# Polar or Distance Form of a Straight Line Equation

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**\(\textbf{Art 10 :} \qquad \boxed{{\text{Polar / Distance form of a line}}}\) **

Sometimes, it is very convenient to write the equation of a straight line in polar / distance form. Suppose we know that the line passes through the fixed point \(P(h,\,k)\) and is at an inclination of \(\theta :\)

For any point \(Q(x,\,y)\) at a distance *r* from *P ***along** this line, we can write the simple relation

\[\boxed{{\frac{{x - h}}{{\cos \theta }} = \frac{{y - k}}{{\sin \theta }} = r}}\]

This is the required equation of the line. The point \(Q(x,\,y),\) at a distance *r* from *P*, has the coordinates

\[Q(x,\,y) \equiv (h + r\cos \theta ,\,k + r\sin \theta ).\]

\[\left\{ \begin{array}{l}{\text{Obviously, there will be another point, say }}Q'(x,y),{\text{ at a distance }}r{\text{ from }}P\\{\text{along this line but on the opposite side of }}Q{\text{; thus }}Q'(x,{\rm{ }}y){\text{

will have the }}\\{\text{coordinates }}Q'(x,\,y) \equiv (h - r\cos \theta ,\,\,\,k - r\sin \theta)\end{array} \right\}\]

**Example – 15**

A line through \(A( - 5,\, - 4)\) meets the lines \(x + 3y = 2,\,\,2x + y + 4 = 0\) and \(x - y - 5 = 0\) at the points *B*, *C* and *D* respectively. If

\[\begin{align}{\left( {\frac{{15}}{{AB}}} \right)^2} + {\left( {\frac{{10}}{{AC}}} \right)^2} = {\left( {\frac{6}{{AD}}} \right)^2},\end{align}\]

find the equation of the line.

**Solution:**

The figure above roughly sketches the situation described in the equation. Let *B*, *C* and *D* be at distances \({r_1},\,{r_2}\) and \({r_3}\) from *A* along the line \(L = 0,\) whose equation we wish to determine. Assume the inclination of *L* to be \(\theta .\) Thus, *B*, *C* and *D* have the coordinates (respectively):

\[\begin{align}B \equiv ( - 5 + {r_1}\cos \theta ,\,\,\,\,\, - 4 + {r_1}\sin \theta )\\C \equiv ( - 5 + {r_2}\cos \theta ,\,\,\,\,\, - 4 + {r_2}\sin \theta )\\D \equiv ( - 5 + {r_3}\cos \theta ,\,\,\,\,\, - 4 + {r_3}\sin \theta )\end{align}\]

Since these three points(respectively) satisfy the three given equations, we have :

**Point B : **\(\begin{align}( - 5 + {r_1}\cos \theta ) + 3( - 4 + {r_1}\sin \theta ) + 2 = 0 \quad \Rightarrow \qquad {r_1} = \frac{{15}}{{\cos \theta + 3\sin \theta }}\end{align}\)

**Point C :** \(\begin{align}2( - 5 + {r_2}\cos \theta ) + ( - 4 + {r_2}\sin \theta ) + 4 = 0 \quad \Rightarrow \qquad {r_2} = \frac{{10}}{{2\cos \theta + \sin \theta }}\end{align}\)

**Point D :** \(\begin{align}( - 5 + {r_3}\cos \theta ) - ( - 4 + {r_3}\sin \theta ) - 5 = 0 \quad \Rightarrow \qquad {r_3} = \frac{6}{{\cos \theta - \sin \theta }}\end{align}\)

It is given that

\[\begin{align}&{\left( {\frac{{15}}{{AB}}} \right)^2} + {\left( {\frac{{10}}{{AC}}} \right)^2} = {\left( {\frac{6}{{AD}}} \right)^2}\\

\text{i.e;}\qquad \qquad \qquad &{\left( {\frac{{15}}{{{r_1}}}} \right)^2} + {\left( {\frac{{10}}{{{r_2}}}} \right)^2} = {\left( {\frac{6}{{{r_3}}}} \right)^2}\\ \Rightarrow \qquad &{(\cos \theta + 3\sin \theta )^2} + {(2\cos \theta + \sin \theta )^2} = {(\cos \theta - \sin \theta )^2}\\ \Rightarrow \qquad &4{\cos ^2}\theta + 9{\sin ^2}\theta + 12\sin \theta \cos \theta = 0\\ \Rightarrow \qquad &{(2\cos \theta + 3\sin \theta )^2} = 0\\ \Rightarrow \qquad &\tan \theta = \frac{{ - 2}}{3}\\ \Rightarrow \qquad &m = \frac{{ - 2}}{3}\end{align}\]

Thus, we obtain the slope of *L *as *\(\begin{align}\frac{{ - 2}}{3}. \end{align}\)* The equation of *L* can now be easily written :

\[\begin{align}&L:y - ( - 4) = \frac{{ - 2}}{3}(x - ( - 5))\\ \Rightarrow \qquad &L:2x + 3y + 22 = 0\end{align}\]

**TRY YOURSELF - I**

**Q1.** A variable straight line drawn through the intersection of the lines\(\begin{align}\frac{x}{a} + \frac{y}{b} = 1\;\;and\;\;\frac{x}{b} + \frac{y}{a} = 1\end{align}\) and meets the axes in *A* and *B*. Show that the locus of the mid-point of *AB* is \(2xy(a + b) = ab(x + y)\)

**Q2.** The line \(bx + ay = ab\) cuts the axes in *A* and *B*. Another variable line cuts the axes in *C* and *D* such that \(OA + OB = OC + OD\) where *O* is the origin. Prove that the locus of the point of intersection of the lines *AD* and *BC* is the line\(\;x + y = a + b\)

**Q3.** A point *P* moves so that the square of its distance from (3, –2) is equal to its distance from the line \(5x - 12y = 13\) Find the locus of *P*.

**Q4.** A line intersects the *x*-axis in *A*(7, 0) and the *y*-axis in *B*(0, –5). A variable line perpendicular to *AB* intersects the *x*-axis in *P* and the *y*-axis in *Q*. If *AQ* and *BP* intersect in *R*, find the locus of *R*.

**Q5. **If the sum of the distances of a point from two perpendicular lines in a plane is 1, prove that its locus is a square.

**Q6. **A vertex of an equilateral triangle is (2, 3) and the opposite side is \(\;x + y = 2\). Find the equations of the other sides.

**Q7. **A ray of light along the line \(x - 2y - 3 = 0\) is incident upon the mirror-line \(3x - 2y - 5 = 0\) Find the equation of the reflected ray.

**Q8**. If the vertices of a triangle have integral coordinates, show that it cannot be equilateral.

**Q9**. Show using coordinate geometry that the angle bisectors of the sides of a triangle are concurrent.

**Q10**. The sides of a triangle are \(4x + 3y + 7 = 0\;,\;5x + 12y - 27 = 0\;\;and\;\;3x + 4y + 8 = 0\) and By explicitly evaluating the medians in this triangle, show that they are concurrent.

Q11. A rod *APB *of constant length meets the axes in *A* and *B*. If *AP* = *b* and *PB* = *a* and the rod slides between the axes, show that the locus of *P* is\(\;{b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}\)

**Q12.** If *p* is the length of the perpendicular from the origin to the line whose intercepts on the axes are *a *and *b*, show that \(\begin{align}\frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}\end{align}\)

**Q13**. The lines \(3x + 4y - 8 = 0\;and\;5x + 12y + 3 = 0\) intersect in *A*. Find the equations of the lines passing through which intersect the given lines at *B* and *C*, such that \(AB = AC\).

**Q14**. The equal sides *AB* and *AC* of an isosceles triangle *ABC* are produced to the points* P* and *Q* such that \(BP.CQ = A{B^2}\) Prove that the line *PQ* always passes through a fixed point.

**Q15**. One side of a square is inclined to the *x*-axis at an angle and one of its extremities is at the origin; prove that the equations to its diagonals are

\[\begin{array}{l}

&y(\cos \alpha - \sin \alpha ) = x(\sin \alpha + \cos \alpha )\\\\

and\qquad &y(\sin \alpha + \cos \alpha ) + x(\cos \alpha - \sin \alpha ) = a

\end{array}\]

where *a* is the length of the side of the sqaure.

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school