Notes on Straight Lines - Part 3 | Grade 11 > Mathematics > Coordinate Geometry straight line | KULLABS.COM

• Note
• Things to remember
• Exercise

### The 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

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

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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#### 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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## ASK ANY QUESTION ON Straight Lines - Part 3

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