Straight Lines Test 23

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Straight Lines Test 23

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of $$x$$ such that $$AB=BC$$ where the coordinates of A, B and C are $$(2,1)$$, $$(x,0)$$ and $$(-2,-1)$$ respectively.

A
$$1$$

B
$$2$$

C
$$-2$$

D
$$0$$

Solution

Distance between two points $$=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$$
Given $$ AB=BC$$
$$\sqrt{(2-x)^{2}+(1)^{2}}=\sqrt{(x+2)^{2}+(1)^{2}}$$
Squaring both sides
$$(2-x)^{2}+(1)=(x+2)^{2}+(1)$$
$$4+x^{2}-4x=x^{2}+4+4x$$
$$x=0$$
Question 2
1 / -0

The condition that the slope of one of the lines represented by $$ax^{2} + 2hxy + by^{2} = 0$$ is twice that of the other is

A
$$h^{2} =ab$$

B
$$2h^{2} = 3ab$$

C
$$8h^{2} = 9ab$$

D
$$4h^{2} = 9ab$$

Solution

We know that ,
for $$ax^2+2hxy+by^2=0$$

$$m_1+m_2=\dfrac{-2h}{b}\\m_1m_2=\dfrac ab$$

Let the Roots be $$m,2m$$

$$2m*m=\dfrac ab\\2m+m=3m=\dfrac{-2h}{b}\Rightarrow m=\dfrac{-2h}{3b}$$

So,
$$\dfrac ab=2m^2\Rightarrow \dfrac ab=2(\dfrac{-2h}{3b})^2\Rightarrow \dfrac ab=\dfrac{8h^2}{9b^2}\Rightarrow 8h^2=9ab$$
Question 3
1 / -0

If $$(3,2)$$ and $$(-3,2)$$ are two vertices of an equilateral triangle which contains within it the origin, what are the coordinates of the third vertex ?

A
$$(2,2\sqrt{2})$$

B
$$(1,2\sqrt{2})$$

C
$$(0,-3\sqrt{3})$$

D
$$None\ of\ these$$

Solution

Let $$\Delta ABC$$ be the equilateral triangle, where $$AB=BC=AC$$.

Vertices of this triangle be,

$$A\left( 3,2 \right),B\left( -3,2 \right)$$ and $$C\left( x,y \right)$$.

The mid-point of the side $$AB$$ is $$M\left( 0,2 \right)$$.

$$ AB=\sqrt{{{\left( 3+3 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}} $$

$$ AB=6\ units $$

Therefore,

$$\Rightarrow AB=BC=AC=6\ units$$

Also,

$$\Rightarrow AM=3\ units$$

As two vertices are $$A\left( 3,2 \right)$$ and $$B\left( -3,2 \right)$$, so third vertex will be at $$y-$$axis.

Therefore,

$$\Rightarrow C\left( 0,y \right)$$

It will be located below the origin as the triangle contains the origin.

Now,

$$ {{y}^{2}}={{\left( AC \right)}^{2}}-{{\left( AM \right)}^{2}} $$

$$ {{y}^{2}}=36-9=27 $$

$$ y=3\sqrt{3}\ units $$

Therefore,

$$\Rightarrow C\left( 0,-3\sqrt{3} \right)$$

Hence, this is the required point.
Question 4
1 / -0

If $$A(a,a), B(-a,-a)$$ are two vertices of an equilateral triangle, then its third vertex is

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

B
$$(-a\sqrt{3},a\sqrt{3})$$

C
$$(a\sqrt{3}, a\sqrt{3})$$

D
$$(-a\sqrt{3},-a\sqrt{3})$$

Solution

The two vertices of an equilateral triangle is $$A\left( {a,a} \right)$$ and $$B\left( { - a, - a} \right)$$.

Let the third vertex be $$C\left( {x,y} \right)$$.

Since the triangle is an equilateral triangle, then,

$$AC = AB = BC$$

$$\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - a} \right)}^2}} = \sqrt {{{\left( { - a - a} \right)}^2} + {{\left( { - a - a} \right)}^2}} = \sqrt {{{\left( {x + a} \right)}^2} + {{\left( {y + a} \right)}^2}} $$

$$\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - a} \right)}^2}} = \sqrt {{{\left( { - 2a} \right)}^2} + {{\left( { - 2a} \right)}^2}} = \sqrt {{{\left( {x + a} \right)}^2} + {{\left( {y + a} \right)}^2}} $$

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

$$\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - a} \right)}^2}} = \sqrt {8{a^2}} = \sqrt {{{\left( {x + a} \right)}^2} + {{\left( {y + a} \right)}^2}} $$

Taking first two equations,

$$\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - a} \right)}^2}} = \sqrt {8{a^2}} $$

Squaring both sides,

$${\left( {x - a} \right)^2} + {\left( {y - a} \right)^2} = 8{a^2}$$

$${x^2} + {a^2} - 2ax + {y^2} + {a^2} - 2ay = 8{a^2}$$

$${x^2} + {y^2} - 2ax - 2ay = 6{a^2}$$ (1)

Taking last two equations,

$$\sqrt {8{a^2}} = \sqrt {{{\left( {x + a} \right)}^2} + {{\left( {y + a} \right)}^2}} $$

Squaring both sides,

$$8{a^2} = {\left( {x + a} \right)^2} + {\left( {y + a} \right)^2}$$

$$8{a^2} = {x^2} + {a^2} + 2ax + {y^2} + {a^2} + 2ay$$

$${x^2} + {y^2} + 2ax + 2ay = 6{a^2}$$ (2)

From equation (1) and (2),

$${x^2} + {y^2} - 2ax - 2ay = {x^2} + {y^2} + 2ax + 2ay$$

$$ - 4ax = 4ay$$

$$x = - y$$

Substituting the value of $$x$$ in equation (1),

$${\left( { - y} \right)^2} + {y^2} - 2a\left( { - y} \right) - 2ay = 6{a^2}$$

$${y^2} + {y^2} + 2ay - 2ay = 6{a^2}$$

$$2{y^2} = 6{a^2}$$

$${y^2} = 3{a^2}$$

$$y = \sqrt 3 a$$

Substituting the value of $$y$$ in $$x = - y$$, then,

$$x = - \sqrt 3 a$$

Therefore, the third vertex is $$\left( { - \sqrt 3 a,\sqrt 3 a} \right)$$.
Question 5
1 / -0

The coordinates of the point which divides the line segment joining points $$A(0 , 0)$$ and $$B(9 , 12)$$ in the ration $$1 : 2$$ are

A
$$(-3 , 4)$$

B
$$(3 , 4)$$

C
$$(3 , -4)$$

D
none of these

Solution

The coordinates of the point which divides the line segment joining$$ (x_{1},y_{1})$$ and $$ (x_{2},y_{2})$$ in the ratio $$m:n$$ is given by
$$\dfrac{mx_{2}+nx_{1}}{m+n}$$
$$(x,y)$$=$$(\dfrac{mx_{2}+nx_{1}}{m+n}$$, $$\dfrac{my_{2}+ny_{1}}{m+n})$$ . ......$$(1)$$
Here
$$ (x_{1},y_{1})$$ =$$(0,0)$$
$$ (x_{2},y_{2})$$ = $$(9,12)$$
$$m:n=1:2$$
Substituting these vaues in the eq $$(1)$$ ,
we get
$$ (x,y)$$=$$(3,4)$$
Question 6
1 / -0

The ratio in which xy-plane divides the line segment joining $$(2,4,5)$$ and $$(3,5,-4)$$ is

A
$$5 : 4$$

B
$$2 : 1$$

C
$$1 : 3$$

D
$$3 : 2$$

Solution

Given points are $$(2,4,5)$$ and $$(3,5,-4)$$
Let the be p:1 and since xy-plane divides the line segment,z-coordinate will be $$ 0$$
the required z-coordinate will be $$\dfrac{5-4{p}}{p+1}=0$$
$$ \ p=\dfrac{5}{4}$$
Ratio$$=5:4$$
Question 7
1 / -0

Find the value of $$x_i$$, if the distance between the points $$(x_i, 2)$$ and $$(3, 4)$$ is $$8$$.

A
$$3\pm 2\sqrt{15}$$

B
$$3\pm \sqrt{15}$$

C
$$2\pm 3\sqrt{15}$$

D
$$2\pm \sqrt{15}$$

Solution

we re given $$ ( x_1 , y_1 ) $$= ( xi, 2) & $$B ( x_2,y_2)=(3,4) $$

let d is the distance between point A & B . d= 8

$$ So, d = \sqrt { (x_2 -x_1)^2 + (y_2 -y_1)^2 } $$

$$ 8 = \sqrt { (3 -x_i )^2 +(4-2)^2 } $$

taking square on both side,

$$ 64 = (3 -x_i)^2 + (4-2)^2 $$

$$ 64 = 9 - 6x_i +(x_i)^2 +(2)^2 $$

$$ (x_i)^2-6x_i - 51 = 0 $$

now, discriminant $$ D = \sqrt { b^2 - 4ac } $$

here $$b = -6$$ , $$a = 1$$ ,$$c= -51$$

so, $$ d= \sqrt { (-6)^2 -4(1)(-51) } = \sqrt {36+ 204 } $$

$$ d= \sqrt {240} > 0 $$

so,
$$\displaystyle x_i = \frac { -b \pm \sqrt d }{2a} = \frac { -(-6) \pm \sqrt {240} }{ 2(1) } = \frac { 6 \pm 4 \sqrt {15} }{2} $$

$$ x_i = 3 \pm 2 \sqrt {15} $$
Question 8
1 / -0

The coordinates of a point on Y-axis which are at a distance of $$5\sqrt 2 $$ from the point $$P(3,-2,5)$$ is -

A
$$(0,6,0)$$ or $$(0,-2,0)$$

B
$$(0,-6,0)$$ or $$(0,2,0)$$

C
$$(0,4,0)$$ or $$(0,-4,0)$$

D
$$(0,5,0)$$ or $$(0,-5,0)$$

Solution

$${\rm{since point is an y - axis}}$$

$${\rm{its x and z coordinate is 0}}$$

$${\rm{Let pt}}{\rm{. on y - axis be A (0,a,0)}}$$

$${\rm{PA = }}\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} $$

$$5\sqrt2 = \sqrt {{{( - 3)}^2} + {{(a + 2)}^2} + {{(0 - 5)}^2}} $$

$$5\sqrt2 = \sqrt {a + {a^2} + 4 + 4a + 25} $$

$$5\sqrt2 = \sqrt {{a^2} + 4a + 38} $$

$$25 \times 2 = {a^2} + 4a + 38$$

$${a^2} + 4a - 12 = 0$$

$${a^2} + 6a - 2(a + b) = 0$$

$$a = 2,a = - 6$$

$$co - ordinate{\rm{ of point (0,2,0)and(0, - 6,0)}}$$
Question 9
1 / -0

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $$(0.-1), (2, 1) and (0, 3)$$

A
$$4$$

B
$$8$$

C
$$1$$

D
$$2$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Also, we know $$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x_1(y_2-y_3)+x_2(y_3+y_1)+x_3(y_1-y_2)]\\=2[0+2(3-1)+0]=8sq\>unit$$
Question 10
1 / -0

C is a point on the line segment joining A(-3, 4) and B (2, 1) such that AC = 2 BC then coordinates of C are

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

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

C
$$\left( {2,7} \right)$$

D
$$\left( {7,2} \right)$$

Solution

AC = 2 BC

$${{AC} \over {BC}} = {2 \over 1}$$

$$x = {{2{x^2} + 1 \times \left( { - 3} \right)} \over {1 + 2}} = {{4 - 3} \over 3} = {1 \over 3}$$

$$y = {{2 \times 1 + 1 \times 4} \over {2 + 1}} = {{2 + 4} \over 3} = 2$$

$$C\left( {{1 \over 3},2} \right)$$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 23

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Straight Lines Test 23

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SHARING IS CARING

Question 1

$$1$$

$$2$$

$$-2$$

$$0$$

Solution

Question 2

$$h^{2} =ab$$

$$2h^{2} = 3ab$$

$$8h^{2} = 9ab$$

$$4h^{2} = 9ab$$

Solution

Question 3

$$(2,2\sqrt{2})$$

$$(1,2\sqrt{2})$$

$$(0,-3\sqrt{3})$$

$$None\ of\ these$$

Solution

Question 4

$$\left(\dfrac{a\sqrt{3}}{2},-\dfrac{a\sqrt{3}}{2}\right)$$

$$(-a\sqrt{3},a\sqrt{3})$$

$$(a\sqrt{3}, a\sqrt{3})$$

$$(-a\sqrt{3},-a\sqrt{3})$$

Solution

Question 5

$$(-3 , 4)$$

$$(3 , 4)$$

$$(3 , -4)$$

none of these

Solution

Question 6

$$5 : 4$$

$$2 : 1$$

$$1 : 3$$

$$3 : 2$$

Solution

Question 7

$$3\pm 2\sqrt{15}$$

$$3\pm \sqrt{15}$$

$$2\pm 3\sqrt{15}$$

$$2\pm \sqrt{15}$$

Solution

Question 8

$$(0,6,0)$$ or $$(0,-2,0)$$

$$(0,-6,0)$$ or $$(0,2,0)$$

$$(0,4,0)$$ or $$(0,-4,0)$$

$$(0,5,0)$$ or $$(0,-5,0)$$

Solution

Question 9

$$4$$

$$8$$

$$1$$

$$2$$

Solution

Question 10

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

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

$$\left( {2,7} \right)$$

$$\left( {7,2} \right)$$

Solution

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