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CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 8 - MCQExams.com
CBSE
Class 11 Engineering Maths
Straight Lines
Quiz 8
In the figure, $$AC = 9, BC = 3$$ and $$D$$ is $$3$$ times as far from $$A$$ as from $$B$$. What is $$BD$$?
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6
0%
9
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12
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15
0%
18
Explanation
Let $$BD=x$$
It is given that
$$AC=9$$,
$$BC=3$$ and $$D$$ is $$3$$ times as far from $$A$$ as from $$B$$, therefore $$AD=3BD$$ which means:
$$AD=3BD$$
$$\Rightarrow 9+3+x=3x$$
$$\Rightarrow 12+x=3x$$
$$\Rightarrow 3x-x=12$$
$$\Rightarrow 2x=12$$
$$\Rightarrow x=6$$
Hence, $$BD=6$$
If the area of the triangle shown is $$30$$ square units, what is the value of $$y$$?
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$$5$$
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$$6$$
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$$8$$
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$$12$$
Explanation
Area of triangle $$=\dfrac 12 \left| \begin {matrix} 1 & 1 & 1 \\ 0 & 0 & 5 \\ y & 0 & 0 \end {matrix}\right|$$
$$\Rightarrow 30=\dfrac 12 \left| 1(0-0) - 1(5y-0) +1(0-0)\right|$$
$$\Rightarrow 60 = |-5y|$$
$$\Rightarrow 60 = 5y$$
$$\Rightarrow y =12$$
If the distance between the two points $$P(a,3)$$ and $$Q(4,6)$$ is $$5$$, then find $$a$$.
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$$0$$
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$$-4$$
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$$8$$
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$$-4$$ and $$0$$
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$$0$$ and $$8$$
Explanation
The distance between points $$(a,3)$$ and $$(4,6)$$ is $$\sqrt { { (a-4) }^{ 2 }+{ (3-6) }^{ 2 } } =\sqrt { { (a-4) }^{ 2 }+9 } $$
Given that the distance is $$5$$
So, we have $$\sqrt { { (a-4) }^{ 2 }+9 } =5$$ $$\Rightarrow {(a-4)}^{2}+9=25$$
$$\therefore a-4 = \sqrt{16} = |4|$$
So, we get $$a = 4+|4| = 0 , 8$$
In the $$xy$$-plane, the vertices of a triangle are $$(-1,3), (6,3)$$ and $$(-1,-4)$$. The area of the triangle is ___ square units.
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$$10$$
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$$17.5$$
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$$24.5$$
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$$35$$
Explanation
If $$(x_1,y_1)$$,
$$(x_2,y_2)$$ and
$$(x_3,y_3)$$
are the vertices of a triangle then its area is given by,
$$A=\left| \dfrac { 1 }{ 2 } (x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 })) \right| $$
Therefore, with the vertices $$(-1,3)$$,
$$(6,3)$$ and
$$(-1,-4)$$.
Area of triangle is given by,
$$A=\left| \dfrac { 1 }{ 2 } (-1(3-(-4))+6(-4-3)+(-1)(3-3)) \right|$$
$$ =\left| \dfrac { 1 }{ 2 } (-7-42) \right| $$
$$=\left| \dfrac { 1 }{ 2 } (-49) \right| $$
$$=\left| -24.5 \right|$$
$$ =24.5$$ square units.
If figure $$\Box ABCD$$
ABCD
is a parallelogram, what is the x-coordinate of point B?
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$$2$$
0%
$$5$$
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$$6$$
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$$8$$
Explanation
Looking at the figure, we can say that the $$y-$$ coordinate of point $$B$$ is $$5$$.
Hence, the coordinates of $$B$$ are $$(x,5)$$.
Since, $$\Box ABCD$$ is a paralleogram,
$$AB = CD$$ ...Opposite sides of parallelogram are congruent
By distance formula,
$$(6-2)^2+(-3+3)^2 = (x-4)^2+(5-5)^2$$
$$\Rightarrow 4^2 = x^2-8x+4^2$$
$$\Rightarrow x^2-8x=0$$
$$x(x-8) = 0$$
$$x=0$$ and $$x=8$$
Now, $$x \neq 0$$ since point B is in the 1st quadrant.
Hence, the $$x-$$ coordinate of B is $$8$$.
In the figure, there is a regular hexagon with sides of length $$6$$. If the coordinate of $$A$$ is $$(9,0)$$, what is the y-coordinate of $$B$$?
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$$0$$
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$$3$$
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$$3\sqrt{2}$$
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$$3\sqrt{3}$$
Explanation
Let the vertex between A and B be C (other than 'A' lying on the x-axis).
Since it is a regular hexagon, hence the length of side AC will be $$6$$.
Therefore, $$AC=6$$. Let the coordinate of C be $$(x,0)$$.
Therefore, application of distance formula gives us
$$9-x=6$$ or $$x=3$$.
Now let the coordinate of B be $$(0,y)$$.
Therefore, $$BC=6$$ $$\Rightarrow \sqrt{3^{2}+y^{2}}=6$$
$$\Rightarrow 9+y^{2}=36$$
$$\Rightarrow y^{2}=27$$
$$\Rightarrow y=3\sqrt{3}$$.
Hence $$B=(0,3\sqrt{3})$$.
Let $$S$$ be the set of points whose abscissas and ordinates are natural numbers. Let $$P \in S$$ such that the sum of the distance of $$P$$ from $$(8,0)$$ and $$(0,12)$$ is minimum among all elements in $$S$$. Then the number of such points $$P$$ in $$S$$ is
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$$1$$
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$$3$$
0%
$$5$$
0%
$$11$$
Explanation
The sum of the distances of $$P$$ from $$A(8,0),B(0,12)$$ is minimum if $$P$$ lies on line joining $$AB$$ and between them
$$AP+PB=AB$$
$$AP'+P'B>AB$$ by triangle property
So let $$P(x,y)$$
The line $$AB$$ is $$\dfrac { x }{ 8 } +\dfrac { y }{ 12 } =1$$
ie $$3x+2y=24$$
$$\Rightarrow 3x=2(12-y)$$
So $$y$$ can be $$3,6,9$$
$$y=3,x=6\rightarrow 1$$ point
$$y=6,x=4\rightarrow 1$$ point
$$y=9,x=2\rightarrow 1$$ point
So in total three points.
Which of the following sets of points is collinear
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$$(1, -1), (-1, 1), (0, 0)$$
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$$(1, -1), (-1, 1), (0, 1)$$
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$$(1, -1), (-1, 1), (1, 0)$$
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$$(1, -1), (-1, 1), (1, 1)$$
Explanation
General equation of a line:
$$y = mx + c $$
Now, equation of line passing through $$( 1,-1)$$ and $$(-1,1)$$
Here, both points $$(1,-1)$$ and $$(-1,1)$$ should satisfies the line.
$$(1,-1)$$ gives $$-1 = m+c$$ ... (i)
$$(-1,1)$$ gives $$1 = -m +c $$ ...(ii)
Solving (i) and (ii), we get
$$m = -1$$ and $$c=0$$
Therefore equation of line is $$y = -x$$
Now,
$$(0,0)$$ satisfies the line but $$(0,1), (1,0)$$ and $$(1,1)$$ does not satisfies it.
Hence, $$(1,-1), (-1,1), (0,0)$$ is a set of collinear points.
Area of the triangle formed by the points $$\left( 0,0 \right) ,\left( 2,0 \right) $$ and $$\left( 0,2 \right) $$ is
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$$1$$ sq. unit
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$$2$$ sq. units
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$$4$$ sq. units
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$$8$$ sq. units
Explanation
Given: $$A=(x_1,y_1)=(0,0)$$
$$B=(x_2,y_2)=(2,0)$$ and
$$C=(x_3,y_3)=(0,2)$$
Area of triangle $$=\dfrac {1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)++x_3(y_1-y_2)]$$
$$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$$
$$=\dfrac {1}{2}[2(2)]$$
$$=2$$ sq. units.
The vertices of $$\triangle {ABC}$$ are $$A(1,8),B(-2,4), C(8,-5)$$. If $$M$$ and $$N$$ are the midpoints of $$AB$$ and $$AC$$ respectively, find the slope of $$MN$$ and hence verify that $$MN$$ is parallel to $$BC$$.
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$$-\cfrac{9}{10}$$
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$$\cfrac{9}{10}$$
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$$-\cfrac{9}{5}$$
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None of these
Explanation
Given,
Vertices of triangle $$ABC$$ i.e. $$A (1, 8), B (–2, 4), C (8, –5)$$
$$M$$ and $$N$$ are $$mid – points$$ of $$AB$$ and $$AC$$.
Finding co–ordinates of $$M$$ and $$N$$,
We know that,
$$M$$ is the $$mid–point$$ of $$AB$$
$$x_1 = 1, x_2 = –2$$
$$y_1 = 8, y_2 = 4$$
Mid–point of $$AB$$
$$=\left(\dfrac{1-2}{2}, \dfrac{8+4}{2}\right)$$
$$=\left(\dfrac{-1}{2}, \dfrac{12}{2}\right)$$
$$=\left(-\dfrac{1}{2}, 6\right)$$
$$N$$ is the $$mid–point$$ of $$AC$$
$$x_1 = 1, x_2 = 8$$
$$y_1 = 8, y_2 = –5$$
Mid–point of $$AC $$
$$=\left(\dfrac{1+8}{2}, \dfrac{8-5}{2}\right)$$
$$=\left(\dfrac{9}{2}, \dfrac{3}{2}\right)$$
Slope of $$MN$$:
Slope of line passing through $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is
$$m=\dfrac{y_2-y_1}{x_2-x_1}$$
So,
$$m_1=\dfrac{\dfrac{3}{2}-6}{\dfrac{9}{2}+\dfrac{1}{2}}$$
$$m_1=\dfrac{-\dfrac{9}{2}}{5}$$
$$m_1=-\dfrac{9}{10}$$
Verification of $$MN$$ and $$BC$$ are parallel:
If $$MN$$ and $$BC$$ are parallel, then their slopes must be equal.
Slope of $$BC$$:
$$B (–2, 4)$$ and $$C (8, –5)$$
Slope of $$BC$$
$$m_2=\dfrac{-5-4}{8+2}$$
$$m_2=-\dfrac{9}{10}$$
$$\Rightarrow m_1=m_2$$
$$\therefore\ $$ Slope of $$MN =$$ Slope of $$BC $$
Hence, $$MN$$ is parallel to $$BC.$$
What is the perimeter of the triangle with vertices $$A (-4, 2), B (0, -1)$$ and $$C (3, 3)$$?
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$$ 7 + 3 \sqrt{2} $$
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$$ 10 + 5 \sqrt{2} $$
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$$ 11 + 6 \sqrt{2} $$
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$$ 5 + 10 \sqrt{2} $$
Explanation
Given :
$$A (-4, 2), B (0, -1)$$ and $$C (3, 3)$$
Now distance between A and B $$=\sqrt { (4)^{ 2 }+(-3)^{ 2 } } = 5$$
Distance between A and C $$=\sqrt { (7)^{ 2 }+(1)^{ 2 } } = 5\sqrt { 2 }$$
Distance between B and C $$=\sqrt { (3)^{ 2 }+(4)^{ 2 } } =5$$
So perimeter $$=AB+BC+AC=10+5\sqrt2$$
Hence, option B is correct.
The vertices of a triangle ABC are A(2, 3, 1), B(-2, 2, 0) and C(0, 1, -1). Find the cosine of angle ABC.
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$$\frac{1}{\sqrt{3}}$$
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$$\frac{1}{\sqrt{2}}$$
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$$\frac{2}{\sqrt{6}}$$
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None of the above
Explanation
Given $$A(2, 3, 1), B(-2, 2, 0)$$ and $$C(0, 1, -1)$$
For $$\triangle ABC,$$
$$a=BC=\sqrt { { (-2-0) }^{ 2 }+{ (2-1) }^{ 2 }+{ (0+1) }^{ 2 } } \\ \therefore a=\sqrt { 6 } \\ b=AC=\sqrt { { (2-0) }^{ 2 }+{ (3-1) }^{ 2 }+{ (1+1) }^{ 2 } } \\ \therefore b=\sqrt { 12 } \\ c=AB=\sqrt { { (2+2) }^{ 2 }+{ (3-2 })^{ 2 }+{ (1) }^{ 2 } } \\ \therefore c=\sqrt { 18 } \\ \cos { \angle ABC } =\cfrac { { a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 } }{ 2ac } \\ =\cfrac { 18+6-12 }{ 2\times 3\sqrt { 2 } \times \sqrt { 6 } } \\ =\cfrac { 1 }{ \sqrt { 3 } } $$
Area of the triangle formed by the points $$(0,0),(2,0)$$ and $$(0,2)$$ is
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$$1$$ sq.units
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$$2$$ sq.units
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$$4$$ sq.units
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$$8$$ sq.units
Explanation
Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
$$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$$
$$=\dfrac{1}{2}\left|0+4+0\right|$$
$$=\dfrac{1}{2}\times 4=2\ sq.\ units$$
Hence, this is the answer.
Find the value of $$a$$ if area of the triangle is $$17$$ square units whose vertices are
$$(0,0), (4,a), (6,4)$$.
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$$a=-2$$
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$$a=-3$$
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$$a=3$$
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none of the above
Explanation
vertices of the triangle are $$A(0,0), B(4,a), C(6,4)$$.
Then area of triangle$$ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$
$$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$$
$$\Rightarrow 34=16-6a$$
$$\Rightarrow a=-\dfrac{18}{6}=-3$$
The slope of a straight line passing through A( -2, 3) is -4/The points on the line that are 10 units away from A are
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(-8, 11), (4, -5)
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(- 7, 9), (17,-1)
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(7, 5) (- 1, -1)
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(6, 10), (3, 5)
Explanation
The equation line will be
$$y-3=\frac{-4}{3}(x+2)$$
$$4x+3y=1$$
Now, let the point $$10$$ unit away is $$(h,k)$$
Then,
$$4h+3k=1$$ ------ $$(1)$$
And
$$\sqrt {(h+2)^2+(k-3)^2}=10$$
$$ {(h+2)^2+(k-3)^2}=10$$ ------ $$(2)$$
Now, from $$(1)$$
$$h=\frac{1-3k}{4}$$
Put in equation $$(2)$$, we get
$$25k^2-150k-1663=0$$
On solving quadric equation
$$K=11$$ and $$k=-5$$
Put $$k=11$$ in equation $$(1)$$
$$h=-8$$
And put $$k=-5$$ in equation $$(1)$$
$$h=4$$
Hence the points are
$$(-8,11),(4,-5)$$
Hence, the correct answer is $$A$$.
If the equation to the locus o points equidistant from the points (-2,3),(6,-5) is $$ax+by+c=0$$ where $$a> 0$$ then, the ascending order of a,b,c is
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a,b,c
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c,b,a
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b,c,a
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a,c,b
Explanation
$$A(-2,3);B(6,-5)$$
$$p(x,y)$$
$$PA^{2}=PB^{2}$$
$$(x+2)^{2}+(y-3)^{2}=(x-6)^{2}+(y+5)^{2}$$
$$4x+4+9-6y=-12x+36+10y+25$$
$$16x-16y-48=0$$
$$x-y-3=0$$
$$ax+by+c=0$$
$$a=1;b=-1;c=-3$$
Ascending order :$$c,b,a $$
The point on the line $$4x+3y=5$$, which is equidistant from $$\left( 1,2 \right)$$ and $$\left( 3,4 \right) $$, is
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$$\left( 7,-4 \right) $$
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$$\left( -10,15 \right) $$
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$$\left( \dfrac { 1 }{ 7 } ,\dfrac { 8 }{ 7 } \right) $$
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$$\left( 0,\dfrac { 5 }{ 4 } \right) $$
Explanation
Let the point $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ be on the line $$4x+3y=5$$.
$$\therefore 4{ x }_{ 1 }+3{ y }_{ 1 }=5$$ ....(i)
Also, we have
$${ \left( { x }_{ 1 }-1 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 }={ \left( { x }_{ 1 }-3 \right) }^{ 3 }+{ \left( { y }_{ 1 }-4 \right) }^{ 2 }$$
$$\Rightarrow { x }_{ 1 }^{ 2 }+1-2{ x }_{ 1 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }={ x }_{ 1 }^{ 2 }+9-6{ x }_{ 1 }+{ y }_{ 1 }^{ 2 }+16-8{ y }_{ 1 }$$
$$\Rightarrow 4{ x }_{ 1 }+4{ y }_{ 1 }=20$$
$$\Rightarrow { x }_{ 1 }+{ y }_{ 1 }=5$$ ....(ii)
From equations (i) and (ii), we get
$${ y }_{ 1 }=15$$ and $${ x }_{ 1 }=-10$$
$$A=(4, 2)$$ and $$B=(2, 4)$$ are two given points and a point $$P$$ on the line $$3x + 2y + 10 = 0 $$ is given then. which of the following is true.
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$$(PA + PB)$$ is minimum when $$P (\dfrac {-14}{5}, \dfrac {-4}{5}) $$
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$$(PA + PB)$$ is maximum when $$P (\dfrac {-14}{5}, \dfrac {-4}{5}) $$
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$$(PA - PB)$$ is maximum when $$ (-22, 28) $$
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$$(PA - PB)$$ is minimum when $$ P (-22, 28) $$
If the slop of one of the lines represented by $$ax^2-6xy+y^2=0$$ is the square of the other,then the value of a is
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$$-27$$ or $$8$$
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$$-3$$ or $$2$$
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$$-64$$ or $$27$$
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$$-4$$ or $$3$$
Explanation
$$m+m^2=6$$
$$m^3=a$$
$$m^3+m^6+3.m^3.6=6^3$$
$$a^2+19a-216=0$$
$$a=-27$$ or $$8$$
Find the area (in square units) of the triangle whose vertices are $$(a, b+c), (a, b-c) $$ and $$(-a, c). $$
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$$2ac$$
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$$2bc$$
0%
$$b(a+c)$$
0%
$$c(a-h)$$
Explanation
Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times |[ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] |$$
Given the vertices of triangle,
$$A (a,b+c) ,B(a,b-c) ,C(-a,c)$$
Therefore, area is given by
$$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$$
If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.
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$$11$$
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$$22$$
0%
$$48$$
0%
$$50$$
Explanation
Area of triangle = $$\dfrac{1}{2} {x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$$
Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$
$$=|\dfrac{1}{2}(-3+16-35)|$$
$$=11$$ sq. Units.
If the straight line $$ax + by + p = 0$$ and $$x\cos \alpha + y \sin \alpha = p$$ enclosed an angle of $$\dfrac {\pi}{4}$$ and the line $$x\sin \alpha - y \cos \alpha = 0$$ meets them at the same point, then $$a^{2} + b^{2}$$ is
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$$4$$
0%
$$3$$
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$$2$$
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$$1$$
Explanation
$$x\sin \alpha-y\cos\alpha =0$$
=> $$x=\dfrac{y\cos\alpha }{\sin\alpha}$$
$$x\cos\alpha+y\sin\alpha=p$$
Put
$$x=\dfrac{y\cos\alpha }{\sin\alpha}$$, we get
=> $$\dfrac{y\cos^2\alpha}{\sin\alpha}+y\sin\alpha=p$$
=>$$y=p\sin\alpha$$
Similiarly, $$x=p\cos\alpha$$
$$ax+by+p=0$$
Put
$$x=p\cos\alpha$$ and
$$y=p\sin\alpha$$
=> $$ap\cos\alpha+bp\sin\alpha+p=0$$
=> $$a\cos\alpha+b\sin\alpha=-1$$ ---- Equation 1
Slope of
$$ax+by+p=0$$, $$m_1=-\dfrac{a}{b}$$
Slope of
$$x\cos\alpha+y\sin\alpha=p$$, $$m_2=-\dfrac{\cos\alpha}{\sin\alpha}$$
Also $$\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$$
=> $$\tan\dfrac{\pi}{4}=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$$
=> $$|m_1-m_2|=|1+m_1m_2|$$
=> $$\left|\dfrac{a}{b}-\dfrac{\cos\alpha }{\sin\alpha}\right|=\left|1+\dfrac{a\cos\alpha}{b\sin\alpha}\right|$$
=> $$|a\sin\alpha-b\cos\alpha|=|b\sin\alpha+a\cos\alpha|$$
=> $$|a\sin\alpha-b\cos\alpha|=1$$ ---- Equation 2
Squaring and adding equation 1 and equation 2, we get
$$(a^2\cos ^2\alpha +b^2\sin ^2\alpha +2ab\sin \alpha \cos \alpha)+(b^2\cos ^2\alpha +a^2\sin ^2\alpha -2ab\sin \alpha \cos \alpha)=1+1$$
=> $$a^2+b^2=2$$
Option $$(C)$$
The distance of the point $$A(a, b, c)$$ from the x-axis is
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$$a$$
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$$\sqrt {b^{2} + c^{2}}$$
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$$\sqrt {a^{2} + b^{2}}$$
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$$a^{2} + b^{2}$$
Explanation
$$\textbf{Step-1: Apply distance formula to get the required unknown.}$$
$$\text{The given point is}$$ $$(a,b,c)$$
$$\text{The distance of the point}$$ $$(a,b,c) $$ $$\text{from x-axis will be}$$ $$\text{the perpendicular distance from point}$$
$$(a,b,c)$$ $$\text{to x-axis whose co-ordinates will be}$$ $$(a,0,0)$$
$$\text{Now, we have to use the distance formula}$$
$$\text{Distance=}$$ $$\sqrt{(a-a)^{2}+(b-0)^{2}+(c-0)^{2}}=\sqrt{b^{2}+c^{2}}$$ $$\textbf{[Using distance formula]}$$
$$\textbf{Hence, the correct option is B}$$
How many points $$(x, y)$$ with integral co-ordinates are there whose distance from $$(1, 2)$$ is two units?
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One
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Two
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Three
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Four
Explanation
$$PQ = 2\Rightarrow PQ^{2} = 4$$
$$(x - 1)^{2} + (y - 2)^{2} = 4$$
possibilities for integral coordinates
$$\Rightarrow (X - 1)^{2} + (y -2)^{2} = 4 + 0$$ or
$$= 0 + 4$$
Check $$(1) (x - 1)^{2} = 4\ (y - 2)^{2} = 0$$
$$\Rightarrow x - 1 = 2, -2\ ,y - 2 = 0$$
$$x = 3, -1\ ,y = 2$$
$$(3, 2)$$ and $$(-1, 2)$$
check $$(2) (x - 1)^{2} = 0\ ,(y - 2)^{2} = 4$$
$$x = 1\ ,y - 2 = 2, -2$$
$$x = 1\ ,y = 4, y = 0$$
$$(1, 4)$$ and $$(1, 0)$$
Total $$4$$ integral coordinates are possible.
The distance between the points $$\left( a\cos { { 48 }^{ o } } ,0 \right) $$ and $$\left( 0,a\cos { { 12 }^{ o } } \right) $$ is $$d$$ then $${ d }^{ 2 }{ a }^{ 2 }=$$
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$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +9 \right) }{ 4 } $$
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$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 4 } $$
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$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 8 } $$
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$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +1 \right) }{ 8 } $$
Explanation
Given $$A=(a\cos 48^o, 0)$$ & $$B=(0, a \cos 12^o)$$
$$\Rightarrow d^2=(a\cos 48-0)^2+(0-a\cos 12)^2$$
$$=a^2[\cos^248+\cos^212]$$
$$\cos^2\theta =\dfrac{1+\cos 2\theta}{2}$$ $$\Rightarrow a^2\left[\left(\dfrac{1+\cos 96}{2}\right)+\left(\dfrac{1+\cos 24}{2}\right)\right]$$
$$\Rightarrow a^2\left[\dfrac{2+\cos 96+\cos 24}{2}\right]$$
$$\left[\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos \left(\dfrac{A-B}{2}\right)\right]$$ $$\Rightarrow a^2\left[\dfrac{2+2\cos 60^o\cos 36^o}{2}\right]$$
$$\Rightarrow \dfrac{a^2}{2}\left[2+\not{2}\times \dfrac{1}{\not{2}}\cos 36^o\right]$$
$$d^2\Rightarrow \dfrac{a^2}{a}\left[2+\cos 36^o\right]$$
We know $$\cos 36=\dfrac{\sqrt{5}+1}{4}$$
$$\therefore d^2a^2=\dfrac{a^4}{2}\left[2+\dfrac{\sqrt{5}+1}{4}\right]=\dfrac{a^4}{2}\left[\dfrac{\sqrt{5}+9}{4}\right]$$
$$\Rightarrow$$ wrong options. [If d then]
$$ABC$$ is an isosceles triangle. If the coordinates of the base are $$B(1,3)$$ and $$C(-2,7)$$. The vertex $$A$$ can be
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$$(1,6)$$
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$$(-1/2,5)$$
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$$(-7,1/6)$$
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$$(5/6,6)$$
Explanation
ABC-Isosceles triangle.
Co-ordinates of base are $$B(1, 3)$$ and $$C(-2, 7)$$
vertex 'A' can be any point on perpendicular bisector of BC, but not 'D'.
Slope of BC$$=\dfrac{7-3}{-2-1}=-\dfrac{4}{3}$$
Slope of Line(L)$$=\dfrac{-1}{\left(-\dfrac{4}{3}\right)}=\dfrac{3}{4}$$
Mid-point of BC$$-$$D$$=\left(\dfrac{-1}{2}, 5\right)$$
$$L: 3x-4y=3\left(-\dfrac{1}{2}\right)-4(5)$$
$$3x-4y=-\dfrac{3}{2}-20$$
$$6x-8y=-3-40$$
$$6x-8y+43=0$$
As, only $$(5/6, 6)$$ satisfies 'L'.
So, 'A' can be $$\left(\dfrac{5}{6}, 6\right)$$.
The point $$P\left( x,y \right)$$ is equidistant from the points $$Q\left( c+d,d-c \right)$$ and $$R\left( c-d,c+d \right)$$ then
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$$cx=dy$$
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$$cx+dy=0$$
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$$dx=cy$$
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$$dx+cy=0$$
Explanation
Point $$P(x,y)$$ is the equidistant from $$Q(c+d,d-x)$$ and $$R(c-d,c+d)$$
$$\therefore PQ=PR$$
$${ PQ }^{ 2 }={ PR }^{ 2 }$$
$${ \left[ x-\left( c+d \right) \right] }^{ 2 }+{ \left[ y-\left( d-c \right) \right] }^{ 2 }={ \left[ x-\left( c-d \right) \right] }^{ 2 }+{ \left[ y-\left( c+d \right) \right] }^{ 2 }$$
$${ x }^{ 2 }+{ \left( c+d \right) }^{ 2 }-2.x\left( c+d \right) +{ y }^{ 2 }+{ \left( d-c \right) }^{ 2 }-2.y(d-c)={ x }^{ 2 }+{ \left( c-d \right) }^{ 2 }-2.x.\left( c-d \right) +{ y }^{ 2 }+{ \left( c+d \right) }^{ 2 }+2.y(c+d)$$
Or $$x(c+d)+y(d-c)=x(c-d)+y(c+d)$$
Or $$(c+d)(x-y)=(c-d)(x+y)$$
Or $$cx-cy+dc-dy=cx+cy-dx-dy$$
Or $$2dx=2cy$$
Or $$dx=cy$$
$$\therefore $$Answer $$dx=cy$$
The area of triangle formed by the lines $$x+y-3=0$$, $$x-3y+9=0$$ and $$3x-2y+1=0$$ is:
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$$\cfrac { 16 }{ 7 } $$ sq. units
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$$\cfrac { 10 }{ 7 } $$ sq. units
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$$4$$ sq. units
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$$9$$ sq. units
Explanation
$$L_1:x+y-3=0$$
$$L_2:x-3y+9=0$$
$$L_3:3x-2y+1=0$$
$$A$$ is intersection point of $$L_1$$ and $$L_2$$ which is $$(0,3)$$ on solving $$L_1$$ and $$L_2$$.
$$B$$ is intersection point of $$L_2$$ and $$L_3$$ which is $$(1,2)$$ on solving $$L_2$$ and $$L_3$$.
$$C$$ is intersection point of $$L_3$$ and $$L_1$$ which is $$(\dfrac{15}{7},\dfrac{26}{7})$$ on solving $$L_3$$ and $$L_1$$.
Now, area of $$\Delta ABC=\dfrac{1}{2}|(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+(x_3y_1-x_1y_3)|$$
$$=\dfrac{1}{2}|(0-3)+(\dfrac{26}{7}-\dfrac{30}{7})+(\dfrac{45}{7}-0)|$$
$$=\dfrac{1}{2}|-3-\dfrac{4}{7}+\dfrac{45}{7}|=\dfrac{1}{2}\times \dfrac{20}{7}$$
$$=\dfrac{10}{7} sq. units$$
A circle passes through the points $$(2, 3)$$ and $$(4, 5)$$. If its centre lies on the line, $$y - 4x + 3 = 0$$, then its radius is equal to
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$$\sqrt {5}$$
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$$1$$
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$$\sqrt {2}$$
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$$2$$
Explanation
Equation of the line through the given points is $$y-3 = x-2 \Rightarrow x-y+1=0$$.
Equation of the perpendicular line through the midpoint $$(3,4)$$ is $$x+y-7=0$$.
This intersects the given line at the center of the circle. So, the center of the circle is found to be $$(2,5)$$.
Clearly, the radius is then$$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} =$$ $$\sqrt{(2-2)^2+(3-5)^2} = 2$$
units.
So the answer is option D.
The angle between the pair of lines whose equation is $$4{ x }^{ 2 }+10xy+m{ y }^{ 2 }+5x+10y=0$$ is
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$$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 8 } \right) }$$
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$$\tan ^{ -1 }{ \dfrac {2 \sqrt { 25-4m } }{ m+4 } }$$
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$$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$
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$$\tan ^{ -1 }{ \dfrac { \sqrt { 25-4m } }{ m+4 } }$$
Explanation
The angle b/w 2 line represented by $$ax^2+2hxy+by^2+2gx+2fy+c=0$$
$$\tan\theta=\Bigg|\dfrac{2\sqrt{h^2-ab}}{a+b}\Bigg|$$
Given the equation of line i.e. $$4x^2+10xy+my^2+5x+10y=0$$
Compare above equation with standard equation
$$a=4,h=5,b=m,g=\dfrac{5}{2},f=5,c=0$$
Hence $$\theta=\tan^{-1}\dfrac{2\sqrt{25-4m}}{m+4}$$
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