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Straight Lines- Part 1

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Notes, Exercises, Videos, Tests and Things to Remember on Straight Lines- Part 1

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  • Note
  • Things to remember
  • Exercise

Straight lines:-

A straight line is a one-dimensional geometry with no fixed ends

Straight lines
Straight lines

Line segment:-

A line segment is a straight line having fixed ends.

Line segment
Line segment

Ray:-

A ray is a straight line with one fixed end.

Ray
Ray

Coordinates of a point on the plane

Coordinates of a point on X-Y plane
Coordinates of a point on X-Y plane

From the figure aside we can conclude that the sign of the coordinates of the point on I, II, III and IV quadrants are (+,+), (-,+), (-,-) and (+,-).

Some Basic Formulae

Distance formula

If A(x1,y1) and B(x2,y2) be a line joining points AB then the distance between them is given by,

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

Slope formula

If A(x1,y1) and B(x2,y2) be a line joining points AB then slope of AB is given by,

$$m=\tan\theta=(\frac{y_2-y_1}{x_2-x_1})$$

Section formula

A(x1,y1) and B(x2,y2) be a line joining points AB and c(x,y) be a point which divides AB in ratio m1:m2 then ,

Internal division:

$$x=(\frac{m_1x_2+m_2x_1}{m_1+m_2})$$, $$y=(\frac{m_1y_2+m_2y_1}{m_1+m_2})$$

External division:-

$$x=(\frac{m_1x_2-m_2x_1}{m_1-m_2})$$, $$y=(\frac{m_1y_2-m_2y_1}{m_1-m_2})$$

Centroid:

IfA(x1,y1), B(x2,y2) and c(x3,y3) be a vertices of a triangle ABC then o(x,y) be its centroid which is given by,

$$x=(\frac{x_1+x_2+x_3}{3})$$

$$y=(\frac{y_1+y_2+y_3}{3})$$

Area of a triangle:

IfA(x1,y1), B(x2,y2) and c(x3,y3) be a vertices of a triangle ABC, then the area of the triangle ABC is given by,

$$Area={\frac{1}{2}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_1-x_1y_3)}$$

Three standard forms of equation of straight lines

slope intercept form

Slope intercept form
Slope intercept form

Let AB be a straight line meeting the X-axis and Y-axis at A and C respectively. Then equation of the line AB is given by,

$$y=mx+c$$

where,

m= slope of the line AB

C= Y-intercept

Double intercept form

Double intercept form
Double intercept form

Let the straight line AB cuts the axes at A and B as shown in the figure aside, then the equation of line AB is given by,

$$\frac{x}{a}+\frac{y}{b}=1$$

Where,

a=X-intercept

b=Y-intercept

Normal form or perpendicular form

Normal form
Normal form

Let AB be a sstraight line on which the length of the perpendicular P from the origin is drawn. Let the angleis made byperpendicular line with the positive X-axis, then the equation of the line AB is given by,

$$x\cos\alpha+y\sin\alpha=p$$

Equation of lines in some special cases

Point slope form

Let a line passing through a given point B(x1,y1)have its slope equal to m Let P(x,y) be any point on the line. Then, equation of the line is given by,

$$y-y_1=m(x-x_1)$$

point slope form
point slope form

Where,

m= slope of the line

Two points form:

Two point form
Two point form

Let a straight line passes through A(x1,y1) and B(x2,y2) and p(x,y) be any point on the line.Then, equation of the line is given by,

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

some point to recall

  • The general equation (linear equation) of first degree in x and y is,

$$Ax+By+C=0$$

Where, A, B and C are constants.

  • Length of the perpendicular line drawn on the line Ax+By+C=0 from p(x1,y1) is given by,

perpendicular length
perpendicular length

$$L=±{\vert\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\vert}$$

Solved problems

1.Prove that the line whose intercepts on the axes of x and y are respectively -2 and 3, passes through the point (2,3).

solution

Here, X-intercept=a= -2

Y-intercept=b= 3

Then the equation of the line is,

$$\frac{x}{a}+\frac{y}{b}=1$$

$$\frac{x}{-2}+\frac{y}{3}=1$$

or,$$-3x+2y=6 ...........(i)$$

Now, put x=2 and y=6 in equation (i),

Then, $$-3*2+2*6=6$$

or, 6=6 which is true.

Hence, the line (i) passes through the point (2,6)

2.If a line with equation 5x+6y=2k together with the coordinate axes form a triangle of area 135 sq. units , find the value of k.

solution

Here, The given equation of the line is,

$$5x+6y=2k$$

Now, reducing the equation to double intercept form, we have,

$$\frac{x}{2k/5}+\frac{y}{2k/6}=1$$

where, X-intercept=$$\frac{2k}{5}$$

Y-intercept=$$\frac{2k}{6}$$

Hence, the area of the triangle$$ =\frac{1}{2}.\frac{2k}{5}.\frac{2k}{6}\\$$

or,$$135=\frac{k^2}{15}$$

∴k=±45

3.Find the equation of the straight line passing through the intersection of the lines 3x-4y-10=0 and 5x+3y-7=0 and making angle 135° with the positive x-axis.

Solution

Here, The given equations of the lines are

$$3x-4y-10=0............(i)$$

$$5x+3y-7=0..............(ii)$$

Solving equations (i) and (ii), we get,

x= 2 and y= -1

∴the point of intersection of the lines (i) and (ii) is (2,-1).

Slope of the line=m=tan 135°= -1

The equation of the line through a given point (x1,y1) is

$$y-y_1=m(x-x_1)$$

∴The equation of the line through(2,-1) and slope -1 is

$$y-(-1)=-1(x-2)$$

$$x+y=1$$

which is required equation of aline.

4.Show that the following set of three lines are concurrent.

x-y+1=0, 2x-y-1=0 and x+3y-11=0

Solution

Here, taking the first two equations,

we get,

$$x-y+1=0..........................(i)$$

$$ 2x-y-1=0.......................(ii)$$

solving equations (i) and (ii),

we get, x= 2 and y= 3

∴The point of intersection of the lines (I) and (II) is (2,3).

put x=2, y=3 in the third equation i.e, $$x+3y-11=0$$ we get,

$$2+3*3-11=0$$

$$i.e, 0=0$$

Which is true.

Hence, the three lines are concurrent.

some important questions

1.Obtain the equation of the straight line passing through the point (3,4) cutting off equal intercepts on the axes.

2.In what ratio is the line joining (1,3) and (2,7) divided by the line 3x+y=9?

3..Show that the following set of three lines are concurrent.

  1. x+2y=0, 3x-4y-10=0 and 5x+3y-7=0
  2. 2x-7y+10=0, 3x-2y=1 and x-12y+21=0

4.find the equation of the straight line passing through the intersection of x+3y+2=0, 2x-y-3=0 and slope is 1/2.

5.P and Q are two points of the line x-y+1=0 and are at distance 5 from the origin. Find the area of the triangle OPQ.

Reference:

(Bajracharya, Shrestha, and Singh)





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

  • $$m=\tan\theta=(\frac{y_2-y_1}{x_2-x_1})$$

    Section formula

  • $$x=(\frac{m_1x_2+m_2x_1}{m_1+m_2})$$, $$y=(\frac{m_1y_2+m_2y_1}{m_1+m_2})$$
  • $$x=(\frac{m_1x_2-m_2x_1}{m_1-m_2})$$, $$y=(\frac{m_1y_2-m_2y_1}{m_1-m_2})$$
  • Length of the perpendicular line drawn on the line Ax+By+C=0 from p(x1,y1) is given by,

              $$L=±{\vert\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\vert}$$

 

.

Questions and Answers

Click on the questions below to reveal the answers

solution

Here, X-intercept=a= -2

Y-intercept=b= 3

Then the equation of the line is,

$$\frac{x}{a}+\frac{y}{b}=1$$

$$\frac{x}{-2}+\frac{y}{3}=1$$

or,$$-3x+2y=6 ...........(i)$$

Now, put x=2 and y=6 in equation (i),

Then, $$-3*2+2*6=6$$

or, 6=6 which is true.

Hence, the line (i) passes through the point (2,6)

 

solution

Here, The given equation of the line is,

$$5x+6y=2k$$

Now, reducing the equation to double intercept form, we have,

$$\frac{x}{2k/5}+\frac{y}{2k/6}=1$$

where, X-intercept=$$\frac{2k}{5}$$

Y-intercept=$$\frac{2k}{6}$$

Hence, the area of the triangle$$ =\frac{1}{2}.\frac{2k}{5}.\frac{2k}{6}\\$$

or,$$135=\frac{k^2}{15}$$

∴k=±45

solution

Here, The given equations of the lines are

$$3x-4y-10=0............(i)$$

$$5x+3y-7=0..............(ii)$$

Solving equations (i) and (ii), we get,

x= 2 and y= -1

∴the point of intersection of the lines (i) and (ii) is (2,-1).

Slope of the line=m=tan 135°= -1

The equation of the line through a given point (x1,y1) is

$$y-y_1=m(x-x_1)$$

∴The equation of the line through(2,-1) and slope -1 is

$$y-(-1)=-1(x-2)$$

$$x+y=1$$

which is required equation of aline.

Solution

Here, taking the first two equations,

we get,

$$x-y+1=0..........................(i)$$

$$ 2x-y-1=0.......................(ii)$$

solving equations (i) and (ii),

we get, x= 2 and y= 3

∴The point of intersection of the lines (I) and (II) is (2,3).

put x=2, y=3 in the third equation i.e, $$x+3y-11=0$$ we get,

$$2+3*3-11=0$$

$$i.e, 0=0$$

Which is true.

Hence, the three lines are concurrent.

Solution:-

Let, x-intercept(a)=k then,

y-intercept(b)=x-intercept(a)=k

Now, we know that,

Equation of line is,

$$\frac{x}{a}+\frac{y}{b}=1$$

$$or,\frac{x}{k}+\frac{y}{k}=1$$

$$x+y=k$$

Now, since the line passes through (3,4), so we get,

$$3+4=k$$

$$\therefore k=7$$

henc, the equation of line is,$$x+y=7$$

Solution:-

let, the intercept on y-axis be k, Then,

the intercept on x-axis=2k

Now, using double intercept form, we get,

$$\frac{x}{a}+\frac{y}{b}=1$$

$$or,\frac{x}{2k}+\frac{y}{k}=1$$

$$or,x+2y=2k...............(i)$$

since the line (i) passes through the point (2,3), so we get,

$$2+2*3=2k$$

$$\therefore k=4$$

Now, putting k=4 in equation (i), we get,

$$x+2y=8$$

                which is the required equation of the line.

Solution:-

say AB is a line divided by point (-5,4) in the ratio 1:2

According to question,

let A(a,0) and B(0,b) are line joining points AB.

Now. by using section formula, we get,

$$(-5,4)=(\frac{1*0+2*a}{1+2},\frac{1*b+2*0}{1+2})$$

$$or,(-5,4)=(\frac{2a}{3},\frac{b}{3})$$

Now,

$$\therefore \frac{2a}{3}=-5$$

$$or,a=\frac{-15}{2}$$

Also,,$$\frac{b}{3}=4$$

$$or,b=12$$

so, A(-15/2,0) and B(0,12) are line joining points AB.

then. equation of the line is,

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

$$or,y-0=\frac{12}{\frac{15}{2}}(x+\frac{15}{2})$$

$$or,y=\frac{4}{5}(2x+15)$$

$$or,5y=8x+60$$

$$\therefore 8x-5y+60=0$$ is the required equation of the line.

Solution:-

Given equation of line is,

$$ 2x+3y+k=0 $$

$$or, 2x+3y=-k$$

Now, reducing it to double intercept form, wee get,

$$\frac{x}{a}+\frac{y}{b}=1$$

$$or,\frac{x}{\frac{-k}{2}}+\frac{y}{\frac{-k}{3}}=1$$

$$\therefore x-intercept(a)=\frac{-k}{2}$$ And, $$y-intercept(b)=\frac{-k}{3}$$

Hence, Area of the traiangle is,

$$A=\frac{1}{2}pb$$

$$or, 27=\frac{1}{2}*\frac{-k}{2}*\frac{-k}{3}$$

$$or, 27=\frac{k^2}{12}$$

$$or,324=k^2$$

$$\therefore k=\pm 18$$

Hence, the required value of k is ±18.

solution:-

here, Equation of the line joining (1,3) and (2,7) is,

 $$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

$$or,y-3=4(x-1)$$

$$or,4x-y-1=0.............(i)$$                              which  is the equation of the line.

we have,

$$3x+y=9................(ii)$$

Now, solving both equations, we get,

$$x=\frac{10}{7}$$ And $$y=\frac{33}{7}$$

Hence, the line is divided at point (10/7, 33/7)

let the ratio be k:1

so, using section formula, taking value of x,

$$x=\frac{m_1x_2+m_2x_1}{m_1+m_2}$$

$$or,\frac{10}{7}=\frac{2k+2}{k+1}$$

$$\therefore k=\frac{3}{4}$$

Hence, putting the value of k in k:1, we get,

$$k:1=3:4$$                     which is the required ratio.

 

Solution:-

Here, Slope of AB= slope of BC= slope of AC, since they are collinear.

Now, we have,

$$Slope Of AB=\frac{y_2-y_1}{x_2-x_1}$$

$$=\frac{16-4}{-3-1}$$

$$=-3$$

Also, $$Slope Of AC=\frac{y_2-y_1}{x_2-x_1}$$

$$=\frac{-2-4}{k-1}$$

$$=\frac{-6}{k-1}$$

Now, we get,

$$\frac{-6}{k-1}=-3$$

$$\therefore k=3$$

Hence, the value of k is 3

 

Solution:-

Given equations of lines are,

$$3x+y-2=0......................(i)$$

$$ ax+2y-3=0 ...............(ii)$$

$$2x-y-3=0.......................(iii)$$

Now, solving (i) and (iii), we get,

$$x=1$$ And,

$$y=-1$$

Hence, the point of concurrence is (1,-1)

Now, putting (1,-1) in equation (ii), we get,

$$ax+2y-3=0 $$

$$or,a-2-3=0$$

$$\therefore a=5$$

Hence, the value of a is 5 And the point of concurrence is (1,-1).

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