Straight Lines Test 33

Result

Straight Lines Test 33

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Question 1
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
Question 2
1 / -0

Area of a triangle whose vertices are $$(a\cos \theta ,b\sin \theta ),(-a\sin \theta ,b\cos \theta)$$ and $$(-a\cos \theta ,-b\sin \theta )$$ is-

A
$$ab\sin \theta \cos \theta $$

B
$$a\cos \theta \sin \theta$$

C
$$\frac 12ab$$

D
ab

Solution
Question 3
1 / -0

Two sides of a rhombus are along the lines, $$x-y+1=0$$ and $$7x-y-5=0$$. If its diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus?

A
(-3, -8)

B
$$(\frac 13,\frac 83)$$

C
$$(-\frac {10}3,-\frac 73)$$

D
(-3, -9)

Solution
Question 4
1 / -0

If three lines $$x-3y=p,ax+2y=q$$ and $$ax+y=r$$ form a right angled triangle, then

A
$${ a }^{ 2 }-9a+18=0$$

B
$${ a }^{ 2 }-6a-12=0$$

C
$${ a }^{ 2 }-6a+9=0$$

D
$${ a }^{ 2 }-9a+12=0$$

Solution
Question 5
1 / -0

If the tangent at $$\theta =\frac { \pi }{ 4 } $$ to the curve $$x=a\cos ^{ 3 }{ \theta } ,y=a\sin ^{ 3 }{ \theta } $$ meets the x and y axes in A and B then the area of the triangle OAB is

A
$$\frac { { a }^{ 2 } }{ 4 } sq.units$$

B
$$\frac { { a }^{ 2 } }{ 2 } sq.units$$

C
$$\frac { { 3a }^{ 2 } }{ 4 } sq.units$$

D
$$\frac { { 5a }^{ 2 } }{ 4 } sq.units$$

Solution
Question 6
1 / -0

Area of a triangle whose vertices are $$ (a \cos \theta, b \sin \theta),(-a \sin \theta, b \cos \theta) $$ and$$ (-a \cos \theta,-b \sin \theta) $$ is $$ - $$

A
a b sin $$ \theta \cos \theta $$

B
$$ a \cos \theta \sin \theta $$

C
$$ \frac{1}{2} a b $$

D
$$ a b $$

Solution
Question 7
1 / -0

If the area of triangle formed by the points (2a , b) (a + b , 2b + a) and (2b , 2a) be $$\lambda$$ , then the area of the triangle whose vertices are (a + b , a b) , (3b a , b + 3a) and (3a b , 3b a) will be

A
$$\dfrac{3}{2}\lambda$$

B
$$3\lambda$$

C
$$4\lambda$$

D
none of these

Solution
Question 8
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 9
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 10
1 / -0

If $$P , Q$$ are two points on the line $$3 x + 4 y + 15 = 0$$ such that $$O P = O Q = 9$$ then the area of $$\Delta O P Q$$ is

A
6$$\sqrt { 2 }$$

B
9$$\sqrt { 2 }$$

C
12$$\sqrt { 2 }$$

D
18$$\sqrt { 2 }$$

Solution