Straight Lines - Class 11 Engineering Maths - Extra Questions

Line perpendicular to $$y-$$ axis is the $$x-$$ axis.
Inclination of $$x-$$ axis is $${ 0 }^{ 0 }$$.
Hence, inclination of a line which is perpendicular to y-axis is $${ 0 }^{ 0 }$$.

Distance $$d$$  between two points $$P(a, b)$$  and $$Q(c ,d)$$ is given by

$$ d = \sqrt{(a - c)^{2} +(b - d)^{2} } $$

Given that :

Length of line i.e $$d = 10$$

Abscissa of second point is $$10$$ 

And let us assume the ordinate of second point be $$y$$

Points  $$(2 , -3)$$ & $$(10 , y)$$

Now apply distance formula

$$ 10 = \sqrt{(2 - 10)^{2} +(-3 - y)^{2} } $$

Squaring both sides

$$ \Rightarrow 100 = ( - 8)^{2} +(-3 - y)^{2} $$

$$ \Rightarrow 100 = 64+(-3 - y)^{2} $$

$$ \Rightarrow 100-64 = (-3 - y)^{2} $$

$$ \Rightarrow 36 = (-3 - y)^{2} $$

Taking square root both sides

$$ \Rightarrow  \pm6 = (-3 - y) $$

or  $$ \Rightarrow  \pm6 = (3 + y) $$  (taking - sign common on both sides)

$$ \Rightarrow  \pm6 -3 =  y $$

Taking both cases

$$ y= +6-3 = 3 $$

$$ y= -6-3 = -9 $$

Area of triangle having angular points
$$(x_{1} , y_{1} )$$   $$(x_{2} , y_{2})$$  $$(x_{3} , y_{3})$$ is given by the formula

$$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$$

$$\triangle=\dfrac{1}{2}\left | 1\times(-2) +3\times16+(-3)\times4-3\times4-(-3)(-2)-1\times16 \right |$$

$$\Rightarrow \triangle=0$$

$$\Rightarrow$$ All three points are in a straight line.

Area of triangle having angular points $$(x_{1} , y_{1} )$$   $$(x_{2} , y_{2})$$  $$(x_{3} , y_{3})$$ is given by the formula

$$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$$

Given that

  $$ P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( a,-8,4 \right) $$

 $$ Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( -3,-5,4 \right) $$

And distance $$PQ=d=5$$

Then, we know that,

Distance of

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

 $$ PQ=\sqrt{{{\left( a+3 \right)}^{2}}+{{\left( -8+5 \right)}^{2}}+{{\left( 4-4 \right)}^{2}}} $$

 $$ 5=\sqrt{{{\left( a+3 \right)}^{2}}+{{\left( -3 \right)}^{2}}} $$

Squaring both side and we get,

  $$ 25={{\left( a+3 \right)}^{2}}+9 $$

 $$ 25-9={{\left( a+3 \right)}^{2}} $$

 $$ {{\left( a+3 \right)}^{2}}=16 $$

 $$ a+3=4 $$

 $$ a=4-3 $$

 $$ a=1 $$

$$Area=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$

Let $$ A (2,0), B (2,3) $$ and $$ C (1,-3) $$
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 BC $$ = \sqrt { \left(1-2 \right) ^{ 2 }+\left( -3-3
\right) ^{ 2 } } = \sqrt { 1 + 37 } = \sqrt { 37 } $$

Distance between the points AC $$ = \sqrt { \left(1 +2 \right) ^{ 2 }+\left( -3 0
\right) ^{ 2 } } = \sqrt { 9 + 9 } = \sqrt { 18 } $$

And similarly distance between points AB = 5 

Since all the sides are of different length, the triangle is a scalene triangle.

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

Given, 

Distance between the points $$ (x,y) ; (5,1) = $$ Distance between $$(x,y) ; (-1,5)$$

$$ \sqrt { \left( 5-x \right) ^{ 2 }+\left( 1- y \right) ^{ 2 } } = \sqrt { \left( -1-x \right) ^{ 2 }+\left( 5 - y \right) ^{ 2 } } $$

$$ \Rightarrow \left( 5-x \right) ^{ 2 }+\left( 1- y \right) ^{ 2 } = \left( -1-x \right) ^{ 2 }+\left( 5 - y \right) ^{ 2 } $$

$$\Rightarrow {5}^{2} + {x}^{2} - 10x + {1}^{2} + {y}^{2} - 2y ={1}^{2} + {x}^{2} + 2x + {5}^{2} + {y}^{2} - 10y $$

$$\Rightarrow -10x -2y = +2x - 10y $$ 

$$\Rightarrow 12x = 8y $$

$$\Rightarrow \dfrac xy = \dfrac 23 $$

Hence, $$p+q = 5$$.

Given: AD be the median of $$\Delta ABC$$

BD = DC

Sice D is the mid-point of BC

Using the mid point theorem to find the coordinates of D,

$$D = [\dfrac{x_2+x_3}{2}, \frac{y_2+y_3}{2}]$$

$$D = [\dfrac{3+5}{2}, \frac{-2+2}{2}]=[4, 0]$$

Area of $$\Delta ABD = \frac{1}{2}\left | [(x_1(y_2-y_4)+x_2(y_4-y_1)+x_4(y_1-y_2))]\right |$$

$$= \frac{1}{2}\left | [(4(-2-0)+3(0-(-6))+4(-6-(-2)))]\right |$$

$$= \frac{1}{2}\left | [-8+18-26]\right |$$

$$= \frac{1}{2}\left | [-6]\right |=3$$ Sq.units

Area of $$\Delta ADC = \frac{1}{2}\left | [(x_1(y_4-y_3)+x_4(y_3-y_1)+x_3(y_1-y_4))]\right |$$

$$= \frac{1}{2}\left | [(4(0-2)+4(2-(-6))+5(-6-0))]\right |$$

$$= \frac{1}{2}\left | [-8+32-30]\right |$$

$$= \frac{1}{2}\left | [-6]\right |=3$$ Sq.units

Area od $$\Delta ABD = \Delta ADC$$

Hence proved that the median of triangle AD divides the triangles into equal areas.

• The equation given is $$4x-5y=12$$
• Which implies $$5y = 4x-12$$ , which gives $$y = \dfrac{4}{5}x-\dfrac{12}{5}$$
• Therefore in slope intercept form , the slope is $$\dfrac {4}5$$ and $$Y$$-intercept is $$\dfrac{-12}5$$

Area of triangle having angular points
$$(x_{1} , y_{1} )$$   $$(x_{2} , y_{2})$$  $$(x_{3} , y_{3})$$ is given by the formula

$$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$$

or $$\triangle=\dfrac{1}{2}\left |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$$

Points given are $$(a,b+c), (a, b-c)$$ and $$(-a,c)$$

Then, $$Area= \triangle=\dfrac{1}{2}\left |a (b-c-c)+a(c-b-c)+(-a)(b+c-(b-c)) \right | $$

$$\Rightarrow Area= \triangle=\dfrac{1}{2}\left | ab-2ac-ab-2ac \right |$$

$$\Rightarrow Area= \triangle=\dfrac{1}{2}\left |-4ac \right | $$

$$\Rightarrow Area= \triangle=2ac $$

Area of triangle having angular points
$$(x_{1} , y_{1} )$$   $$(x_{2} , y_{2})$$  $$(x_{3} , y_{3})$$ is given by the formula

$$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$$

$$ \Rightarrow \triangle = \dfrac{1}{2} a^{2}\left | m_{1}(m_{2}^{2}  -m_{3}^{2}   ) +m_{2} ( m_{3}^{2}-m_{1}^{2}    ) + m_{3}(m_{1}^{2}-m_{2}^{2}) \right |$$

$${\textbf{Step  - 1: Plotting the two points and finding the distance between them}}{\text{.}}$$

                    $${\text{The two points A and B can be plotted as above,where}}$$

                    $$A(\cos \phi ,\sin \phi ){\text{ and }}B(\cos \theta ,\sin \theta )$$

                    $${\text{Using distance formula we can calculate the distance between the two points}}{\text{.}}$$

                    $${\text{So the distance between two points }}A({x_1},{y_1}),({x_2},{y_2}){\text{can be written as}}$$

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

                    $${\text{So we can write }}AB = \sqrt {{{(\cos \theta  - \cos \phi )}^2} + {{(\sin \theta  - \sin \phi )}^2}} $$

$${\textbf{Step  - 2: Finding the perpendicular distance between the centre }}O{\textbf{ and the line }}AB{\text{.}}$$

                    $${\text{We can see that }}\Delta AOD{\text{ is a right - angled triangle}}$$

                    $${\text{According to the Pythagoras theorem, }}O{D^2} + D{A^2} = A{O^2}$$

                    $${\text{Also we can write }}AO = \sqrt {{{\cos }^2}\phi  + {{\sin }^2}\phi } $$

                    $$OD = \sqrt {A{O^2} - A{D^2}} $$

                    $$OA = \sqrt {A{O^2} - {{\left( {\dfrac{{AB}}{2}} \right)}^2}} $$

                    $${AB = 2AD..{\text{as any perpendicular from the centre of the circle bisects the chord}}}$$

                    $$OD = \sqrt {{{\left( {\sqrt {{{\cos }^2}\phi  + {{\sin }^2}\phi } } \right)}^2} - {{\left( {\dfrac{{\sqrt {{{(\cos \theta  - \cos \phi )}^2} + {{(\sin \theta  - \sin \phi )}^2}} }}{2}} \right)}^2}} $$

                    $$OD = \sqrt {{{\cos }^2}\phi  + {{\sin }^2}\phi  - \dfrac{{{{(\cos \theta  - \cos \phi )}^2} + {{(\sin \theta  - \sin \phi )}^2}}}{4}} $$

                    $$OD = \sqrt {\dfrac{{4 - ({{\cos }^2}\theta  + {{\cos }^2}\phi  - 2\cos \theta \cos \phi  + {{\sin }^2}\theta  + {{\sin }^2}\phi  - 2\sin \theta \sin \phi )}}{4}} $$$$\boldsymbol{\mathbf{{\textbf{(As si}}{{\text{n}}^2}x + \mathbf{{\cos ^2}x = 1)}}}$$

                    $$OD = \dfrac{{\sqrt {4 - 2 + 2\sin \theta \sin \phi  + 2\cos \theta \cos \phi } }}{2}$$

                    $$OD = \sqrt {\dfrac{{1 + \sin \theta \sin \phi  + \cos \theta \cos \phi }}{2}} $$

                    $$OD = \sqrt {\dfrac{{1 + \cos (\theta  - \phi )}}{2}} $$$${\textbf{(As cos(a - b) = cos a cos b + sin a sinb)}}$$

                    $$OD = \cos \left( {\dfrac{{\theta  - \phi }}{2}} \right)$$$$\mathbf{\left( {{\text{As cos2}}\theta {\text{ = 2co}}{{\text{s}}^2}\theta  - 1} \right)}$$

$$\boldsymbol{\mathbf{{\textbf{Thus,the distance d is }}\cos \left( {\dfrac{{\theta  - \phi }}{2}} \right).}}$$

A square is quadrilateral whose all sides are equal and all angles are right angles.

Distance between two points $$=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$$

Slope of line $$=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

AB $$=\sqrt{(4-1)^{2}+(2-7)^{2}}$$

      $$=\sqrt{(3)^{2}+(5)^{2}}$$

      $$=\sqrt{34}$$

Slope of AB  $$=\dfrac{2-7}{4-1}$$  $$=\dfrac{-5}{3}$$

BC $$=\sqrt{(4+1)^{2}+(2+1)^{2}}$$     $$=\sqrt{34}$$

Slope of BC $$=\dfrac{-1-2}{-1-4}$$  $$=\dfrac{3}{5}$$

CD $$=\sqrt{(-4+1)^{2}+(4+1)^{2}}$$     $$=\sqrt{9+25}$$ $$=\sqrt{34}$$

Slope of CD $$=\dfrac{4+1}{-4+1}=\dfrac{-5}{3}$$

AD $$=\sqrt{(7-4)^{2}+(-4-1)^{2}}$$  $$=\sqrt{34}$$

Slope of AD $$=\dfrac{4-7}{-4-1}=\dfrac{3}{5}$$

Slope of (AB x BC) $$=-1$$

Slope of (BC x CD)  $$=-1$$

Slope of (CD x AD) $$=-1$$

Slope of (AD x AB) $$=-1$$

If product of two slopes are perpendicular it means they are perpendicular.

All adjacent sides are perpendicular.

Opposite sides are parallel and All sides are equal (length $$=\sqrt{34}$$ units )

$$\therefore $$ It is a square.

We know that the diagonal of a parallelogram divides the parallelogram into $$2$$ equal triangles.

So, Area of parallelogram,

$$ABCD=2\times $$ area of $$\Delta ABC$$   ......... (1) 

We know that, 

Area of $$\Delta ABC$$  with vertices $$\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$$  is

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

So, Area of $$\Delta ABC$$ with vertices $$[A\left( 2,4 \right),B\left( 2+\sqrt{3},5 \right),C\left( 2,6 \right)]$$ is

  $$ A=\dfrac{1}{2}\times \left[ 2\left( 5-6 \right)+\left( 2+\sqrt{3} \right)\left( 6-4 \right)+2\left( 4-5 \right) \right] $$

 $$ A=\dfrac{1}{2}\times \left[ -2+2\left( 2+\sqrt{3} \right)-2 \right] $$

 $$ A=\dfrac{1}{2}\times \left[ -2+4+2\sqrt{3}-2 \right] $$

 $$ A=\sqrt{3} $$

Put this value in equation (1), we get,

Area of parallelogram,

$$ABCD=2\times$$ area of $$\Delta ABC$$

$$ABCD=2\sqrt{3}$$ sq units

Hence, area of the parallelogram is $$2\sqrt{3}$$ sq units.

$$\textbf{Step -1: Write down given information and relevant formula}$$       

                  $${\text{The two points are (x, - 1) and (3,2) and the distance between them is 5.}}$$

                  $${\text{Let (x, - 1) be (}}{{\text{x}}_1}{\text{,}}{{\text{y}}_1}{\text{) and (3,2) be (}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{)}}$$

                  $${\text{As we know that distance between two points (}}{{\text{x}}_{\text{1}}}{\text{,}}{{\text{y}}_{\text{1}}}{\text{) and (}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{) is}}$$

                  $${\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}} $$

$${\textbf{Step -2: Substitute the given values in the formula}}$$

                  $${\text{5 = }}\sqrt {{{(x - 3)}^2} + {{( - 1 - 2)}^2}} $$

                  $$ \Rightarrow \sqrt {{{(x - 3)}^2} + 9}  = 5$$

                  $$ \Rightarrow {(x - 3)^2} + 9 = 25$$

                  $$ \Rightarrow {(x - 3)^2} = 16$$

                  $$ \Rightarrow \left( {x - 3} \right) =  \pm 4$$

                  $$ \Rightarrow {\text{x = 7 or x =  - 1}}$$

$${\textbf{Therefore, the value of x is 7 or  - 1.}}$$