Coordinate Geometry Test

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Coordinate Geometry Test - 47

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Question 1
1 / -0

$$\triangle ABC$$ has vertices $$A(-11, 4), B(-3, 8)$$, and $$C(3, -10)$$. The coordinates of the center of the circle circumscribed about $$\triangle ABC$$ are

A
$$(-2, -3)$$

B
$$(-3, -2)$$

C
$$(3, 2)$$

D
$$(2, 3)$$
Question 2
1 / -0

Define the $$x$$-coordinate of the midpoint between $$(-1,1)$$ and $$(4,1)$$ as $$m$$ and define the $$y$$-coordinate of the midpoint between $$(4,1)$$ and $$(4,3)$$ as $$n$$. What are the $$x$$- and $$y$$- coordinates of $$(m,n)$$?

A
$$\left( \cfrac { 5 }{ 2 } ,\cfrac { 3 }{ 2 } \right) $$

B
$$\left( \cfrac { 5 }{ 2 } ,2 \right) $$

C
$$\left( \cfrac { 3 }{ 2 } ,\cfrac { 3 }{ 2 } \right) $$

D
$$\left( \cfrac { 3 }{ 2 } ,2 \right) $$

E
$$(4,1)$$

Solution
Question 3
1 / -0

The vertices of $$\triangle ABC$$ have coordinates $$A(-7, 3), B(-1, 0)$$, and $$C(-2, 8)$$. The coordinates of the center of the circumscribed circle are

A
$$\left (-3\dfrac {1}{3}, 3\dfrac {2}{3}\right )$$

B
$$\left (-2\dfrac {5}{6}, 3\dfrac {5}{6}\right )$$

C
$$\left (-1 \dfrac {1}{2}, 4\right )$$

D
$$\left (4, 1\dfrac {1}{2}\right )$$

E
$$\left (-4\dfrac {1}{2}, 5\dfrac {1}{2}\right )$$

Solution
Question 4
1 / -0

The area of the triangle whose co-ordinates are $$(2012, 7), (2014, 7)$$ and $$(2014, a)$$ is $$1 \,sq$$ unit. The sum of possible values of $$a$$ is

A
$$6$$

B
$$8$$

C
$$14$$

D
$$16$$

Solution
Question 5
1 / -0

If $$A = (-3,4) , B =(-1,-2) , C=(5,6) D= (x,-4) $$ are the vertices of a quadrilateral such that area triangle $$ABD= 2 \times$$ (area of a triangle $$ACD$$), then $$x =$$

A
6

B
9

C
69

D
96

Solution
Question 6
1 / -0

If (5, 7, 10), (1, 9, 6) are the extremities of the hypotenuse of a right angled isosceles triangle, then the third vertex is

A
(4, 6, 6)

B
(4, 7, 7)

C
(7, 4, 7)

D
(4, 7, 7)

Solution
Question 7
1 / -0

A line L passes through the points $$ (1,1) $$and $$ (2,0) $$ and another line $$ L^{\prime} $$ passes through $$ \left(\frac{1}{2}, 0\right) $$ and perpendicular to L.Then the area of the triangle formed by the lines $$ L, L^{\prime} $$ and $$ y- $$ axis, is

A
$$15 / 8 $$

B
$$25 / 4 $$

C
$$25 / 8 $$

D
$$25 / 16 $$

Solution

$$\textbf{Hint: Product of slopes of two perpendicular lines is -1.}$$

$$\textbf{Step 1: Find the equation of two lines.}$$

$$\text{Using two points form, equation of line L is given by,}$$

$$\Rightarrow y-0=\dfrac{0-1}{2-1}(x-2)$$

$$\Rightarrow y=-(x-2)$$

$$\Rightarrow x+y=2.......(1)$$

$$\text{Slope of L = - 1}$$

$$\text{L and }$$ $$\text{L' are perpendicular to each other}$$

$$\text{Slope of L' =}$$ $$\dfrac{-1}{-1}=1$$

$$\text{Using point slope form, equation of line L' is given by,}$$

$$\Rightarrow y-0=1(x-\dfrac{1}{2})$$

$$\Rightarrow 2x-2y=1 ......(2)$$

$$\textbf{Step 2: Find the required area.}$$

$$\text{Let, L and L' intersect each other at B.}$$

$$\text{Solving eq(1) and eq(2) , we get,}$$

$$\Rightarrow A(x,y)=\left(\dfrac{5}{4},\dfrac{3}{4}\right)$$

$$\text{The intersection points of lines L and L' with y-axis are B(0,2) and C}$$$$\left(0,-\dfrac{1}{2}\right)$$ $$\text{respectively.}$$

$$\text{Using distance formula,}$$
$$AB=\sqrt{\left(0-\dfrac{5}{4}\right)^2+\left(2-\dfrac{3}{4}\right)^2}=\dfrac{5}{4}\sqrt2$$

$$\text{And,}$$
$$AC=\sqrt{\left(\dfrac{5}{4}-0\right)^2+\left(\dfrac{3}{4}+\dfrac{1}{2}\right)^2}=\dfrac{5}{4}\sqrt2$$

$$\text{Area of the triangle}$$ $$=\dfrac{1}{2}\times AB\times AC=\dfrac{1}{2}\times\dfrac{5}{4}\sqrt2\times\dfrac{5}{4}\sqrt2\ sq.unit $$

$$=\dfrac{25}{16}\ sq.unit$$

$$\textbf{Hence, the correct option is D.}$$
Question 8
1 / -0

The area of the triangle formed by the lines $$x=0;y=0$$ and $$x\sin { { 18 }^{ 0 } } +y\cos { { 36 }^{ 0 } } +1=0$$ is

A
1

B
2

C
3

D
4

Solution
Question 9
1 / -0

One diagonal of a square is along the line $$8x-15y=0$$ and one of its vertex is $$(1,2)$$. Then the equations of the sides of the square passing through this vertex, are

A
$$23x+7y=9,7x+23y=53$$

B
$$23x-7y+9=0,7x+23y+53=0$$

C
$$23x-7y-9=0,7x+23y-53=0$$

D
None of these

Solution
Question 10
1 / -0

if the points A(z),B(-z) and C(z+1) are vertices of an equilateral triangle then ---- test question

A
area of triangle is $$\left( \sqrt { 3 } /4 \right) $$sq. units

B
Re(z)=1/2

C
perimeter of the triangle is 3 units

D
Re(z)=1/4

Solution