Straight Lines Test 26

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Straight Lines Test 26

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Weekly Quiz Competition

Question 1
1 / -0

If a circle $$ { x }^{ 2 }+{ y }^{ 2 }=20$$ meet the parabola $$ { y }^{ 2 }=8x$$ at $$P$$ and $$Q$$. Then $$PQ=$$__

A
$$4$$

B
$$8$$

C
$$16$$

D
$$ \sqrt { 5 }$$

Solution
Question 2
1 / -0

The coordinates of base $$BC$$ of an isosceles triangles $$ABC$$ are given by $$B (1,3)$$ and $$C (-2,7)$$. Which of the following points can be the possible coordinates of the vertex $$A$$ ?

A
$$\left(-7, \dfrac{1}{8}\right) $$

B
$$(0,4,6)$$

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

D
$$\left(-\dfrac{5}{6},6\right)$$

Solution

Given that ABC is an isosceles triangle.
The two vertices are B (1,3) = (x₂,y₂) and C (-2,7) = (x₂,y₂)
We have to find the co-ordinates of the vertex A
Since the triangle is an isosceles triangle So,
AB = AC ............. (1)
And also vertex A lies on y-axis.
Let A be (0,y) = (x₁,y₁)
Equation (1) ⇒ AB = AC
√ [(x₂ - x₁)² + (y₂ - y₁)²] = √ [(x₂ - x₁)² + (y₂ - y₁)²]
(1 - 0)² + (3 - y)² = (- 2 - 0)² + (7 - y)²
1 + (3 - y)² = 2 + (7 - y)²
1 + 9 + y² - 6y = 2 + 49 + y² -14y
10 - 6y = 51 - 14y
14y - 6y = 51 - 10
9y = 41
$$y = \dfrac{41}{9}$$
y = 4.6
So, co-ordinates of the vertex A are A (0,4.6)
Question 3
1 / -0

Point P(2,3) lines on the $$4x + 3y = 17$$. Then find the co-ordinated of points farthest from the line which are at 5 units distance from the P.

A
$$(6,6)$$

B
$$(6 , -6)$$

C
$$(2 , 0)$$

D
$$(-2 , 0)$$

Solution

Let point $$(x,y)$$ from the line.
$$\left|\dfrac{4x+3y-17}{\sqrt{4^2+3^2}}=5\right |$$
$$\Rightarrow \boxed{4x+3y-17=\pm 25}$$
but given point Only $$(A)$$ option is satisfying line.
So, Answer should be $$(A)$$.
Question 4
1 / -0

The sum of the lengths of tangents and subtangent at a point of $$y=a\log{(x^{2}-a^{2})}$$,$$(a>0)$$ is porportional to

A
$$\left| x\right|$$

B
$$\left| y\right|$$

C
$$\left| xy\right|$$

D
$$\left| x/y\right|$$

Solution

$$y=a\log(x^2-a^2).a>0.$$
$$\dfrac{dy}{dx}=\dfrac{2ax}{x^2-a^2}$$
Length of tangent $$=y\sqrt{1+\dfrac{1}{\left(\dfrac{dy}{dx}\right)^2}}$$
$$= \sqrt{1+\dfrac{(x^2-a^2)^2}{(2ax)^2}}$$
$$=\dfrac{|y(x^2+a^2)|}{|2ax|}$$
Length of subtangent $$=\left|y\dfrac{dx}{dy}\right|=\left|y\times \dfrac{1}{\dfrac{dy}{dx}}\right|$$
$$=\dfrac{y(x^2-a^2)}{|2ax|}$$
Sum $$=\dfrac{|x^2y|}{|2ax|}=\dfrac{|xy|}{2a}$$
Proportional to $$|xy|$$
Question 5
1 / -0

The points $$(-1,5),(-2,-3),(5,1),(6,9)$$ taken on order are the vertices of a

A
Parallelogram

B
Rhombus

C
Rectangle

D
Suare

Solution

$$A(-1,5)$$,$$B(-2,-3)$$,$$C(5,1)$$ and $$D(6,9)$$
Slopes of sides of quadrilateral =>
$${ m }_{ ab }=8\\ { m }_{ bc }=\cfrac { 4 }{ 7 } \\ { m }_{ cd }=8\\ { m }_{ da }=\cfrac { 4 }{ 7 } $$
We see, $${ m }_{ ab }=8={ m }_{ cd }$$ & $${ m }_{ bc }={ m }_{ da }=\cfrac { 4 }{ 7 } $$=> ABCD is Parallelogram as opposite sides are Parallel.
Question 6
1 / -0

Find the point on the x-axis which is equidistant from $$(2, -5)$$ and $$(-2, 9)$$.

A
$$(1,2)$$

B
$$(-7,0)$$

C
$$(3,-5)$$

D
$$(8,8)$$

Solution
Question 7
1 / -0

If the points (0,0), $$(3,\sqrt{3})$$ ,(p,q) from an equilateral triangle and $$q_{1},q_{2}$$ are the two values of q then $$q_{1}+q_{2}$$ =

A
$$2\sqrt{3}$$

B
$$\sqrt{3}$$

C
$$-\sqrt{3}$$

D
$$0$$

Solution

Consider the given point

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 0,0 \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,\sqrt{3} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( p,q \right) $$

Then,

We know that it is an equilateral triangle,

$$ AB=BC $$

$$ \Rightarrow \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}} $$

$$ \Rightarrow \sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 0-\sqrt{3} \right)}^{2}}}=\sqrt{{{\left( 3-p \right)}^{2}}+{{\left( \sqrt{3}-q \right)}^{2}}} $$

$$ \Rightarrow \sqrt{12}=\sqrt{9+{{p}^{2}}-6p+3+{{q}^{2}}-2\sqrt{3}q} $$

$$ \text{Squaring both side and we get,} $$

$$ \Rightarrow 12={{p}^{2}}+{{q}^{2}}-6p-2\sqrt{3}q+12 $$

$$ \Rightarrow {{p}^{2}}+{{q}^{2}}-6p-2\sqrt{3}q=0$$ ....... (1)

Again,

$$ AB=CA $$

$$ \Rightarrow \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{3}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{3}}-{{y}_{1}} \right)}^{2}}} $$

$$ \Rightarrow \sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 0-\sqrt{3} \right)}^{2}}}=\sqrt{{{\left( p-0 \right)}^{2}}+{{\left( q-0 \right)}^{2}}} $$

$$ \Rightarrow \sqrt{12}=\sqrt{{{p}^{2}}+{{q}^{2}}} $$

$$ \text{squaring both side and we get,} $$

$${{p}^{2}}+{{q}^{2}}=12$$…… (2)

By equation (1) and (2) to, and we get,

$$ 12-6p-2\sqrt{3}q=0 $$

$$ \Rightarrow 6p+2\sqrt{3}q=12 $$

$$ \Rightarrow 3p+\sqrt{3}q=6 $$

$$ \Rightarrow 3p=6-\sqrt{3}q $$

$$ \Rightarrow p=\dfrac{6-\sqrt{3}q}{3} $$

Put the value of p in equation (2) and we get,

$$ {{p}^{2}}+{{q}^{2}}=12 $$

$$ \Rightarrow {{\left( \dfrac{6-\sqrt{3}q}{3} \right)}^{2}}+{{q}^{2}}=12 $$

$$ \Rightarrow 36+3{{q}^{2}}-12\sqrt{3}q+9{{q}^{2}}=108$$

$$ \Rightarrow 12{{q}^{2}}-12\sqrt{3}q=72 $$

$$\begin{align}
$$ \Rightarrow 12\left( {{q}^{2}}-\sqrt{3}q-6 \right)=0 $$
$$ \Rightarrow {{q}^{2}}-\sqrt{3}q-6=0 $$
$$ \Rightarrow {{q}^{2}}-2\sqrt{3}q+\sqrt{3}q-6=0 $$
$$ \Rightarrow q\left( q-2\sqrt{3} \right)+\sqrt{3}\left( q-2\sqrt{3} \right)=0 $$
$$ \Rightarrow \left( q-2\sqrt{3} \right)\left( q+\sqrt{3} \right)=0 $$
Then, $$q=2\sqrt{3},\,\,q=-\sqrt{3}$$
Now, According to given question,
$$ q={{q}_{1}}+{{q}_{2}} $$
$$ q=2\sqrt{3}-\sqrt{3} $$
$$ q=\sqrt{3} $$
Hence, this is the answer.
Question 8
1 / -0

Three points $$\left( {0,0} \right),\left( {3,\sqrt 3 } \right),\left( {3,\lambda } \right)$$ from an equilateral triangle, then $$\lambda $$ is equal to

A
$$2$$

B
$$-3$$

C
$$-4$$

D
$$ - \sqrt 3 $$

Solution

Given:An equilateral triangle $$ABC$$ whose coordinates are $$A\left(0,0\right),\,B\left(3,\sqrt{3}\right)$$ and $$C\left(3,\lambda\right)$$
Since the triangle is equilateral, we have
$${AB}^{2}={AC}^{2}$$
So,$${\left(3-0\right)}^{2}+{\left(\sqrt{3}-0\right)}^{2}={\left(3-0\right)}^{2}+{\left(\lambda-0\right)}^{2}$$
$$\Rightarrow\,{\left(\sqrt{3}-0\right)}^{2}={\left(\lambda-0\right)}^{2}$$
$$\Rightarrow\,{\lambda}^{2}=3$$
$$\therefore\,\lambda=\pm \sqrt{3}$$
Question 9
1 / -0

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are $$(2, 1)$$, $$(-2, 3)$$, $$(4, -3)$$.

A
$$1.5$$ sq. units

B
$$3$$ sq. units

C
$$6$$ sq. units

D
$$20$$ sq. units
Question 10
1 / -0

Area of the triangle formed by $$(x_{1,}y_{1}),(x_{2},y_{2}), (3y_{2},(-2y_{1}))$$

A
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

B
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

C
$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

D
$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

We have,

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right) $$

We know that the area of triangle is

$$ Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}+2{{y}_{1}} \right)+{{x}_{2}}\left( -2{{y}_{1}}-{{y}_{1}} \right)+3{{y}_{2}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$
Hence, the is the answer

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Straight Lines Test 26

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Straight Lines Test 26

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SHARING IS CARING

Question 1

$$4$$

$$8$$

$$16$$

$$ \sqrt { 5 }$$

Solution

Question 2

$$\left(-7, \dfrac{1}{8}\right) $$

$$(0,4,6)$$

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

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

Solution

Question 3

$$(6,6)$$

$$(6 , -6)$$

$$(2 , 0)$$

$$(-2 , 0)$$

Solution

Question 4

$$\left| x\right|$$

$$\left| y\right|$$

$$\left| xy\right|$$

$$\left| x/y\right|$$

Solution

Question 5

Parallelogram

Rhombus

Rectangle

Suare

Solution

Question 6

$$(1,2)$$

$$(-7,0)$$

$$(3,-5)$$

$$(8,8)$$

Solution

Question 7

$$2\sqrt{3}$$

$$\sqrt{3}$$

$$-\sqrt{3}$$

$$0$$

Solution

Question 8

$$2$$

$$-3$$

$$-4$$

$$ - \sqrt 3 $$

Solution

Question 9

$$1.5$$ sq. units

$$3$$ sq. units

$$6$$ sq. units

$$20$$ sq. units

Question 10

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

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