Differentiation (Part-38) | S N De

Straight Line | Part-4 | Ex-2B 

In mathematics, a straight line is a geometric figure that extends infinitely in both directions. It is the shortest path between two points and can be represented by an equation of the form y = mx + b, where m is the slope of the line and b is the y-intercept (the point where the line intersects the y-axis). A straight line is characterized by having a constant slope and a constant rate of change. It can be used to model various real-world phenomena and is an essential concept in geometry and algebra.

 In addition to its mathematical definition, a straight line has several important properties. It has a constant slope, meaning that the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line remains the same. This slope determines whether the line is steep (large slope) or shallow (small slope).

The y-intercept, denoted by b in the equation y = mx + b, represents the point where the line intersects the y-axis. It indicates the value of y when x is zero, giving the initial position of the line on the vertical axis.

Moreover, two distinct points uniquely determine a straight line. By connecting these points, we obtain the line that passes through both of them. Similarly, if we know the slope and a point on the line, we can find the equation of the line using the point-slope form or slope-intercept form.

Straight lines have a variety of applications in mathematics and other fields. They are extensively used in geometry for constructing shapes, calculating angles, and solving geometric problems. In algebra, straight lines serve as the foundation for linear equations and systems of equations. They are also employed in physics to represent motion, such as the path of an object moving at a constant speed.

Understanding the properties and equations of straight lines is fundamental for grasping more advanced mathematical concepts and solving practical problems in various disciplines.

11. Find the distance of the line $~3x+y+4=0~$ from the point $~(2,5)~$ measured parallel to the line $~3x-4y+8=0.$ 

Solution.

Any straight line parallel to the line $~3x-4y+8=0~$ can be written as $~3x-4y+k=0~~(k \neq 0).$ If this straight line passes through the point $~(2,5)~$ then 

$3 \times 2-4 \times 5+k=0 \Rightarrow k=14.$

So, the equation of straight line is $~3x-4y+14=0 \rightarrow(1)$

Now, we need to find the point of intersection of the straight lines $~3x+y+4=0\rightarrow(2)~$ and $~3x-4y+14=0.$

By $~(1)~$ and $~(2)~$ we get,

$~3x-4y+14-(3x+y+4)=0 \\ \text{or,}~~ -5y+10=0 \\ \therefore~ y=\frac{-10}{-5}=2.$

Then putting the value of $~y~$ in $~(2)~$ we get,

$~ 3x+2+4=0 \Rightarrow y=\frac{-6}{3}=-2.$

Hence, the point of intersection of $~(1)~$ and $~(2)~$ is $~(-2,2).$

Hence, the required distance = the distance between $~(2,5)~$ and $~(-2,2)~$

$=\sqrt{(2+2)^2+(5-2)^2}\\=\sqrt{4^2+3^2}\\=\sqrt{16+9}\\=\sqrt{25}\\=5~~\text{unit.}$

Read More : Plane-Ex-5B (S .N. De ) | Complete Solution with PDF link

12.  Let $~A(2,-5)~$ and $~B(6,-1)~$ be two given points . Find the length of orthogonal projection of the line-segment $~AB~$ upon the line $~x-y=1.$

Solution.

The given equation of straight line is $~x-y=1\rightarrow(1).$ Now, the equation of straight line joining the points $~A(2,-5)~$ and $~B(6,-1)~$ is 

$y+1=\frac{-1+5}{6-2}(x-6) \\ \text{or,}~~ x-y=7\rightarrow(2)$

Now , the straight lines $~(1)~$ and $~(2)~$ are parallel to each other. So,  length of orthogonal projection of the line-segment $~AB~$ upon the line $~x-y=1~$ is equal to  the distance between the points $~A(2,-5)~$ and $~B(6,-1)~$ 

$=\sqrt{(6-2)^2+(-1+5)^2}\\=\sqrt{4^2+4^2}\\=4\sqrt{2}~\text{unit.}$

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13.  The equations of the sides $~BC,~CA~$ and $~AB~$ of the $~\Delta ABC~$  are $~2x+y+1=0,~~ 2x+3y+1=0~$ and $~3x+4y+3=0~$ respectively. Find the equation of the perpendicular drawn from $~A~$ on the side $~BC.$

Solution.

$~BC: 2x+y+1=0 \rightarrow(1),\\~CA : 2x+3y+1=0 \rightarrow(2),\\~AB: 3x+4y+3=0 \rightarrow(3).$

Solving $~(2),~(3)~$ we get, $~A \equiv (-5,3),~$ and slope of the side $~BC~$ is $~-2.$

So, slope of any sl perpendicular to $~BC~$ is $~\frac 12.$

Now, the equation of any sl  passing  through the point $~A(-5,3)~$ and having slope $~\frac 12~$ is given by 

$~y-3=\frac 12 [x-(-5)] \\ \text{or,}~~ 2(y-3)=x+5 \\ \therefore~ x-2y+11=0.$

14.  $~A(-2,7),~B(7,15),~C(-1,-5)~$ and $~D(h,k)~$ are the vertices of a parallelogram and $~BC~$ is one of its diagonals . Find $~(h,k)~$ and the angle between its diagonals.

Solution.

Since $~BACD~$  is a parallelogram , so the midpoint of $~BC~=$the midpoint of $~AD.$ 

$\therefore~ \left(\frac{7-1}{2},\frac{15-5}{2}\right)=\left(\frac{h-2}{2},\frac{k+7}{2}\right) \\ \text{or,}~~  (3,5)=\left(\frac{h-2}{2},\frac{k+7}{2}\right) \\ \therefore~ \frac{h-2}{2}=3\Rightarrow h=8,\\~\frac{k+7}{2}=5 \Rightarrow k=3.\\ \therefore~ (h,k)=(8,3).$

Now, the slope of $~BC~,$

$m_1=\frac{15+5}{7+1}=\frac{20}{8}=\frac 52.$

and  the slope of $~AD~,$

$m_2=\frac{k-7}{h+2}=\frac{3-7}{8+2}=-\frac{4}{10}=-\frac 25.$

$\therefore~ m_1 \times m_2=\frac 52 \times (-2/5)=-1.$

So, the angle between its diagonals is $~90^{\circ}.$

15.  Show that the points whose co-ordinates are $~(1,4),~(3,-2)~$ and $~(-3,16)~$ are on the same straight line on which the points lie. Verify that this line is perpendicular  to the line whose equation is $~2x-6y+13=0.$

Solution.

The equation of the straight line joining the points $~(1,4)~$ and $~(3,-2)~$ is 

$~y-4=\frac{-2-4}{3-1}(x-1) \\ \text{or,}~~ y-4=-3(x-1) \\ \therefore~3x+y=7 \rightarrow(1)$

Clearly, the point $~(-3,16)~$ satisfies the equation $~(1)~$ and so the aforementioned point lies on the straight line $~(1).~$ So, the three points are collinear.

The slope of the straight line $~2x-6y+13=0~$ is $~(m_1)=\frac{-2}{-6}=\frac 13$ and 

the slope of the straight line $~(1)~$ is $~(m_2)=-3.$

Now, since $~m_1 \times m_2= \frac 13 \times (-3)=-1,~$ this line represented by $~(1)~$ is perpendicular  to the line whose equation is $~2x-6y+13=0.$

16.  A straight line $~AB~$ intersects the $~y-$axis at $~B~$ and the perpendicular to $~AB~$ at $~B~$ intersects the $~x-$axis at $~C.~$ If the equation of the straight line $~AB~$ is $~\frac x3-\frac y4=-1,~$ find the co-ordinates of the point $~C.$ 

Solution.

The equation of the straight line $~AB~$ is $~ \frac x3-\frac y4=-1 \rightarrow(1).$

It intersects $~y-$axis at $~B(0,4).$

The equation of the straight line perpendicular to the straight line$~(1)~$ is given by 

$~ \frac x4+\frac y3=k \rightarrow(2)$

The straight line $~(2)~$ passes through the point $~B(0,4)~.$

$\therefore~ \frac 04+\frac 43=k \\ \text{or,}~~ k=\frac 43.$

Now, putting the value of $~k~$ in $~(2)~$ we get,

$~\frac x4+\frac y3=\frac 43 \\ \text{or,}~~ \frac{x}{16/3}+\frac y4=1 \rightarrow(3)$

clearly, the straight line $~(3)~$ intersects $~x-$axis at $~\left(\frac{16}{3},0\right)~.$

$~\therefore ~ C \equiv ~\left(\frac{16}{3},0\right).$

17.  The slope of the straight line is $~7.~$ Find the slopes of sl which are inclined at an angle of $~45^{\circ}~$ with this line.

Solution.

Let the slopes of straight lines are $~m.$

So, by question,

$~\tan 45^{\circ}=\left|\frac{m-7}{1+7m}\right| \\ \text{or,}~~ (m-7)^2=(1+7m)^2 \\ \text{or,}~~ m^2-14m+7^2=1+14m+(7m)^2 \\ \text{or,}~~ 0=49m^2-m^2+14m+14m+1-49 \\ \text{or,}~~48m^2+28m-48=0 \\ \text{or,}~~ 4(12m^2+7m-12)=0 \\ \text{or,}~~ 12m^2+7m-12=0 \\ \text{or,}~~(4m-3)(3m+4)=0\\ \text{or,}~~ 4m-3=0,~~ 3m+4=0 \\ \therefore~ m=\frac 34,~-\frac 43.$

18.  Find the equations of  straight lines which are perpendicular to the sl $~4x-3y+7=0~$ and at a distance of $~3~$ unit from the origin.

Solution.

Equation of any  straight line which is perpendicular to the straight line $~4x-3y+7=0~$ can be written as $~3x+4y+k=0 \rightarrow(1).$

Now, the perpendicular to distance of the straight line $~(1)~$ from the origin is

$=\frac{|3 \times 0+4 \times 0+k|}{\sqrt{3^2+4^2}}=\frac{|k|}{5}.$

By question,

$~\frac{|k|}{5}=3 \Rightarrow k=\pm 15.$

So, the equation of the required  straight line is 

$~3x+4y \pm 15=0 \\ \text{or,}~~ 3x+4y=\pm 15.$

19.  Find the equation of the straight line through the origin and perpendicular to the line joining $~(4,0)~$ and $~(0,4)~$ . Hence, show that the point $~(4,0)~$ is the image of the point $~(0,4)~$ with respect to the line.

Solution.

The slope of the straight line joining $~(4,0)~$ and $~(0,4)~$ is $=\frac{4-0}{0-4}=-1.$

So, the slope of any straight line perpendicular to the straight line having slope $~-1~$ is $~1.$

Now, the equation of any straight line passing through the origin and having slope $~1~$ is $~ y=x.$

The midpoint of the points $~(4,0)~$ and $~(0,4)~$ is $~\left(\frac{0+4}{2},\frac{4+0}{2}\right)=(2,2).$

Clearly the point $~(2,2)~$ satisfies the equation of straight line $~y=x~$ and so the point $~(2,2)~$ lies on $~y=x.$

Hence, we can conclude that the point $~(4,0)~$ is the image of the point $~(0,4)~$ with respect to the line.

20.  Find the area of the parallelogram formed by the line $~y=mx,~y=mx+1,~y=nx~$ and $~y=nx+1.$

Solution.

The given equations of the straight lines are 

$~y=mx \rightarrow(1),\\~y=mx+1 \rightarrow(2),\\~y=nx\rightarrow(3),\\~y=nx+1 \rightarrow(4).$

The point of intersection of $~(1),~(3)~$ is $~A(0,0),~$ the point of intersection of $~(1),~(4)~$ is $~B\left(\frac{1}{m-n},\frac{m}{m-n}\right)~$ and the point of intersection of $~(2),~(3)~$ is $~D\left(\frac{1}{n-m},\frac{n}{n-m}\right).$

So, the area of $~\Delta~ ABD~$ 

#$= \left|\frac 12\begin{vmatrix} 0 &0 &1 \\ \frac{1}{m-n} &\frac{m}{m-n} &1 \\ \frac{1}{n-m}&\frac{n}{n-m} &1 \\ \end{vmatrix}\right|\\=\frac 12\left| \left[-\frac{n}{(m-n)^2}+\frac{m}{(m-n)^2}\right]\right|\\=\left|\frac{1}{2(m-n)}\right|$

$\therefore~$ the area of   the parallelogram $~ABCD~$ is 

$=2 \times \text{area of}~~\Delta ABD\\=\frac{1}{|m-n|}~~\text{sq. unit.}$