Straight Lines Test 31

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Straight Lines Test 31

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Question 1
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A vertical line $$l$$ passes through the point $$(2,3)$$. A horizontal line $$m$$ passes through the point $$(-1,6)$$. Where do lines $$l$$ and $$m$$ intersect?

A
$$(0,5)$$

B
$$(2,6)$$

C
$$(6,2)$$

D
$$(-1,3)$$

E
$$(3,-1)$$

Solution

Given that the vertical line $$l$$ passes through the point $$(2,3)$$
Which implies that the equation of line $$l$$ is $$x=2$$
Given that the horizontal line $$m$$ passes through the point $$(-1,6)$$
Which implies that the equation of line $$l$$ is $$y=6$$
Therefore the point of intersection of lines $$l$$ and $$m$$ is $$(2,6)$$
Question 2
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In which adjacent figures, the path is the shortest path?

A

B

C

D

Solution

A. The total path length assuming each side of the square measures one unit is given by $$1 + \sqrt{5} + \sqrt{10} + \sqrt{8}$$
B. Similarly, total length is $$1 + 2 + \sqrt{10} + \sqrt{8}$$
C. Here, total length $$= \sqrt{5} + 2 + \sqrt{5} + \sqrt{5}$$
D. Total length $$= 1 + 2 + \sqrt{13} + \sqrt{5}$$
Clearly, Option A is greater than option B and so option A can be neglected for further comparison.
Comparing option B and option D,
$$1 + 2 + \sqrt{10} + \sqrt{8}$$ (inequality sign) $$1 + 2 + \sqrt{13} + \sqrt{5}$$
i.e. $$\sqrt{10} + \sqrt{8}$$ (inequality sign) $$\sqrt{13} + \sqrt{5}$$
Squaring both sides, we get $$18 + 2\sqrt{80}$$ (inequality sign) $$18 + 2\sqrt{65}$$
Since $$80 > 65$$, option B is greater than option D.
Comparing C and D,
$$2\sqrt{5}$$ (inequality sign) $$1 + \sqrt{13}$$
Squaring both sides, $$20$$ (inequality sign) $$14 + 2\sqrt{13}$$
i.e. $$3$$ (inequality sign) $$\sqrt{13}$$
Since $$3 < \sqrt{13}$$, option C is the shortest path
Question 3
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The line $$\displaystyle 3x+2y=24$$ meets x-axis at A and y-axis at B. The perpendicular bisector of $$\displaystyle \overline { AB } $$ meets the line through (0, -1) and parallel to x-axis at C. Find the area of $$\displaystyle \Delta ABC$$.

A
85 $$\displaystyle { units }^{ 2 }$$

B
87 $$\displaystyle { units }^{ 2 }$$

C
90 $$\displaystyle { units }^{ 2 }$$

D
91 $$\displaystyle { units }^{ 2 }$$

Solution
Question 4
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Area of the regular hexagon whose diagonal is the join of $$\left( 2,4 \right)$$ and $$\left( 6,7 \right)$$ is

A
$$\dfrac { 75\sqrt { 3 } }{ 8 }$$

B
$$\dfrac { 75\sqrt { 3 } }{ 16 }$$

C
$$\dfrac { 25\sqrt { 3 } }{ 16 }$$

D
$$\dfrac { 6\sqrt { 3 } }{ 5 }$$

Solution

Here , $$AB = \sqrt {(6-2)^2+ (7-4)^2 }$$

= $$\sqrt {(16)+ (9) }$$

= $$\sqrt 25 = 5$$.

$$AO = \dfrac {5}{2}$$

and we know that the area of regular hexagon is = $$ 6 \times \dfrac {\sqrt 3}{4} \times a^2$$

= $$ 6 \times \dfrac {\sqrt 3}{4} \times \dfrac {5^2}{2^2}$$

= $$ 6 \times \dfrac {\sqrt 3}{4} \times \dfrac {25}{4}$$

= $$\dfrac {75 \sqrt 3}{8}$$ sq.units.
Question 5
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An equilateral triangle has one vertex at $$(3, 4)$$ and another at $$(-2, 3) $$. Find the coordinates of the third vertex.

A
$$(\frac{1+\sqrt3}{2} , \frac{7+\sqrt3}{2})$$

B
$$(\frac{1-\sqrt3}{2} , \frac{7-\sqrt3}{2})$$

C
$$(\frac{1+\sqrt2}{3} , \frac{7+5\sqrt3}{2})$$

D
both $$A$$and $$B$$

Solution
Question 6
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If the area of the triangle whose vertices are $$(b,c),(c,a)$$ and $$(a,b)$$ is $$\triangle$$, then the area of the triangle whose vertices are $$(ac-{b}^{2},ab-{c}^{2}),(ab-{c}^{2},bc-{a}^{2}) and $$(cb-{a}^{2},ca-{b}^{2})$$ is

A
$$(a+b+c)\triangle$$

B
$$(a+b+c)^{2}\triangle$$

C
$$2(a+b+c)\triangle$$

D
$$None\ of\ these$$

Solution
Question 7
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What is the angle between the straight lines $$\left( { m }^{ 2 }-mn \right) y=\left( mn+{ n }^{ 2 } \right) x+{ n }^{ 3 }$$ and $$\left( mn+{ m }^{ 2 } \right) y=\left( mn-{ n }^{ 2 } \right) x+{ m }^{ 3 }$$, where $$m> n$$?

A
$$\tan ^{ -1 }{ \left( \cfrac { 2mn }{ { m }^{ 2 }+{ n }^{ 2 } } \right) } $$

B
$$\tan ^{ -1 }{ \left( \cfrac { 4{ m }^{ 2 }{ n }^{ 2 } }{ { m }^{ 4 }-{ n }^{ 4 } } \right) } $$

C
$$\tan ^{ -1 }{ \left( \cfrac { 4{ m }^{ 2 }{ n }^{ 2 } }{ { m }^{ 4 }+{ n }^{ 4 } } \right) } $$

D
$${ 45 }^{ o }$$

Solution

Given $$(m^{2} - mn)y = (mn + n^{2})x + n^{3}$$
$$(mn + m^{2})y = (mn - n^{2})x + m^{3}$$
Slope of line $$(1) = \dfrac {mn + n^{2}}{m^{2} -mn}$$

Slope of line $$(2) \rightarrow \dfrac {mn - n^{2}}{m^{2} + mn}$$
By comparing with $$y = mx + c$$

$$\tan \theta = \pm \left (\dfrac {m_{1} - m_{2}}{1 + m_{1}m_{2}}\right ), m_{1}$$ and $$m_{2}$$ are slope

$$\therefore \tan \theta = \dfrac {\left (\dfrac {mn + n^{2}}{m^{2} -mn} - \dfrac {mn - n^{2}}{m^{2} + mn}\right )}{\dfrac {1 + (mn + n^{2}(mn - n^{2})}{(m^{2} - mn)(m^{2} + mn)}}$$

$$\tan \theta = \dfrac {(mn + n^{2}) (m^{2} + mn) - (mn - n^{2})(m^{2} - mn)}{m^{4} - (mn)^{2} + (mn)^{2} - n^{4}}$$

$$\tan \theta = \dfrac {m^{3}n + (mn)^{2} + (mn)^{2} + mn^{3} - m^{3}n + (m^{2}n^{2}) + (mn)^{2} - mn^{3}}{m^{4} - n^{4}}$$

$$\tan \theta = \dfrac {4m^{2}n^{2}}{m^{4}- n^{4}}$$

$$\therefore \theta = \tan^{-1} \left (\dfrac {4m^{2} n^{2}}{m^{4} - n^{4}}\right )$$.
Question 8
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The line through point $$(m, -9)$$ and $$(7, m)$$ has slope m. The y-intercept of this line, is?

A
$$-18$$

B
$$-6$$

C
$$6$$

D
$$18$$

Solution

We know that the slope $$m$$ of a line passing through two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$m=\dfrac{y_2-y_1}{x_2-x_1}$$.

Let $$(m,-9)=(x_1,y_1)$$ and $$(7,m)=(x_2,y_2)$$, then the slope of the line can be determined as:

$$m=\dfrac { y_{ 2 }-y_{ 1 } }{ x_{ 2 }-x_{ 1 } } \\ \Rightarrow m=\dfrac { m-(-9) }{ 7-m } \\ \Rightarrow m=\dfrac { m+9 }{ 7-m } \\ \Rightarrow m(7-m)=m+9$$
$$\Rightarrow { -m }^{ 2 }+7m=m+9\\ \Rightarrow { m }^{ 2 }-7m+m+9=0\\ \Rightarrow { m }^{ 2 }-6m+9=0\\ \Rightarrow { (m-3) }^{ 2 }=0\\ \Rightarrow m-3=0\\ \Rightarrow m=3$$

Therefore, the points are $$(3,-9)$$ and $$(7,3)$$.

We also know that the equation of the line passing through a point $$(x_1,y_1)$$ is $$(y-y_1)=m(x-x_1)$$ where $$m$$ is the slope of the line.

Thus, the equation of the line passing through the point $$(7,3)$$ is as follows:

$$(y-y_{ 1 })=m(x-x_{ 1 })\\ \Rightarrow (y-3)=3(x-7)\\ \Rightarrow y-3=3x-21\\ \Rightarrow 3x-21-y+3=0\\ \Rightarrow 3x-y-18=0\\ \Rightarrow 3x-y=18$$

Now, put $$x=0$$ in the above equation to get the y-intercept as:

$$(3\times 0)-y=18\\ \Rightarrow 0-y=18\\ \Rightarrow y=-18$$

Hence, the y-intercept of the line is $$y=-18$$.
Question 9
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The area of the triangle formed by the positive $$x -$$axis, and the normal and tangent to the circle $${x}^{2}+{y}^{2}=4$$ at $$(1,\ \sqrt{3})$$ is

A
$$3\sqrt {2}\ sq$$. unit

B
$$2\sqrt {3}\ sq$$. unit

C
$$\sqrt {6}\ sq$$. unit

D
$$None\ of\ these$$

Solution

Consider the given equation.
$$ { x }^{ 2 }+{ y }^{ 2 }=4 $$

We know that the equation of the tangent to the circle $$ { x }^{ 2 }+{ y }^{ 2 }=C\\$$ at point $$ (x_1, y_1)$$ is is given by
$$ xx_1 + yy_1 = C $$

So, the equation of the tangent at $$ (1, \sqrt {3} )$$ is
$$ x $$ + $$ \sqrt {3} $$ $$ y = 4 $$ .......... (1)
$$ y = \dfrac { 4 }{ \sqrt { 3 } } - \dfrac { x }{ \sqrt { 3 } } $$, it cuts the $$x$$ axis at $$ (4, 0) $$.

Now, equation of normal to the circle is
$$ (y - \sqrt {3} ) $$ = Slope of normal $$ \times (x -1 ) $$.

Slope of normal = $$ \dfrac {- 1 } { Slope \ of \ tangent} $$

So,
Slope of normal = $$ \sqrt {3} $$

Now, equation of normal is
$$ (y -\sqrt {3} ) =\sqrt {3} $$ $$ ( x -1 ) $$
$$y= \sqrt {3}x$$ .......... (2)

Therefore, in figure 1, the area formed by tangent, normal and x axis will be,

$$A=\int_{0}^{1}\sqrt3xdx$$ + $$ \int_{1}^{4}(\dfrac{- x}{\sqrt 3} + \dfrac {4}{\sqrt 3 })dx $$
$$ A = { \sqrt { 3 } \left[ \dfrac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }\\ $$ + $$ { \dfrac { 1 }{ \sqrt { 3 } } \left[ -\dfrac { { x }^{ 2 } }{ 2 } +4x \right] }_{ 1 }^{ 4 }\\ $$
$$ A = \dfrac { \sqrt { 3 } }{ 2 } $$ + $$ { \dfrac { 1 }{ \sqrt { 3 } } \left[ -\dfrac { 16 }{ 2 } +16+\dfrac { 1 }{ 2 } -4 \right] } $$
$$ A = \dfrac { \sqrt { 3 } }{ 2 } $$ + $$ { \dfrac { 1 }{ \sqrt { 3 } } \left[ \dfrac { 9 }{ 2 } \right] }$$
$$ A = \dfrac { \sqrt { 3 } }{ 2 } +\dfrac { 3\sqrt { 3 } }{ 2 } $$
$$ A = \dfrac { 4\sqrt { 3 } }{ 2 } $$
$$ A = 2\sqrt { 3 } $$

Hence, option b is correct.
Question 10
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The points A $$(2a, 4a)$$, B $$(2a, 6a)$$ and C$$(2a+\sqrt{3a}, 5a)$$ (when $$a > 0$$) are vertices of?

A
An obtuse angled triangle

B
An equilateral triangle

C
An isosceles obtuse angled triangle

D
A right angled triangle

Solution