Straight Lines Test 17

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Straight Lines Test 17

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Question 1
1 / -0

In the coordinate plane, the points $$F(-2, 1), G(1, 4)$$ and $$H(4, 1)$$ lie on a circle with center $$P$$. What are the coordinates of point $$P$$ ?

A
$$(0, 0)$$

B
$$(1, 1)$$

C
$$(1, 2)$$

D
$$(1, -2)$$

Solution

Let $$H\equiv(x_1,y_1)=(4,1)$$
$$G\equiv(x_2,y_2)=(1,4)$$
$$F\equiv(x_3,y_3)=(-2,1)$$
$$P\equiv (x,y)$$
slope of $$m_r=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{1-4}=-1$$
slope of $$m_t=\dfrac{y_3-y_2}{x_3-x_2}=\dfrac{1-4}{-2-1}=1$$
Formula for x-coordinate of the center of the circle is given as:
$$x=\dfrac{m_rm_t(y_3-y_1)+m_r(x_2+x_3)-m_t(x_1+x_2)}{2(m_r-m_t)}$$
$$\therefore x=\dfrac{(-1)(1)(1-1)+-1(1+(-2))-1(4+1)}{2(-1-1)}$$
$$\therefore x=\dfrac{0+1-5}{-4}=1$$
$$y=-\dfrac{1}{m_r}\bigg[x-\dfrac{x_1+x_2}{2}\bigg]+\dfrac{y_1+y_2}{2}$$
$$\therefore y=-\dfrac{1}{-1}\bigg[1-\dfrac{4+1}{2}\bigg]+\dfrac{1+4}{2}$$
$$\therefore y=\bigg[\dfrac{2-5}{2}\bigg]+\dfrac{5}{2}=1$$
$$\therefore P\equiv(1,1)$$
Hence, option $$B$$ is correct.
Question 2
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The line $$L$$ crosses the $$X$$-axis at $$(-8, 0)$$. The area of the shaded region is $$16$$. What is the slope of the line $$L$$?

A
$$-\dfrac {1}{2}$$

B
$$\dfrac {1}{2}$$

C
$$2$$

D
$$4$$

Solution

Given line L in graph intersect x-axis at point$$A(-8,0)$$
Let line L intersecting y-axis at point $$B(0,y)$$
Then $$area=16$$
Area of triangle$$=\left |\dfrac{1}{2}\times xy \right |$$
$$16=\left |\dfrac{1}{2}\times (-8)y \right |$$
$$16=4y\Rightarrow y=4$$
Hence $$B(0,4)$$
Slope of line l from A and B is $$m=\dfrac{4-0}{0+8}$$
Slope of line l $$m=\dfrac{4}{8}\Rightarrow m=\dfrac{1}{2}$$
Question 3
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If $$QR = 5\ units$$, identify the co-ordinates of Q.

A
$$(1, 5)$$

B
$$(3, 4)$$

C
$$(2, 4)$$

D
$$(1, 4)$$

Solution

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
$$\therefore y=4$$
$$\rightarrow Q=(1,4)$$
Question 4
1 / -0

In the figure above, line $$\iota $$ (not shown) is perpendicular to segment $$AB$$ and bisects segment $$AB$$. Which of the following points lies on line $$\iota $$?

A
$$(0, 2)$$

B
$$(1, 3)$$

C
$$(3,1)$$

D
$$(3, 3)$$

E
$$(3, 6)$$

Solution

Given line l bisects AB and perpendicular to AB.
To find point lie on the line
Slope of the line perpendicular to AB
$${ m }_{ 1 }{ m }_{ 2 }=-1\\ -1m_{ 2 }=-1\\ m_{ 2 }=1$$
And it will pass through the midpoint of AB
Let coordinates of midpoint be $$(x,y)$$
$$x=\cfrac { 0+2 }{ 2 } =1\\ y=\cfrac { 0+2 }{ 2 } =1\\ (x,y)=(1,1)$$
So equation of line passing through $$(1,1)$$ and having slope $$m=1$$
$$(y-1)=1\times (x-1)\\ y-1=x-1\\ y=x$$
Hence $$(3,3)$$ be the point that lie on this line.
Question 5
1 / -0

In the rectangle shown, find the value of $$a - b$$

A
$$-3$$

B
$$-1$$

C
$$3$$

D
$$1$$

Solution

From the figure,
$$A{ D }^{ 2 }-A{ B }^{ 2 }$$ i.e, distance is same
$$(9-5)^{ 2 }+(2-5)^{ 2 }=(9-15)^{ 2 }+(2-b)^{ 2 }\\ 16+9=36+4-4b+{ b }^{ 2 }\\ -15=-4b+b^{ 2 }\\ \Rightarrow { b }^{ 2 }-4b+15=0\\ \therefore b=12.2\\ D{ C }^{ 2 }=B{ C }^{ 2 }\\ (5-a)^{ 2 }+(5-13)^{ 2 }=(a-15)^{ 2 }+(13-2)^{ 2 }\\ 25-10a+{ a }^{ 2 }+64={ a }^{ 2 }-30a+225+121\\ 20a=225+121-64-25\\ \therefore a=13\\ \therefore a-b=1\\ $$
Question 6
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The vertices of a triangle are $$A(2,2), B(-4,4), C(5,-8)$$. Then, find the length of the median through $$C$$.

A
$$\sqrt{157}$$

B
$$\sqrt{15}$$

C
$$\sqrt{57}$$

D
None of these

Solution

'D' is mid point
$$\therefore D=\left(\dfrac{2-4}{2},\dfrac{2+4}{2}\right)$$
$$D=(-1,3)$$
$$\therefore CD=\sqrt{[5-(-1)]^{2}+(-8-3)^{2}}$$
$$=\sqrt{6^{2}+11^{2}}=\sqrt{157}$$
Question 7
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Find the third vertex of an equilateral triangle whose two vertices are $$(2,4)$$ and $$(2,6)$$.

A
$$(2+\sqrt{3},5)$$ or $$(2-\sqrt{3},5)$$

B
$$(2+\sqrt{3},5)$$ or $$(2-\sqrt{7},5)$$

C
$$(2+\sqrt{7},5)$$ or $$(2-\sqrt{3},5)$$

D
None of these

Solution

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 $$(2-\sqrt3,5)$$.
Question 8
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Find the perimeter of the triangle formed by $$(0,0),(1,0)$$ and $$(0,1)$$.

A
$$2+\sqrt{2}$$ units

B
$$2-\sqrt{2}$$ units

C
$$3+\sqrt{2}$$ units

D
None of these

Solution

$$AB=\sqrt{OA^{2}+OB^{2}}$$
$$\Rightarrow AB=\sqrt{2}$$
$$\therefore S=1+1+\sqrt{2}=2+\sqrt{2}$$ units
Question 9
1 / -0

$$A,B,C$$ are three collinear points. The coordinates of $$A$$ and $$B$$ are $$(3,4)$$ and $$(7,7)$$ respectively and $$AC=10$$ units. Find the co-ordinates of $$C$$.

A
$$(11,10)$$

B
$$(1,10)$$

C
$$(11,14)$$

D
None of these

Solution
Question 10
1 / -0

Find the mid points of $$(6,2)$$ and $$(-1,3)$$.

A
$$(2,-1)$$

B
$$(-2,-1)$$

C
$$(2,1)$$

D
None of these

Solution