Straight Lines Test 10

A median of a triangle is a line segment that joins the vertex of a triangle to

the midpoint of the opposite side.

Median through C will pass through midpoint of AB.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y

}_{ 2 }) $$ is  calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{

x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$

Using this formula,

mid point of AB $$ = D = \left( \dfrac { 2-4 }{ 2 } ,\dfrac { 2-4 }{ 2 } 

\right) \quad =\quad (-1,-1) $$

Distance between two points $$

\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }

\right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{

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


Distance between the points C$$ (5,-8) $$ and D $$ (-1,-1) = \sqrt { \left( -1-5

\right) ^{ 2 }+\left( -1 +8 \right) ^{ 2 } } = \sqrt { 36 + 49} = \sqrt {85 }

$$

Distance

between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( {

x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt {

\left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 }

\right) ^{ 2 } } $$

Distance between $$ (a,2) $$ and $$ (3,4) =  8

$$

$$ \sqrt { \left( 3-a \right) ^{ 2 }+\left( 4-2 \right) ^{ 2 } } = 8 $$

$$ \sqrt { 9 + { a }^{ 2 } - 6a + 4 } = 8 $$

$$ \sqrt { { a }^{ 2 } - 6a + 13 } = 8 $$

$$ { a }^{ 2 }- 6a + 13 = 64 $$

$$ { a }^{ 2 } - 6a - 51  = 0 $$

$$ a=  \dfrac { -(-6)\pm \sqrt { \left( { -6 }^{ 2 } \right) -4(1)(-51) }  }{ 2(1) } $$

$$ => a = \dfrac {6 \pm \sqrt{36 + 204}}{2} $$
$$ => a = 3 \pm 2\sqrt {15} $$

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$  and $$({ x }_{ 3 },{ y }_{ 3 })$$  is:

Let the points be $$ A(0,0) B(4,0), C(4,3), D(0,3) $$

So, the length of the diagonal will be distance between A and C.

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Hence, Length of diagonal AC $$ = \sqrt { \left( 4-0 \right) ^{ 2 }+\left(3 -0 \right) ^{ 2 } } = \sqrt { 16 + 9 } = \sqrt { 25 } = 5 $$

Straight lines $$x+y -4 = 0, 3x+y-4=0, x+3y-4=0$$ form a triangle.

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$

We can see that distance between $$ (8,0) $$ and $$(0,-8) = { \left( 0-8 \right) ^{ 2 }+\left(-8-0 \right) ^{ 2 } } = \sqrt { 64 + 64 } = 8\sqrt { 2} $$.
Hence, $$ A $$ cannot be $$ (0,-8) $$

Here it is given that the circle with center ar
origin passes through the point (-8, -6)
The distance from origin (0, 0) to point (-8, -6) will be
the radius of the given circle
So by distance formula
Radius of circle $$\displaystyle =\sqrt{\left ( -8-0 \right )^{2}+\left ( -6-0 \right )^{2}}$$
$$\displaystyle =\sqrt{8^{2}+6^{2}}=\sqrt{64+34}=\sqrt{100}=10$$
Now any point that lies on the circle will be ar
the same distance as that of radius
Hence check the options whether any point has
the same distance from the origin as radius of
the circle
Option (D) gives
$$\displaystyle \sqrt{\left ( 9-0 \right )^{2}+\left ( \sqrt{19}-0 \right )^{2}}=\sqrt{81+19}=\sqrt{100}=10$$
Hence the distance of point $$\displaystyle \left ( 9,\sqrt{19} \right )$$ from the
origin is same as that of radius equal to 10
The circle given passes through the point $$\displaystyle \left ( 9,\sqrt{19} \right )$$

Let $$ A (2,5), B (3,-4) $$ and $$ C (7,10) $$

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y}_{ 1 } \right) ^{ 2 } } $$

Distance between the points A and B $$ AB = \sqrt { \left( 3+2 \right) ^{ 2 }+\left( -4-5\right) ^{ 2 } } = \sqrt { 25 + 81 } = \sqrt { 106 } $$

Distance between the points B and C $$ BC = \sqrt { \left(7-3 \right) ^{ 2 }+\left( 10+4\right) ^{ 2 } } = \sqrt { 16 + 196 } = \sqrt { 212 } $$

Distance between the points A and C $$AC = \sqrt { \left(7+2 \right) ^{ 2 }+\left(10-5\right) ^{ 2 } } = \sqrt { 81 + 25 } = \sqrt { 106 } $$

Since, $$ {(\sqrt { 212}) }^{2} = {(\sqrt { 106 }) }^{2} + {(\sqrt { 106 })}^{2} $$, the triangle is a right angled triangle.

Let the point on the x-axis be $$ (x,0) $$

Distance between $$ (x,0) $$ and $$ (7,6) = \sqrt { \left( 7-x \right) ^{ 2

}+\left( 6 - 0 \right) ^{ 2 } } = \sqrt { { 7}^{ 2 }+ { x }^{ 2 } - 14x + 36 }

= \sqrt { { x }^{ 2 } - 14x + 85 } $$

Distance between $$ (x,0) $$ and $$ (-3,4) = \sqrt { \left( -3-x \right) ^{ 2

}+\left( 4 - 0 \right) ^{ 2 } } = \sqrt { { 3}^{ 2 }+ { x }^{ 2 } + 6x + 16 } =

\sqrt { { x }^{ 2 } + 6x + 25 } $$

As the point $$ (x,0) $$ is equidistant from the two points, both the distances

calculated are equal.

$$ \sqrt { { x }^{ 2 } - 14x + 85 } = \sqrt { { x }^{ 2 } + 6x + 25 } $$

$$ => { x }^{ 2 } - 14x + 85  = { x }^{ 2 } + 6x + 25  $$

$$ 85 - 25 = 6x  + 14x $$

$$ 60 = 20x $$

$$ x = 3 $$

Thus, the point is $$ (3,0) $$