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Straight Lines - Part 3

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  • Note
  • Things to remember
  • Exercise

The two sides of a line

Two sides of a line
Two sides of a line

Let Ax+By+C=0 be a given line and P(x1,y1) and Q(x2,y2) be two given points.

Let R divides join of P and Q in the ratio m1:m2 then co-ordinates of R is given by,

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

since the point R lies on the line Ax+By+C=0, so we get,

$$A(\frac{m_1x_2+m_2x_1}{m_1+m_2})+B(\frac{m_1y_2+m_2y_1}{m_1+m_2})+C$$

or,$$Am_1x_2+Am_2x_1+Bm_1y_2+Bm_2y_1+Cm_1+Cm_2=0$$

or,$$m_1(Ax_2+By_2+C)+m_2(Ax_1+By_1+C)=0$$

or,$$m_2(Ax_1+By_1+C)=-m_1(Ax_2+By_2+C)$$

or,$$\frac{Ax_1+By_1+C}{Ax_2+By_2+C}=-\frac{m_1}{m_2}...................(i)$$

This above relation (i) shows that,

If linesAx1+By1+C andAx2+By2+C have same sign, then P and Q lies on the same side of the lineAx+By+C=0.

If linesAx1+By1+C andAx2+By2+C have opposite sign,then P and Q lies on the opposite side of the lineAx+By+C=0.

Length of the perpendicular from a point on a straight line xCosα+ySinα=p

Length of the perpendicular
Length of the perpendicular

Let equation of a line bexCosα+ySinα=p and P(x1,y1) be a given point.

Let PR be the perpendicular length drawn from P(x1,y1) to the line,

$$x\cos\alpha+y\sin\alpha=p...........(i)$$

Draw a line through point P(x1,y1) which should be parallel with line (i),

Let ON=p'

Then equationof PN(new line) is given by,

$$x\cos\alpha+y\sin\alpha=p'...............(ii)$$

sinceline (ii) passes throughP(x1,y1), so we get,

$$x_1\cos\alpha+y_1\sin\alpha=p'...............(iii)$$

Now,

$$PR=\pm(p'-p)$$

or,$$PR=\pm(x_1\cos\alpha+y_1\sin\alpha-p).............from (iii)$$

Since the given line is in the formAx+By+C=0,

Then, it is reduced into normal form as,

If C>0 then,

$$\cos\alpha=\frac{-A}{\sqrt{A^2+B^2}}$$

$$\sin\alpha=\frac{-B}{\sqrt{A^2+B^2}}$$

$$p=\frac{C}{\sqrt{A^2+B^2}}$$

If C<0 then,

$$\cos\alpha=\frac{A}{\sqrt{A^2+B^2}}$$

$$\sin\alpha=\frac{B}{\sqrt{A^2+B^2}}$$

$$p=\frac{-C}{\sqrt{A^2+B^2}}$$

Now,

$$\therefore PR=\pm(\frac{Ax_1}{\sqrt{A^2+B^2}}+\frac{By_1}{\sqrt{A^2+B^2}}+\frac{C}{\sqrt{A^2+B^2}})$$

$$\therefore PR=\pm(\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}})$$

Which is the required length of perpendicular drawn from a point on a straight line xCosα+ySinα=p.

Bisectors of the angles between two lines A1x+B1y+C1=0 and A2x+B2y+C2=0

Bisectors of the angles between two lines
Bisectors of the angles between two lines

Let B1and B2 are bisectors of the angle between two given lines L1and L2 whose equations areA1x+B1y+C1=0 and A2x+B2y+C2=0.

We have to find the equations of angular bisectors B1and B2

Now,

Let us take any point P(h,k) on any one of the bisectors.

say on bisectorB1

Then,

$$PM=PN$$

since point lying on anangular bisector of two lines is equidistant from both the lines.

$$\frac{A_1h+B_1k+C_1}{\sqrt{A_1^2+B_1^2}}=\pm\frac{A_2h+B_2k+C_2}{\sqrt{A_2^2+B_2^2}}$$

Since the above relation is satisfied by every point on the angular bisector, so equation of angular bisectors of the linesL1and L2 is given by,

$$\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm\frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$$

Method to identify

Note 1: To identify the bisector of acute and obtuse angle between the given lines , we proceed as follows,

  • we choose one of the given line (say L1) and one of the bisector (say B1)
  • we find thier slopes (m1 and m2)
  • If, $$\lvert\tan\theta\rvert=\lvert\frac{m_1-m_2}{1+m_1m_2}\rvert<1$$, then chosen bisector B1is of acute angle.
  • If, $$\lvert\tan\theta\rvert=\lvert\frac{m_1-m_2}{1+m_1m_2}\rvert>1$$, thenchosen bisector B1is of obtuse angle.
  • If, $$\lvert\tan\theta\rvert=1$$, then both bisectors bisects the right angle between the lines.

Note 2: To identify the bisector of the angle between the lines which contains origin, we proceed as follows,

  • we make C1and C2positive in the equation of the given lines.
  • Then we have to find the angle bisectors and taking positive sign, we get bisector containing an origin



  • If linesAx1+By1+C andAx2+By2+C have same sign, then P and Q lies on the same side of the lineAx+By+C=0.
  • If linesAx1+By1+C andAx2+By2+C have opposite sign,then P and Q lies on the opposite side of the lineAx+By+C=0.
  • Equation of angular bisectors of the linesL1and L2 is given by,

    $$\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm\frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$$

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Questions and Answers

Click on the questions below to reveal the answers

Solution:-

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

We know that,

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

$$or, p=\pm\frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$$

$$or,p^2=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}$$

$$\therefore \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}$$

proved...

Solution:-

Given equations of the lines are,

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

$$2y-11x=-6...............(ii)$$.

Now,

Equation of the bisectors of the angles between the lines is given by,

$$\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm\frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$$

$$or, \frac{x-2y}{\sqrt{1^2+(-2)^2}}=\pm\frac{2y-11x+6}{\sqrt{2^2+(-11)^2}}$$

$$or, \frac{x-2y}{\sqrt{5}}=\pm\frac{2y-11x+6}{\sqrt{125}}$$

$$or,5x-10y=\pm(2y-11x+6)$$

Nox-xy+-wy+, Taking positive sign,we get,

$$8x-6y+3=0$$

And taking negative sign, we get,

$$3x+4y-3=0$$

Hence,8x-6y+3=0 and3x+4y-3=0 are the required equation of the bisectors.

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