MCQ Questions for 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

0%
$\dfrac{25}{2}$ sq. units
0%
$\dfrac{23}{2}$ sq. units
0%
$\dfrac{19}{2}$ sq. units
0%
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.

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.

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

$(10, 5)$ ?

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

$(3, -1)$ ?

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.

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

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

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

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

$BCEF$ .

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.

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

$(5, 7)$ .

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?

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

$\overline{GH}$ .

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

$BD$ .

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

$k$ .

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

$(0, 3)$ .

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

$(0,6)$ .

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?

0%
One
0%
Two
0%
Three
0%
Four
0%
More than four

Find the value of

$x$ , so that the three points,

$(2, 7), (6, 1), (x, 0)$ are collinear.

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

$\Longrightarrow \dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] =0\\ =\dfrac { 1 }{ 2 } \left[ 2(1-0)+6(0-7)+x(7-1) \right] =0\\ \dfrac { 1 }{ 2 } \left[ 2-42+6x \right] =0\\ -40+6x=0\Longrightarrow x=\dfrac { 40 }{ 6 } =6\dfrac { 2 }{ 3 } =6.666$

$QR = 5\ units$ , identify the co-ordinates of Q.

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

$a - b$

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:

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

$C$ .

0%
$\sqrt{157}$
0%
$\sqrt{15}$
0%
$\sqrt{57}$
0%
None of these

Explanation

'D' is mid point

$\therefore D=\left(\dfrac{2-4}{2},\dfrac{2+4}{2}\right)$

$D=(-1,3)$

$\therefore CD=\sqrt{[5-(-1)]^{2}+(-8-3)^{2}}$

$=\sqrt{6^{2}+11^{2}}=\sqrt{157}$

1085343_516769_ans_b02ecf4aa7f64d8db34028125fbb7a03.png

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?

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

$(2,6)$ .

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

$(0,1)$ .

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

1085320_516778_ans_def933897387436d889bae9caa569f67.png

In the diagram,

$PQR$ , is an isosceles triangle and

$QR=5$ units.
The coordinates of

$Q$ are:

0%
$(4,5)$
0%
$(3,4)$
0%
$(2,4)$
0%
$(1,4)$

Explanation

$PQ+QR$

Let coordinates of

$'Q'$ be

$(x,y)$

$\sqrt{(x+2)^2+y^2}=\sqrt{(x-4)^2+y^2}$

Squaring both sides

$(x+2)^2+y^2=(x-4)^2+y^2$

$\Rightarrow x^2+4+4x=x^2+16-8x$

$\Rightarrow 12x=12\Rightarrow \boxed{x=1}$

$QR=5$ units

$\Rightarrow \sqrt{(x-4)^2+y^2=5}$

$\Rightarrow (1-4)^2+y^2=25$

$\Rightarrow y^2=16\Rightarrow y=\pm 4$

$Q$ is above x-axis

$y=4$

$Q=(1,4)$

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:

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?

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,

$x=\dfrac{-3}{2}\,and\, y=1.\xi$

2114965_529609_ans_065389d2a11a4ea6b4a304b23d00a972.png

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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