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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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0%
25
2
sq. units
0%
23
2
sq. units
0%
19
2
sq. units
0%
None of the above
Explanation
Area of triangle
=
1
2
[
x
1
(
y
2
−
y
3
)
+
x
2
(
y
3
−
y
1
)
+
x
3
(
y
1
−
y
2
)
]
=
1
2
[
0
(
4
+
9
)
+
0
(
−
9
−
9
)
+
(
−
5
)
(
9
−
4
)
]
=
1
2
[
−
5
×
5
]
=
−
25
2
The area cannot be negative.
∴
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
0%
False
0%
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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0%
10 square units
0%
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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0%
94
square units
0%
96
square units
0%
97
square units
0%
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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0%
2.3 square units
0%
4.5 square units
0%
4.1 square units
0%
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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0%
1
0%
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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0%
(2, 6)
0%
(4, 14)
0%
(1, 2)
0%
(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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0%
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}
0%
3\sqrt {2}
0%
6 + 3\sqrt {2}
0%
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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0%
\sqrt {10}
0%
\sqrt {13}
0%
\sqrt {85}
0%
\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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0%
20
0%
23
0%
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)
.
Report Question
0%
6
0%
7
0%
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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0%
(1, 2)
0%
(1, 5)
0%
(3, 1)
0%
(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}
.
Report Question
0%
5.408
0%
5.454
0%
5.568
0%
5.590
0%
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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0%
1.118
0%
1.414
0%
1.803
0%
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
.
Report Question
0%
-10
0%
-9
0%
-5
0%
5
0%
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)
.
Report Question
0%
15.37
0%
17.75
0%
21.87
0%
25.61
0%
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)
.
Report Question
0%
\sqrt 3
0%
\sqrt {13}
0%
\sqrt {14}
0%
\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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0%
One
0%
Two
0%
Three
0%
Four
0%
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.
Report Question
0%
7
0%
4 \dfrac{1}{2}
0%
10
0%
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.
Report Question
0%
(1, 5)
0%
(3, 4)
0%
(2, 4)
0%
(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
Report Question
0%
-3
0%
-1
0%
3
0%
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:
Report Question
0%
15
0%
10
0%
7.5
0%
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
.
Report Question
0%
\sqrt{157}
0%
\sqrt{15}
0%
\sqrt{57}
0%
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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0%
\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right)
0%
\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right)
0%
\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right)
0%
\displaystyle (0,0)
Find the third vertex of an equilateral triangle whose two vertices are
(2,4)
and
(2,6)
.
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0%
(2+\sqrt{3},5)
or
(2-\sqrt{3},5)
0%
(2+\sqrt{3},5)
or
(2-\sqrt{7},5)
0%
(2+\sqrt{7},5)
or
(2-\sqrt{3},5)
0%
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)
.
Report Question
0%
2+\sqrt{2}
units
0%
2-\sqrt{2}
units
0%
3+\sqrt{2}
units
0%
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:
Report Question
0%
(4,5)
0%
(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:
Report Question
0%
(21,13)
0%
(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?
Report Question
0%
(3.5, -6.5)
0%
(3.5, 6.5)
0%
(-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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