Easy Practice — Set 1

Quick Recap — Points and Straight Lines

• Distance $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$; midpoint $=\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)$.
• Section (internal, $m:n$): $\left(\tfrac{mx_2+nx_1}{m+n},\tfrac{my_2+ny_1}{m+n}\right)$; centroid $=\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right)$.
• Slope $=\tfrac{y_2-y_1}{x_2-x_1}$; line $y=mx+c$; parallel $\Rightarrow m_1=m_2$; perpendicular $\Rightarrow m_1m_2=-1$.
• Circle: $(x-h)^2+(y-k)^2=r^2$; general $x^2+y^2+2gx+2fy+c=0$ has centre $(-g,-f)$, radius $\sqrt{g^2+f^2-c}$.

Beyond-NCERT JEE Formulae

The one high-yield corner of Coordinate Geometry that NCERT barely touches is the Pair of Straight Lines (POSL) — a single second-degree equation that secretly stores two lines. These results also power the fast homogenisation trick.

1. Homogeneous Pair Through the Origin

The equation $ax^2+2hxy+by^2=0$ always represents two straight lines through the origin (real and distinct if $h^2>ab$, coincident if $h^2=ab$, not real if $h^2<ab$). Substituting $y=mx$ gives $bm^2+2hm+a=0$, so the two slopes satisfy $m_1+m_2=-\dfrac{2h}{b}$ and $m_1m_2=\dfrac{a}{b}$. The angle $\theta$ between the lines is $$\tan\theta=\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|.$$

• Perpendicular iff $a+b=0$ (coefficient of $x^2$ plus coefficient of $y^2$ is zero).
• Coincident iff $h^2=ab$.

[JEE Tip] The angle depends only on $a,h,b$, so the same formula gives the angle between the two lines of any second-degree equation — the $x$, $y$ and constant terms never affect it.

2. General Second-Degree as a Pair of Lines

The conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ breaks into two straight lines iff $$abc+2fgh-af^2-bg^2-ch^2=0,$$ equivalently the $3\times3$ determinant $$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}=0.$$ When it holds, the two linear factors are found by solving the equation as a quadratic in $x$ (treating $y$ as a constant).

3. Homogenisation

To get the joint equation of the two lines joining the origin to the points where a curve $S=0$ meets a line, first write the line as $\dfrac{lx+my}{n}=1$; then make every term of $S$ second-degree by multiplying each linear term by $\dfrac{lx+my}{n}$ and the constant term by $\left(\dfrac{lx+my}{n}\right)^2$. The result is a homogeneous pair $ax^2+2hxy+by^2=0$.

[JEE Tip] After homogenising, "the two lines are perpendicular" collapses to (coefficient of $x^2$) plus (coefficient of $y^2$) equal to $0$ — a one-line finish to a whole class of problems.

4. Angle Bisectors of Two Lines

The pair of bisectors of the angles between $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is $$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}.$$ The two bisectors are always mutually perpendicular. If $c_1,c_2$ are first made positive, the $+$ sign gives the bisector of the angle that contains the origin.

5. Distance Between Parallel Lines

For two parallel lines written with identical $x$- and $y$-coefficients, $ax+by+c_1=0$ and $ax+by+c_2=0$, $$d=\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}.$$ Caution: equalise the coefficients first — e.g. rewrite $6x+8y+15=0$ as $3x+4y+\dfrac{15}{2}=0$ before comparing it with $3x+4y-5=0$.

Solved Examples — Beyond-NCERT Formulae

Example 1 (Angle of a homogeneous pair). Find the acute angle between the two lines $6x^2-xy-y^2=0$.

Here $a=6$, $2h=-1$ so $h=-\dfrac{1}{2}$, and $b=-1$, giving $h^2-ab=\dfrac{1}{4}-(6)(-1)=\dfrac{25}{4}$. Hence $$\tan\theta=\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|=\left|\dfrac{2\cdot\dfrac{5}{2}}{6-1}\right|=\dfrac{5}{5}=1,$$ so $\theta=45^\circ$. As a check, $6x^2-xy-y^2=(3x+y)(2x-y)$, i.e. lines of slopes $-3$ and $2$, and $\left|\dfrac{-3-2}{1+(-3)(2)}\right|=1$ confirms $45^\circ$.

Example 2 (Perpendicular pair). Show that $3x^2+8xy-3y^2=0$ is a pair of perpendicular lines and name them.

The coefficient test gives $a+b=3+(-3)=0$, so the lines are perpendicular. Factorising, $3x^2+8xy-3y^2=(3x-y)(x+3y)$, so the two lines are $3x-y=0$ and $x+3y=0$; their slopes $3$ and $-\dfrac{1}{3}$ multiply to $-1$, exactly as the perpendicularity test predicts.

Example 3 (Coincident pair). Find the values of $k$ for which $9x^2+kxy+4y^2=0$ represents a pair of coincident lines.

Coincident lines require $h^2=ab$. With $a=9$, $b=4$ and $h=\dfrac{k}{2}$, this becomes $\dfrac{k^2}{4}=36$, so $k^2=144$ and $k=\pm12$. For $k=12$ the equation is $9x^2+12xy+4y^2=(3x+2y)^2$, the single repeated line $3x+2y=0$.

Example 4 (Does a conic split into lines?). Decide whether $2x^2+3xy+y^2+x+2y-3=0$ is a pair of straight lines, and if so find them.

Matching $ax^2+2hxy+by^2+2gx+2fy+c=0$ gives $a=2$, $b=1$, $c=-3$, $h=\dfrac{3}{2}$, $g=\dfrac{1}{2}$, $f=1$. The test value is $$abc+2fgh-af^2-bg^2-ch^2=(2)(1)(-3)+2(1)\left(\dfrac{1}{2}\right)\left(\dfrac{3}{2}\right)-2(1)^2-1\left(\dfrac{1}{2}\right)^2-(-3)\left(\dfrac{3}{2}\right)^2,$$ which simplifies to $-6+\dfrac{3}{2}-2-\dfrac{1}{4}+\dfrac{27}{4}=0$. So it is a pair of lines, and factoring the quadratic in $x$ gives $2x^2+3xy+y^2+x+2y-3=(2x+y+3)(x+y-1)$, namely $2x+y+3=0$ and $x+y-1=0$. Had the constant been $-1$ instead of $-3$, the test value would be $-\dfrac{1}{2}\neq0$ and no such split could exist.

Example 5 (Homogenisation). Find the pair of lines joining the origin to the points where the line $x+3y=5$ cuts the circle $x^2+y^2-2x-6y+5=0$, and hence show they are perpendicular.

Write the line as $\dfrac{x+3y}{5}=1$. Make the circle homogeneous by multiplying the linear terms by $\dfrac{x+3y}{5}$ and the constant by $\left(\dfrac{x+3y}{5}\right)^2$: $$x^2+y^2-(2x+6y)\cdot\dfrac{x+3y}{5}+5\left(\dfrac{x+3y}{5}\right)^2=0.$$ Since $2x+6y=2(x+3y)$, the last two terms combine to $-\dfrac{(x+3y)^2}{5}$, leaving $x^2+y^2-\dfrac{(x+3y)^2}{5}=0$. Multiplying by $5$ and simplifying $5x^2+5y^2-(x^2+6xy+9y^2)=0$ gives $$4x^2-6xy-4y^2=0,\qquad\text{i.e.}\qquad 2x^2-3xy-2y^2=0.$$ The coefficient of $x^2$ plus the coefficient of $y^2$ is $2+(-2)=0$, so the two lines (which are $2x+y=0$ and $x-2y=0$) are perpendicular.

Example 6 (Angle bisectors). Find the bisectors of the angles between $x+2y-3=0$ and $2x+y+3=0$.

Each line has $\sqrt{a^2+b^2}=\sqrt{5}$, so the bisectors are $$\dfrac{x+2y-3}{\sqrt{5}}=\pm\dfrac{2x+y+3}{\sqrt{5}}\quad\Longrightarrow\quad x+2y-3=\pm(2x+y+3).$$ The $+$ sign gives $x+2y-3=2x+y+3$, i.e. $x-y+6=0$; the $-$ sign gives $x+2y-3=-(2x+y+3)$, i.e. $3x+3y=0$, or $x+y=0$. The bisectors $x-y+6=0$ and $x+y=0$ have slopes $1$ and $-1$, so they are perpendicular, as any pair of angle bisectors must be.

Example 7 (Distance between parallel lines). Find the distance between $2x+3y-4=0$ and $4x+6y+5=0$.

The lines are parallel, but their coefficients differ, so first divide the second by $2$ to obtain $2x+3y+\dfrac{5}{2}=0$. Now with $c_1=-4$ and $c_2=\dfrac{5}{2}$, $$d=\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\dfrac{\left|-4-\dfrac{5}{2}\right|}{\sqrt{2^2+3^2}}=\dfrac{\dfrac{13}{2}}{\sqrt{13}}=\dfrac{\sqrt{13}}{2}.$$