-6$
$ \Rightarrow a=3 a-6 $
$ \Rightarrow 2 a=6$
$ \Rightarrow a=3 .$

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Mathematics [1 Mark Question Answer]

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