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CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Maths
Straight Lines
Quiz 7
The area of triangle with vertices $$A(0,\,9),\,B(0,\,4)$$ and $$C(-5,\,-9)$$ is
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$$\dfrac{25}{2}$$ sq. units
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$$\dfrac{23}{2}$$
sq. units
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$$\dfrac{19}{2}$$
sq. units
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None of the above
Explanation
Area of triangle $$=\dfrac{1}{2}\begin{bmatrix}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\end{bmatrix}$$
$$=\dfrac{1}{2}\begin{bmatrix}0(4+9)+0(-9-9)+(-5)(9-4)\end{bmatrix}$$
$$=\dfrac{1}{2}\begin{bmatrix}-5\times5\end{bmatrix}$$
$$=-\dfrac{25}{2}$$
The area cannot be negative.
$$\therefore$$ It must be $$\dfrac{25}{2}$$ sq. units
So, option A is correct.
Are the points $$(5,\,5),\,(8,\,2)$$ and $$(3,\,-4)$$ are vertices of right angled triangle.
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0%
True
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False
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Cant say
0%
None
Explanation
Let the points $$A(5,\,5),\,B(8,\,2)$$ and $$C(3,\,-4)$$ are vertices of triangle.
$$AB^2=(5-8)^2+(5-2)^2=9+9=18$$
$$BC^2=(8-3)^2+(2+4)^2=25+36=61$$
$$CA^2=(3-5)^2+(-4-5)^2=4+81=85$$
$$AB^2+AC^2\neq BC^2$$
The vertices are not points of a right angled triangle.
$$(9, 2), (5, -1) $$ and $$ (7, -5)$$ are the vertices of the triangle. Find its area.
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10 square units
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11 square units
0%
12 square units
0%
13 square units
Explanation
Area of triangle with vertices $$(x_1,y_1)$$,
$$(x_2,y_2)$$ and
$$(x_3,y_3)$$ is:
Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 9$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = -1$$, $$x_{3} = 7$$ and $$y_{3} = -5$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[9(-1 + 5) + 5(-5 - 2) + 7(2 + 1)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 35 + 21]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 22\right|$$
$$=$$ $$\left|11 \right|$$
Area always in absolute value.
So, area of the triangle $$= 11$$ square units.
What is the area of the triangle whose vertices are: $$(-3, 15), (6, -7) $$ and $$(10, 5)$$?
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$$94$$ square units
0%
$$96$$ square units
0%
$$97$$ square units
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$$98$$ square units
Explanation
Area of triangle with vertices $$(x_1,y_1)$$,
$$(x_2,y_2)$$ and
$$(x_3,y_3)$$
is $$A=\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
Here $$x_{1} = -3$$, $$y_{1} = 15$$, $$x_{2} = 6$$, $$y_{2} = -7$$, $$x_{3} = 10$$ and $$y_{3} = 5$$
Substituting the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[-3(-7 - 5) + 6(5 - 15) + 10(15 + 7)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 60 + 220]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 196\right|$$
$$=$$ $$\left|98 \right|$$
Area is always in absolute value.
So, area of the triangle $$= 98$$ square units.
What is the area of the triangle for the following points $$(6, 2), (5, 4)$$ and $$(3, -1)$$?
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2.3 square units
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4.5 square units
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4.1 square units
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3.6 square units
Explanation
Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 6$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = -1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[6(4 + 1) + 5(-1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[30 - 15 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 9\right|$$
$$=$$ $$\left|4.5 \right|$$
Area always in absolute value.
So, area of the triangle $$= 4.5$$ square units.
The area of the triangle whose vertices are $$(0, 1), (1, 4)$$ and $$(1, 2)$$ is ___ square units.
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1
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2
0%
3
0%
4
Explanation
Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 0$$, $$y_{1} = 1$$, $$x_{2} = 1$$, $$y_{2} = 4$$, $$x_{3} = 1$$ and $$y_{3} = 2$$
Substitute the values, we get
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[0(4 - 2) + 1(2 - 1) + 1(1 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[0 + 1 - 3]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -2\right|$$
$$=$$ $$\left|- 1 \right|$$
Area always in absolute value.
So, area of the triangle is $$ 1$$ square units.
The point on the line $$4x - y - 2 = 0$$ which is equidistant from the points $$(-5, 6)$$ and $$(3, 2)$$ is
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$$(2, 6)$$
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$$(4, 14)$$
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$$(1, 2)$$
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$$(3, 10)$$
Explanation
Let the point on line $$4x-y-2=0$$ be $$P(x,y)$$.
Let $$A \equiv (-5,6)$$ and $$B \equiv (3,2)$$
$$4x-y-2=0$$ ....(1)
Point P is equidistant from points A and B ....Given
$$\therefore AP = PB$$
By distance formula,
$$(x+5)^2 + (y-6)^2 = (x-3)^2 + (y-2)^2$$
$$x^2+10x+25 + y^2 - 12y + 36 = x^2 - 6x +9 + y^2 - 4y + 4$$
$$16x - 8y + 48=0$$
$$4x-2y+12=0$$ ....(2)
Subtract eq(2) from eq (1), we get
$$y=14$$
Substitute in eq(1), we get
$$x=4$$
So, the point on the line is $$(4, 14)$$.
The equations of the lines through $$(1,\,1)$$ and making angles of $$45^{\circ}$$ with the line $$x+y=0$$ are
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$$x-1=0,\,x-y=0$$
0%
$$x-y=0,\,y-1=0$$
0%
$$x+y-2=0,\,y-1=0$$
0%
$$x-1=0,\,y-1=0$$
Explanation
$$m=1,\,y-1=\dfrac{m\pm\,tan\,45^{\circ}}{1+\mp\,m\,tan\,45^{\circ}}(x-1)$$,
$$\Rightarrow\,y-1=\dfrac{(-1)\pm1}{1\pm1}(x-1)$$
$$\Rightarrow\,y=1,\,x=1$$
The perimeter of the triangle with vertices $$(1,3), (1,7)$$ and $$(4,4)$$ is
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0%
$$3 + \sqrt {2}$$
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$$3\sqrt {2}$$
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$$6 + 3\sqrt {2}$$
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$$9 + \sqrt {2}$$
Explanation
Assume $$A(1, 3), B(1, 7), C(4, 4)$$
Formula of distance $$=$$ $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}$$
Distance of $$AB =$$ $$\sqrt{(1-1)^2+(7-3)^2}$$
$$=$$ $$\sqrt{0+4^2}$$
$$= 4$$
Likewise calculate the distance for $$BC$$ and $$CA$$.
Distance of $$BC =$$ $$\sqrt{(4-1)^2+(4-7)^2} = 3\sqrt{2}$$
Distance of $$CA =$$ $$\sqrt{(1-4)^2+(3-4)^2} = 2$$
Perimeter $$=$$ distance $$AB$$ $$+$$ distance $$BC$$ $$+$$ distance $$CA$$
$$=$$ $$4+ 3\sqrt{2}+2$$
$$=$$ $$6+3\sqrt{2}$$
From the above figure, calculate the length of $$AG$$, if point $$G$$ is the center of rectangle $$BCEF$$.
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$$\sqrt {10}$$
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$$\sqrt {13}$$
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$$\sqrt {85}$$
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$$\sqrt {97}$$
0%
$$11$$
Explanation
$$G$$ will be the midpoint of $$BE$$.
Hence, $$G=\dfrac{6+12}{2},\dfrac{4+0}{2}=9,2$$
Hence, $$AG=\sqrt{(9-0)^{2}+(2-0)^{2}}$$
$$=\sqrt{81+4}$$
$$=\sqrt{85}$$
In figure, if the midpoints of segments $$\overline{GH}, \overline{JK}$$, and $$\overline{LM}$$ are connected, calculate the area of the resulting triangle.
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$$20$$
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$$23$$
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$$26$$
0%
$$33$$
Explanation
Midpoint of $$GH$$ is $$\left (0, \dfrac {15}{4}\right)$$
Midpoint of $$JK$$ is $$(-5,-2)$$
Midpoint of $$ML$$ is $$(3,-2)$$
If the coordinates of triangle is $$(a,b) , (c,d) , (e,f)$$ then the area formed by the coordinates of triangle is $$\dfrac{1}{2} \times |a(d-f) + c(f-b) + e(b-d)|$$
Therefore, the area enclosed by those midpoints will be
$$ \dfrac12 \times \left |0(-2+2) -5\left (-2-\dfrac {15}{4}\right) +3\left (\dfrac {15}{4}+2\right)\right| $$
$$= \dfrac {1}{2} \times |0+46| $$
$$= 23$$
Calculate the area of a triangle with vertices $$(1, 1), (3, 1)$$ and $$(5, 7)$$.
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$$6$$
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$$7$$
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$$9$$
0%
$$10$$
Explanation
We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of triangle
$$=\dfrac 12 \left|1(21-5)-1(7-5)+1(1-3) \right|$$
$$=\dfrac 12 \left| 16-2-2\right| = 6$$
Hence, area of the triangle with the given coordinates is $$6$$.
In the XY-coordinate plane, point P is a distance of $$4$$ from the point $$(1, 1)$$. Which of the following could be P?
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$$(1, 2)$$
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$$(1, 5)$$
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$$(3, 1)$$
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$$(4, 1)$$
Explanation
It may help a lot to make a figure so you can visualize what is going on:
The figure above shows the point $$(1, 1)$$ along with the five possible answers for point P.
Only answer B, point $$(1, 5)$$, is a distance of $$4$$ from point $$(1, 1)$$: the y value increases by $$4$$ units to go from $$(1, 1)$$ to $$(1, 5)$$, and the x-value doesn't change.
In the figure, calculate the distance from the midpoint to $$\overline{EF}$$ to the midpoint of $$\overline{GH}$$.
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$$5.408$$
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$$5.454$$
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$$5.568$$
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$$5.590$$
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$$5.612$$
Explanation
Mid point of point $$E (-3,3)$$ and $$F (-2,-4)$$ is $$\left ( \dfrac{-3-2}{2} \right ),\left ( \dfrac{3-4}{2} \right )=\dfrac{-5}{2},\dfrac{-1}{2}=-2.5,-0.5$$ ....(hint: using mid-point formula)
And mid point of $$G (1,-2)$$ and $$H ( 5,3) $$ is $$\left ( \dfrac{1+5}{2} \right ),\left ( \dfrac{3-2}{2} \right )=\dfrac{6}{2},\dfrac{1}{2}= 3,0.5$$
Then distance between mid point $$EF$$ to mid point $$GH =$$ $$\sqrt{(-2.5-3)^{2}+(-0.5-0.5)^{2}}=\sqrt{(5.5)^{2}+(-1)^{2}}=\sqrt{30.25+1}=\sqrt{31.25}=5.590$$
In Figure 1, calculate the distance from the midpoint of segment $$AC$$ to the midpoint of segment $$BD$$.
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1.118
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1.414
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1.803
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2.236
0%
2.828
Explanation
Midpoint of $$AC$$ is $$\left (0 , \dfrac {3}{2}\right)$$.......[hint: using mid-point formula i.e, $$ x = (x_1 + x_2)/2\ ,\ y = (y_1 + y_2)/2$$ ]
Midpoint of $$BD$$ is $$(-1,1/2)$$
Distance between those two midpoints is $$\sqrt { { (1) }^{ 2 }+{ \left (\dfrac {3}{2} - \dfrac {1}{2}\right) }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } = 1.414$$
The area of the triangle with coordinates $$(1, 2), (5, 5)$$ and $$(k, 2)$$ is $$15$$ square units. Calculate a possible value for $$k$$.
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$$-10$$
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$$-9$$
0%
$$-5$$
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$$5$$
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$$6$$
Explanation
Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
Area of triangle is $$15 = \dfrac {1}{2} \times |1(5-2) + 5(2-2) + k(2-5)| $$
$$ \Rightarrow \dfrac {1}{2} \times |3+0-3k| = \dfrac {1}{2} \times |3-3k| = 15$$
$$\Rightarrow |3-3k|=30$$ => $$|1-k| = 10$$
$$\Rightarrow 1-k = 10$$ and $$1-k=-10$$
$$\Rightarrow k=-9$$ and $$k=11$$
Find the area of a triangle whose vertices are $$(0, 6\sqrt {3}), (\sqrt {35}, 7)$$, and $$(0, 3)$$.
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$$15.37$$
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$$17.75$$
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$$21.87$$
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$$25.61$$
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$$39.61$$
Explanation
Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle whose vertices are $$(0,6\sqrt{3}),(\sqrt{35},7) ,(0,3)$$
$$=\dfrac{1}{2}\left [ 0(7-3)+\sqrt{35}(3-6\sqrt{3})+0(6\sqrt{3}-7)) \right ]$$
$$=\dfrac{1}{2}\left [ \sqrt{35}(3-6\sqrt{3}) \right ]$$
$$=\dfrac{1}{2}\left [ 5.91(3-6\times 1.73) \right ]$$
$$=\dfrac{1}{2}\times 5.91\times 7.40$$
$$=\dfrac{1}{2}\times 43.74$$
$$=21.87$$
Find the distance between the points $$(2,3)$$ and $$(0,6)$$.
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$$\sqrt 3$$
0%
$$\sqrt {13}$$
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$$\sqrt {14}$$
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$$\sqrt 5$$
Explanation
We’ll use
the distance formula
which can be used to calculate the distance d between any two points in a coordinate plane.
This formula is given as follows:
$$d = \sqrt {(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
$$d =\sqrt {(0-2 )^2 + (6-3)^2}$$
$$d = \sqrt{(-2 )^2 + (3)^2}$$
$$d = \sqrt{4 + 9}$$
$$d=\sqrt {13}$$.
In the XY-coordinate plane, how many points are at the distance
of 4 units from the origin?
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One
0%
Two
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Three
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Four
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More than four
Explanation
There are an infinite number of points that are a distance of 4 from the origin.
That’s a circle centered on the origin with a radius of 4.
Then answer is option (E) more then four
Find the value of $$x$$, so that the three points, $$(2, 7), (6, 1), (x, 0)$$ are collinear.
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$$7$$
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$$4 \dfrac{1}{2}$$
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$$10$$
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$$6 \dfrac{2}{3}$$
Explanation
The given points are $$(2,7) , (6.1)$$ and $$(x,0)$$
of the points are collinear , they will lie on the same line ,i.e, they will not form triangle .
Area of $$\triangle$$ABC = 0
$$\Longrightarrow \dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] =0\\ =\dfrac { 1 }{ 2 } \left[ 2(1-0)+6(0-7)+x(7-1) \right] =0\\ \dfrac { 1 }{ 2 } \left[ 2-42+6x \right] =0\\ -40+6x=0\Longrightarrow x=\dfrac { 40 }{ 6 } =6\dfrac { 2 }{ 3 } =6.666$$
If $$QR = 5\ units$$, identify the co-ordinates of Q.
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$$(1, 5)$$
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$$(3, 4)$$
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$$(2, 4)$$
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$$(1, 4)$$
Explanation
By using distance formula
$$P{ Q }^{ 2 }=R{ Q }^{ 2 }$$
Let the point Q be taken as (x,y)
$$(x+2)^{ 2 }+(y-0)^{ 2 }=(x-4)^{ 2 }+(y-0)^{ 2 }\\ { x }^{ 2 }+4x+4+{ y }^{ 2 }={ x }^{ 2 }-8x+16+{ y }^{ 2 }\\ 4x+4=-8x+16\\ 12x=12\Rightarrow x=1$$
Distance of QR is given as 5 units
$$\therefore y=4$$
$$\rightarrow Q=(1,4)$$
In the rectangle shown, find the value of $$a - b$$
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$$-3$$
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$$-1$$
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$$3$$
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$$1$$
Explanation
From the figure,
$$A{ D }^{ 2 }-A{ B }^{ 2 }$$ i.e, distance is same
$$(9-5)^{ 2 }+(2-5)^{ 2 }=(9-15)^{ 2 }+(2-b)^{ 2 }\\ 16+9=36+4-4b+{ b }^{ 2 }\\ -15=-4b+b^{ 2 }\\ \Rightarrow { b }^{ 2 }-4b+15=0\\ \therefore b=12.2\\ D{ C }^{ 2 }=B{ C }^{ 2 }\\ (5-a)^{ 2 }+(5-13)^{ 2 }=(a-15)^{ 2 }+(13-2)^{ 2 }\\ 25-10a+{ a }^{ 2 }+64={ a }^{ 2 }-30a+225+121\\ 20a=225+121-64-25\\ \therefore a=13\\ \therefore a-b=1\\ $$
In fig., the area of triangle ABC (in sq. units) is:
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$$15$$
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$$10$$
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$$7.5$$
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$$2.5$$
Explanation
Given: Coordinates of Point $$A (1,3) ,B (-1,0)$$ and $$C (4,0)$$
Construction: Drop a perpendicular from $$A$$ on $$x-$$ axis, which meets x-axis at $$D\equiv(1,0)$$
Now in $$\Delta ADC, AD = 3, DC = 3$$
Area of $$\Delta ADC = \dfrac12\times DC\times AD$$
$$= \dfrac12\times3\times3 = \dfrac92 \ cm^2$$
Now in $$\Delta ADB, AD = 3, DB = 2$$
Area of $$\Delta ADB = \dfrac12\times DB\times AD$$
$$= \dfrac12\times2\times3 = 3 \ cm^2$$
Area of $$\Delta ABC =$$ Ara of $$\Delta ADC + $$ Area of $$\Delta ABD$$
$$ = \dfrac92 + 3 = \dfrac{15}2 = 7.5\ cm^2$$
The vertices of a triangle are $$A(2,2), B(-4,4), C(5,-8)$$. Then, find the length of the median through $$C$$.
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$$\sqrt{157}$$
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$$\sqrt{15}$$
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$$\sqrt{57}$$
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None of these
Explanation
'D' is mid point
$$\therefore D=\left(\dfrac{2-4}{2},\dfrac{2+4}{2}\right)$$
$$D=(-1,3)$$
$$\therefore CD=\sqrt{[5-(-1)]^{2}+(-8-3)^{2}}$$
$$=\sqrt{6^{2}+11^{2}}=\sqrt{157}$$
The area of a triangle is 5 and its two vertices are A(2, 1) and B(3, -2). The third vertex lies on $$\displaystyle y=x+3$$. What is the third vertex?
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$$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) $$
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$$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) $$
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$$\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right) $$
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$$\displaystyle (0,0)$$
Find the third vertex of an equilateral triangle whose two vertices are $$(2,4)$$ and $$(2,6)$$.
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$$(2+\sqrt{3},5)$$ or $$(2-\sqrt{3},5)$$
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$$(2+\sqrt{3},5)$$ or $$(2-\sqrt{7},5)$$
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$$(2+\sqrt{7},5)$$ or $$(2-\sqrt{3},5)$$
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None of these
Explanation
Length of side of equilateral triangle is $$AB=BC=CA=2$$.
Let's take the third vertex to be $$C(x,y)$$
Then $$AC=BC=>AC^2=BC^2$$.
By distance formula:
$$\sqrt((x-2)^2+(y-4)^2)=\sqrt((x-2)^2+(y-6)^2)$$
$$=>-4y+4=-12y+36$$
$$=>y=5$$
Now $$AC^2=4=((x-2)^2+(y-4)^2)$$.
Putiing $$y=5$$:
$$=>x^2-4x+1=0$$.
So, solving the quadratic we get:
$$x=2+\sqrt3$$ or $$2-\sqrt3$$.
So, required point is $$(2+\sqrt3,5)$$ and $$(2-\sqrt3,5)$$.
Find the perimeter of the triangle formed by $$(0,0),(1,0)$$ and $$(0,1)$$.
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$$2+\sqrt{2}$$ units
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$$2-\sqrt{2}$$ units
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$$3+\sqrt{2}$$ units
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None of these
Explanation
$$AB=\sqrt{OA^{2}+OB^{2}}$$
$$\Rightarrow AB=\sqrt{2}$$
$$\therefore S=1+1+\sqrt{2}=2+\sqrt{2}$$ units
In the diagram, $$PQR$$, is an isosceles triangle and $$QR=5$$ units.
The coordinates of $$Q$$ are:
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$$(4,5)$$
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$$(3,4)$$
0%
$$(2,4)$$
0%
$$(1,4)$$
Explanation
As $$PQ+QR$$
Let coordinates of $$'Q'$$ be $$(x,y)$$
$$\sqrt{(x+2)^2+y^2}=\sqrt{(x-4)^2+y^2}$$
Squaring both sides
$$(x+2)^2+y^2=(x-4)^2+y^2$$
$$\Rightarrow x^2+4+4x=x^2+16-8x$$
$$\Rightarrow 12x=12\Rightarrow \boxed{x=1}$$
As $$QR=5$$ units
$$\Rightarrow \sqrt{(x-4)^2+y^2=5}$$
$$\Rightarrow (1-4)^2+y^2=25$$
$$\Rightarrow y^2=16\Rightarrow y=\pm 4$$
As $$Q$$ is above x-axis
So $$y=4$$
$$Q=(1,4)$$
In the diagram $$MN$$ is a straight line on a Cartesian plane. The coordinates of $$N$$ are $$(12,13)$$ and $${MN}^{2}=9$$ units. The coordinates of $$M$$ are:
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$$(21,13)$$
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$$(12,22)$$
0%
$$(12,4)$$
0%
$$(3,13)$$
Explanation
(As MN is parallel to 'x' axis \therefore 'y' coordinate not change)
Let $$M=(x,13)$$
$$\sqrt[6]{(x-12)^{2}+(13-13)^{2}}=9$$
$$\Rightarrow (x-12)^{2}=9^{2}\Rightarrow x+12=\pm 9$$
$$\Rightarrow x=3,21$$ (As 'M' is left of N )
$$\therefore x=3$$
Two vertices of a triangle are (2, 1) and (3, -2). Its third vertex is (x, y) such that $$\displaystyle y=x+3$$. If its area is 5 sq. units, what are the co-ordinates of the third vertex?
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(3.5, -6.5)
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(3.5, 6.5)
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(-1.5, -1.5)
0%
(1.5, -1.5)
Explanation
Given third vertex such that,
$$y=(x+3)$$
since, Area of triangle $$ABC=55q$$ units
$$\pm \dfrac{1}{2}{x(1+2)+2(-2-y)+3(y-1)}=\xi$$
$$\Rightarrow \pm \dfrac{1}{2}{x+2x-4-2y+3y-3}=\xi$$
$$\Rightarrow {3x+y-7}=\pm10$$
$$\Rightarrow 3x+y-17=0$$ ------------(1)
and $$3x+y+3=0$$ ------------(2)
Given that $$A(x,y)$$ lies any $$=(x+3)$$ ------------(3)
from equation (L) and (3),
$$x=\dfrac{7}{2},y=\dfrac{13}{2}$$
$$\Rightarrow \boxed{x=3.\xi\,and \, y=6.\xi}$$
from equation (2) and (3) we get,
$$x=\dfrac{-3}{2}\,and\, y=1.\xi $$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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