Cartesian System of Rectangular Co-ordinates and Straight Lines: Basics for IITJEE Mains

I. Results regarding points in a plane:

1a) Distance Formula:

The distance between two points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ is given by $PQ=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$ . The distance from the origin $O(0,0)$ to the point $P(x_{1},y_{1})$ is $OP=\sqrt{x_{1}^{2}+y_{1}^{2}}$ .

1b) Section Formula:

If $R(x,y)$ divides the join of $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ in the ratio $m:n$ with $m>0, n>0, m \neq n$ , then

$x = \frac{mx_{2} \pm nx_{1}}{m \pm n}$ , and $y = \frac{my_{2} \pm ny_{1}}{m \pm n}$

The positive sign is taken for internal division and the negative sign for external division. The mid-point of $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ is $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ which corresponds to internal division, when $m=n$ . Note that for external division $m \neq n$ .

1c) Centroid of a triangle:

If $G(x,y)$ is the centroid of the triangle with vertices $A(x_{1},y_{1})$ , $B(x_{2},y_{2})$ and $C(x_{3},y_{3})$ then $x=\frac{x_{1}+x_{2}+x_{3}}{3}$ and $y=\frac{y_{1}+y_{2}+y_{3}}{3}$

1d) Incentre of a triangle:

If $I(x,y)$ is the incentre of the triangle with vertices $A(x_{1},y_{1})$ , $B(x_{2},y_{2})$ , and $C(x_{3},y_{3})$ , then

$x=\frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c}$ , $y=\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c}$ , a, b and c being the lengths of the sides BC, CA and AB, respectively of the triangle ABC.

1e) Area of triangle:

ABC with vertices $A(x_{1},y_{1})$ , $B(x_{2},y_{2})$ , and $C(x_{3},y_{3})$ is $\frac{1}{2}\left|\begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$ ,

and is generally denoted by $\triangle$ . Note that if one of the vertex $(x_{3},y_{3})$ is at $O(0,0)$ , then $\triangle = \frac{1}{2}|x_{1}y_{2}-x_{2}y_{1}|$ .

Note: When A, B, and C are taken as vertices of a triangle, it is assumed that they are not collinear.

1f) Condition of collinearity:

Three points $A(x_{1},y_{1})$ , $B(x_{2},y_{2})$ , and $C(x_{3},y_{3})$ are collinear if and only if

$\left | \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right |=0$

1g) Slope of a line:

Let $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ with $x_{1} \neq x_{2}$ be any two points. Then, the slope of the line joining A and B is defined as

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \tan{\theta}$

where $\theta$ is the angle which the line makes with the positive direction of the x-axis, $0 \deg \leq \theta \leq 180 \deg$ , except at $\theta=90 \deg$ . Which is possible only if $x_{1}=x_{2}$ and the line is parallel to the y-axis.

1h) Condition for the points $Z_{k}=x_{k}+iy_{k}$ , $(k=1,2,3)$ to form an equilateral triangle is

$Z_{1}^{2}+Z_{2}^{2}+Z_{3}^{2}=Z_{1}Z_{2}+Z_{2}Z_{3}+Z_{3}Z_{1}$

II) Standard Forms of the Equation of a Line:

An equation of a line parallel to the x-axis is $y=k$ and that of the x-axis itself is $y=0$ .
An equation of a line parallel to the y-axis is $x=h$ and that of the y-axis itself is $x=0$ .
An equation of a line passing through the origin and (a) making an angle $\theta$ with the positive direction of the x-axis is $y=x\tan{\theta}$ , and (b) having a slope m is $y=mx$ , and (c) passing through the point $x_{1}y=y_{1}x$ .
Slope-intercept form: An equation of a line with slope m and making an intercept c on the y-axis is $y=mx+c$ .
Point-slope form: An equation of a line with slope m and passing through $(x_{1},y_{1})$ is $y-y_{1}=x-x_{1}$ .
Two-point form: An equation of a line passing through the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $\frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}$ .
Intercept form: An equation of a line making intercepts a and b on the x-axis and y-axis respectively, is $\frac{x}{a} + \frac{y}{b}=1$ .
Parametric form: An equation of a line passing through a fixed point $A(x_{1},y_{1})$ and making an angle $\theta$ with $0 \leq \theta \leq \pi$ with $\theta \neq \pi/2$ with the positive direction of the x-axis is $\frac{x-x_{1}}{\cos{\theta}}= \frac{y-y_{1}}{\sin{\theta}} = r$ where r is the distance of any point $P(x,y)$ on the line from the point $A(x_{1},y_{1})$ . Note that $x=x_{1}+r\cos{\theta}$ and $y=y_{1}+r\sin{\theta}$ .
Normal form: An equation of a line such that the length of the perpendicular from the origin on it is p and the angle which this perpendicular makes with the positive direction of the x-axis is $\alpha$ , is $x\cos{\alpha}+y\sin{\alpha}=p$ .
General form: In general, an equation of a straight line is of the form $ax+by+c=0$ , where a, b, and c are real numbers and a and b cannot both be zero simultaneously. From this general form of the equation of the line, we can calculate the following: (i) the slope is $-\frac{a}{b}$ (ii) the intercept on the x-axis is $-\frac{c}{a}$ with $a \neq 0$ and the intercept on the y-axis is $-\frac{c}{b}$ with $b \neq 0$ (iii) $p=\frac{|c|}{\sqrt{a^{2}+b^{2}}}$ and $\cos{\alpha} = \pm \frac{|a|}{\sqrt{a^{2}+b^{2}}}$ and $\sin{\alpha}=\pm \frac{|b|}{\sqrt{a^{2}+b^{2}}}$ , the positive sign being taken if c is negative and vice-versa (iv) If $p_{1}$ denotes the length of the perpendicular from $(x_{1},y_{1})$ on this line, then $p_{1}=\frac{|ax_{1}+by_{1}+c|}{|\sqrt{a^{2}+b^{2}}|}$ and (v) the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ lie on the same side of the line if the expressions $ax_{1}+by_{1}+c$ and $ax_{2}+by_{2}+c$ have the same sign, and on the opposite side if they have the opposite signs.

III) Some results for two or more lines:

Two lines given by the equations and are
- parallel (that is, their slopes are equal) if $ab^{'}=a^{'}b$
- perpendicular (that is, the product of their slopes is -1) if $aa^{'}+bb^{'}=0$
- identical if $ab^{'}c^{'}=a^{'}b^{'}c=a^{'}c^{'}b$
- not parallel, then
  - angle $\theta$ between them at their point of intersection is given by $\tan{\theta}= \pm \frac{m-m^{'}}{1+mm^{'}} = \pm \frac{a^{'}b-ab^{'}}{aa^{'}+bb^{'}}$ where $m, m^{'}$ being the slopes of the two lines.
  - the coordinates of their points of intersection are $(\frac{bc^{'}-c^{'}b}{ab^{'}-a^{'}b}, \frac{ca^{'}-c^{'}a}{ab^{'}-a^{'}b})$
  - An equation of any line through their point of intersection is $(ax+by+c) + \lambda (a^{'}x+b^{'}y+c^{'})=0$ where $\lambda$ is a real number.
An equation of a line parallel to the line $ax+by+c=0$ is $ax+by+c^{'}=0$ , and the distance between these lines is $\frac{|c-c^{'}|}{\sqrt{a^{2}+b^{2}}}$
The three lines $a_{1}x+b_{1}y+c_{1}=0$ , $a_{2}x+b_{2}y+c_{2}=0$ and $a_{3}x+b_{3}y+c_{3}=0$ are concurrent (intersect at a point) if and only if $\left | \begin{array}{ccc} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} \right|=0$
Equations of the bisectors of the angles between two intersecting lines $ax+by+c=0$ and $a^{'}x+b^{'}y+c^{'}=0$ are $\frac{ax+by+c}{\sqrt{a^{2}+b^{2}}}=\pm \frac{a^{'}x+b^{'}y+c^{'}}{\sqrt{a^{'2}+b^{'2}}}$ . Any point on the bisectors is equidistant from the given lines. If $\phi$ is the angle between one of the bisectors and one of the lines $ax+by+c=0$ such that $|\tan{\phi}|<1$ , that is, $-\frac{\pi}{4} < \phi < \frac{\pi}{4}$ , then that bisector bisects the acute angle between the two lines, that is, it is the acute angle bisector of the two lines. The other equation then represents the obtuse angle bisector between the two lines.
Equations of the lines through $(x_{1},y_{1})$ and making an angle $\phi$ with the line $ax+by+c=0$ , $b \neq 0$ are $y-y_{1}=m_{1}(x-x_{1})$ where $m_{1}=\frac{\tan{\theta}-\tan{\phi}}{1+\tan{\theta}\tan{\phi}}$ and $y-y_{1}=m_{2}(x-x_{1})$ where $m_{2}=\frac{\tan{\theta}+\tan{\phi}}{1-\tan{\theta}\tan{\phi}}$ where $\tan{\theta}=-\frac{a}{b}$ is the slope of the given line. Note that $m_{1}=\tan{(\theta-\phi)}$ and $m_{2}=\tan{(\theta + \phi)}$ and when $b=0$ , $\theta=\frac{\pi}{2}$ .

IV) Some Useful Points:

To show that A, B, C, D are the vertices of a

parallelogram: show that the diagonals AC and BD bisect each other.
rhombus: show that the diagonals AC and BD bisect each other and a pair of adjacent sides, say, AB and BC are equal.
square: show that the diagonals AC and BD are equal and bisect each other, a pair of adjacent sides, say AB and BC are equal.
rectangle: show that the diagonals AC and BD are equal and bisect each other.

V) Locus of a point:

To obtain the equation of a set of points satisfying some given condition(s) called locus, proceed as follows:

Let $P(h,k)$ be any point on the locus.
Write the given condition involving h and k and simplify. If possible, draw a figure.
Eliminate the unknowns, if any.
Replace h by x and k by y and obtain an equation in terms of $(x,y)$ and the known quantities. This is the required locus.

VI) Change of Axes:

Rotation of Axes: if the axes are rotated through an angle $\theta$ in the anti-clockwise direction keeping the origin fixed, then the coordinates $(X,Y)$ of a point $P(x,y)$ with respect to the new system of coordinates are given by $X=x\cos{\theta}+y\sin{\theta}$ and $Y=y\cos{\theta}-x\sin{\theta}$ .
Translation of Axes: the shifting of origin of axes without rotation of axes is called translation of axes. If the origin $(0,0)$ is shifted to the point $(h,k)$ without rotation of the axes then the coordinates $(X,Y)$ of a point $P(x,y)$ with respect to the new system of coordinates are given by $X=x-h$ and $Y=y-k$ .

I hope to present some solved sample problems with solutions soon.

Nalin Pithwa.

Mathematics Hothouse