## Category Archives: KVPY

### Theory of Equations: III: IITJEE maths: algebra

may overlap a bit with previous lecture(s)…

### Applications of Derivatives: Training for IITJEE Maths: Part V

Question I:

Show that the equation of the tangent to the curve $x=a \frac{f(t)}{h(t)}$ and $y=a \frac{g(t)}{h(t)}$ can be represented in the form:

$\left| \begin{array}{ccc} x & y & a \\ f(t) & g(t) & h(t) \\ f^{'}(t) & g^{'}(t) & h^{'}(t) \end{array} \right|=0$

Question 2:

Show that the derivative of the function $f(x) = x \sin{\frac{\pi}{x}}$, when $x>0$ and $f(x)=0$, when $x=0$ vanishes on an infinite set of points of the interval $latex (0,1), Hint: Use Rolle’s theorem. Question 3: Prove that $\frac{1}{1+x}<\log (1+x) < x$ for $x>0$. Use Lagrange’s theorem. Question 4: Find the largest term in the sequence $a_{n}=\frac{n^{2}}{n^{3}+200}$. Hint: Consider the function $f(x)=\frac{x^{2}}{x^{3}+200}$ in the interval $[1,\infty)$. Question 5: A point P is given on the circumference of a circle with radius r. Chords QR are drawn parallel to the tangent at P. Determine the maximum possible area of the triangle PQR. Question 6: Find the polynomial $f(x)$ of degree 6, which satisfies $\lim_{x \rightarrow 0} (1+\frac{f(x)}{x^{3}})=e^{2}$ and has a local maximum at $x=1$ and local minimum at $x=0$ and 2. Question 7: For the circle $x^{2}+y^{2}=r^{2}$, find the value of r for which the area enclosed by the tangents drawn from the point $(6,8)$ to the circle and the chord of contact is maximum. Question 8: Suppose that f has a continuous derivative for all values of x and $f(0)=0$, with $|f^{'}(x)| <1$ for all x. Prove that $|f(x)| \leq |x|$. Question 9: Show that $(e^{x}-1)>(1+x)\log {1+x}$, if $x \in (0,\infty)$. Question 10: Let $-1 \leq p \leq 1$. Show that the equations $4x^{3}-3x-p=0$ has a unique root in the interval $[1/2,1]$ and identify it. ### Applications of Derivatives IITJEE Maths tutorial: practice problems part IV Question 1. If the point on $y = x \tan {\alpha} - \frac{ax^{2}}{2u^{2}\cos^{2}{\alpha}}$, where $\alpha>0$, where the tangent is parallel to $y=x$ has an ordinate $\frac{u^{2}}{4a}$, then what is the value of $\alpha$? Question 2: Prove that the segment of the tangent to the curve $y=c/x$, which is contained between the coordinate axes is bisected at the point of tangency. Question 3: Find all the tangents to the curve $y = \cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$. Question 4: Prove that the curves $y=f(x)$, where $f(x)>0$, and $y=f(x)\sin{x}$, where $f(x)$ is a differentiable function have common tangents at common points. Question 5: Find the condition that the lines $x \cos{\alpha} + y \sin{\alpha} = p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$. Question 6: Find the equation of a straight line which is tangent to one point and normal to the point on the curve $y=8t^{3}-1$, and $x=4t^{2}+3$. Question 7: Three normals are drawn from the point $(c,0)$ to the curve $y^{2}=x$. Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the two other normals are perpendicular to each other. Question 8: If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2} + p_{2}^{2}=a^{2}$. Question 9: Show that the curve $x=1-3t^{2}$, and $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$. If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t= \tan {\psi} +\sec {\psi}$. If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles. Question 10: Find the condition that the curves $ax^{2}+by^{2}=1$ and $a^{'}x^{2} + b^{'}y^{2}=1$ intersect orthogonality and hence show that the curves $\frac{x^{2}}{(a^{2}+b_{1})} + \frac{y^{2}}{(b^{2}+b_{1})} = 1$ and $\frac{x^{2}}{a^{2}+b_{2}} + \frac{y^{2}}{(b^{2}+b_{2})} =1$ also intersect orthogonally. More later, Nalin Pithwa. ### Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths “Easy” questions: Question 1: Find the slope of the tangent to the curve represented by the curve $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$ at the point $(2,-1)$. Question 2: Find the co-ordinates of the point P on the curve $y^{2}=2x^{3}$, the tangent at which is perpendicular to the line $4x-3y+2=0$. Question 3: Find the co-ordinates of the point $P(x,y)$ lying in the first quadrant on the ellipse $x^{2}/8 + y^{2}/18=1$ so that the area of the triangle formed by the tangent at P and the co-ordinate axes is the smallest. Question 4: The function $f(x) = \frac{\log (\pi+x)}{\log (e+x)}$, where $x \geq 0$ is (a) increasing on $(-\infty, \infty)$ (b) decreasing on $[0, \infty)$ (c) increasing on $[0, \pi/e)$ and decreasing on $[\pi/e, \infty)$ (d) decreasing on $[0, \pi/e)$ and increasing on $[\pi/e, \infty)$. Fill in the correct multiple choice. Only one of the choices is correct. Question 5: Find the length of a longest interval in which the function $3\sin(x) -4\sin^{3}(x)$ is increasing. Question 6: Let $f(x)=x e^{x(1-x)}$, then $f(x)$ is (a) increasing on $[-1/2, 1]$ (b) decreasing on $\Re$ (c) increasing on $\Re$ (d) decreasing on $[-1/2, 1]$. Fill in the correct choice above. Only one choice holds true. Question 7: Consider the following statements S and R: S: Both $\sin(x)$ and $\cos (x)$ are decreasing functions in the interval $(\pi/2, \pi)$. R: If a differentiable function decreases in the interval $(a,b)$, then its derivative also decreases in $(a,b)$. Which of the following is true? (i) Both S and R are wrong. (ii) Both S and R are correct, but R is not the correct explanation for S. (iii) S is correct and R is the correct explanation for S. (iv) S is correct and R is wrong. Indicate the correct choice. Only one choice is correct. Question 8: For which of the following functions on $[0,1]$, the Lagrange’s Mean Value theorem is not applicable: (i) $f(x) = 1/2 -x$, when $x<1/2$; and $f(x) = (1/2-x)^{2}$, when $x \geq 1/2$. (ii) $f(x) = \frac{\sin(x)}{x}$, when $x \neq 0$; and $f(x)=1$, when $x=0$. (iii) $f(x)=x |x|$ (iv) $f(x)=|x|$. Only one choice is correct. Which one? Question 9: How many real roots does the equation $e^{x-1}+x-2=0$ have? Question 10: What is the difference between the greatest and least values of the function $f(x) = \cos(x) + \frac{1}{2}\cos(2x) -\frac{1}{3}\cos(3x)$? More later, Nalin Pithwa. ### Circles and System of Circles: IITJEE Mains: some solved problems I Part I: Multiple Choice Questions: Example 1: Locus of the mid-points of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the centre is (a) $x+y=2$ (b) $x^{2}+y^{2}=1$ (c) $x^{2}+y^{2}=2$ (d) $x-y=0$ Answer 1: C. Solution 1: Let O be the centre of the circle $x^{2}+y^{2}=4$, and let AB be any chord of this circle, so that $\angle AOB=\pi /2$. Let $M(h,x)$ be the mid-point of AB. Then, OM is perpendicular to AB. Hence, $(AB)^{2}=(OA)^{2}+(AM)^{2}=4-2=2 \Longrightarrow h^{2}+k^{2}=2$. Therefore, the locus of $(h,k)$ is $x^{2}+y^{2}=2$. Example 2: If the equation of one tangent to the circle with centre at $(2,-1)$ from the origin is $3x+y=0$, then the equation of the other tangent through the origin is (a) $3x-y=0$ (b) $x+3y=0$ (c) $x-3y=0$ (d) $x+2y=0$. Answer 2: C. Solution 2: Since $3x+y=0$ touches the given circle, its radius equals the length of the perpendicular from the centre $(2,-1)$ to the line $3x+y=0$. That is, $r= |\frac{6-1}{\sqrt{9+1}}|=\frac{5}{\sqrt{10}}$. Let $y=mx$ be the equation of the other tangent to the circle from the origin. Then, $|\frac{2m+1}{\sqrt{1+m^{2}}}|=\frac{5}{\sqrt{10}}=25(1+m^{2})=10(2m+1)^{2} \Longrightarrow 3m^{2}+8m-3=0$, which gives two values of m and hence, the slopes of two tangents from the origin, with the product of the slopes being -1. Since the slope of the given tangent is -3, that of the required tangent is 1/3, and hence, its equation is $x-3y=0$. Example 3. A variable chord is drawn through the origin to the circle $x^{2}+y^{2}-2ax=0$. The locus of the centre of the circle drawn on this chord as diameter is (a) $x^{2}+y^{2}+ax=0$ (b) $x^{2}+y^{2}+ay=0$ (c) $x^{2}+y^{2}-ax=0$ (d) $x^{2}+ y^{2}-ay=0$. Answer c. Solution 3: Let $(h,k)$ be the centre of the required circle. Then, $(h,k)$ being the mid-point of the chord of the given circle, its equation is $hx+ky-a(x+h)=h^{2}+k^{2}-2ah$. Since it passes through the origin, we have $-ah=h^{2}+k^{2}-2ah \Longrightarrow h^{2}+k^{2}-ah=0$. Hence, locus of $(h,k)$ is $x^{2}+y^{2}-ax=0$. Quiz problem: A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) $m(m+n)$ (b) $m+n$ (c) $n(m+n)$ (d) $(1/2)(m+n)$. To be continued, Nalin Pithwa. ### Circles and Systems of Circles: IITJEE mains co-ordinate geometry basics Section I: Definition of a Circle: A circle is the locus of a point which moves in a plane so that it’s distance from a fixed point in the plane is always constant.The fixed point is called the centre of the circle and the constant distance is called its radius. Section II: Equations of a circle: • An equation of a circle with centre $(h,k)$ and radius r is $(x-h)^{2}+(y-k)^{2}=r^{2}$. • An equation of a circle with centre $(0,0)$ and radius r is $x^{2}+y^{2}=r^{2}$. • An equation of a circle on the line segment joining $(x_{1},y_{1})$ and $(x_{2},y_{2})$ as diameter is $(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$. • General equation of a circle is :$x^{2}+y^{2}+2gx+2fy+c=0$ where g, f, and c are constants • centre of this circle is $(-g,-f)$ • Its radius is $\sqrt{g^{2}+f^{2}-c}$, $g^{2}+f^{2}\geq c$ • Length of the intercept made by this circle on the x-axis is $2\sqrt{g^{2}-c}$ if $g^{2}-c \geq 0$ and that on the y-axis is $\sqrt{f^{2}-c}$ if $f^{2}-c \geq 0$. • General equation of second order degree $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ in x and y represents a circle if and only if: • coefficient of $x^{2}$ equals coefficient of $y^{2}$, that is, $a=b \neq 0$ • coefficient of $xy$ is zero, that is , $h=0$ • $g^{2}+f^{2}-ac \geq 0$ Section III: Some results regarding circles: • Position of a point with respect to a circle: Point $P(x_{1},y_{1})$ lies outside, on or inside a circle $S \equiv x^{2}+y^{2}+2gx+2fy+c=0$, according as $S_{1} \equiv x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c>, =, \hspace{0.1in} or \hspace{0.1in}< 0$ • Parametric coordinates of any point on the circle $(x-h)^{2}+(y-k)^{2}=r^{2}$ are given by $(h+r\cos{\theta},k+r\sin{\theta})$ with $0 \leq \theta < 2\pi$. In particular, parametric coordinates of any point on the circle. • An equation of the tangent to the circle $x^{2}+y^{2}+2gx+2fy+c=0$ at the point $(x_{1},y_{1})$ on the circle is $xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0$ • An equation of the normal to the circle $x^{2}+y^{2}+2gx+2fy+c=0$ at the point $(x_{1},y_{1})$ on the circle is $\frac{y-y_{1}}{y_{1}+f} = \frac{x-x_{1}}{x_{1}+g}$ • Equations of the tangent and normal to the circle $x^{2}+y^{2}=r^{2}$ at the point $(x_{1},y_{1})$ on the circle are, respectively, $xx_{1}+yy_{1}=r^{2}$ and $\frac{x}{x_{1}}=\frac{y}{y_{1}}$ • The line $y=mx+c$ is a tangent to the circle $x^{2}+y^{2}=r^{2}$ if and only if $c^{2}=r^{2}(1+m^{2})$. • The lines $y=mx \pm r\sqrt{(1+m^{2})}$ are tangents to the circle $x^{2}+y^{2}=r^{2}$, for all finite values of m. If m is infinite, the tangents are $x \pm r=0$. • An equation of the chord of the circle $S=x^{2}+y^{2}+2gx+2fy+c=0$, whose mid-point is $(x_{1},y_{1})$ is $T=S_{1}$, where $T \equiv xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c$ and $S_{1}=x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c$. In particular, an equation of the chord of the circle $x^{2}+y^{2}=r^{2}$, whose mid-point is $(x_{1},y_{1})$ is $xx_{1}+yy_{1}=x_{1}^{2}+y_{1}^{2}$. • An equation of the chord of contact of the tangents drawn from a point $(x_{1},y_{1})$ outside the circle $S=0$ is $T=0$.(S and T are as defined in (8) above). • Length of the tangent drawn from a point $(x_{1},y_{1})$ outside the circle $S=0$, to the circle, is $\sqrt{S_{1}}$. (S and $\sqrt{S_{1}}$) are as defined in (8) above.) • Two circles with centres $C_{1}(x_{1},y_{1})$ and $C_{2}(x_{2},y_{2})$ and radii $r_{1}$, $r_{2}$ respectively, (i) touch each other externally if $C_{1}C_{2}|=r_{1}+r_{2}$. the point of contact is $(\frac{r_{1}x_{2}+r_{2}x_{1}}{r_{1}+r_{2}},\frac{r_{1}y_{2}+r_{2}y_{1}}{r_{1}+r_{2}})$ and (ii) touch each other internally if $|C_{1}C_{2}|=|r_{1}-r_{2}|$, where $r_{1} \neq r_{2}$; the point of contact is $(\frac{r_{1}x_{2}-r_{2}x_{1}}{r_{1}-r_{2}} , \frac{r_{1}y_{2}-r_{2}y_{1}}{r_{1}-r_{2}})$ • An equation of the family of circles passing through the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) +\lambda(F)=0$, where $F=\left|\begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right|$ • An equation of the family of circles which touch the line $y-y_{1}=m(x-x_{1})$ at $(x_{1}, y_{1})$ for any finite value of m is $(x-x_{1})^{2}+(y-y_{1})^{2}+\lambda ((y-y_{1})-m(x-x_{1}))=0$. If m is infinite, the equation becomes $(x-x_{1})^{2}+(y-y_{1})^{2}+\lambda (x-x_{1})=0$. • Let QR be a chord of a circle passing through the point $P(x_{1},y_{1})$ and let the tangents at the extremities Q and R of this chord intersect at the point $L(h,k)$. Then, locus of L is called the polar of P with respect to the circle, and P is called the pole of its polar. • Equation of the polar of $P(x_{1},y_{1})$ with respect to the circle $S \equiv x^{2}+y^{2}+2gx+2fy+c=0$ is $T=0$, where T is defined as above. • If the polar of P with respect to a circle passes through Q, then the polar of Q with respect to the same circle passes through P. Two such points P and Q, are called conjugate points of the same circle. • If lengths of the tangents drawn from a point P to the two circles $S_{1} \equiv x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $S_{2} \equiv x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ are equal, then the locus of P is called the radical axis of the two circles $S_{1} =0$ and $S_{2}=0$, and its equation is $S_{1}-S_{2}=0$, that is, $2(g_{1}-g_{2})+2(f_{1}-f_{2})y+(c_{1}-c_{2})=0$ • Radical axis of two circles is perpendicular to the line joining their circles. • Radical axes of three circles, taken in pairs, pass through a fixed point called the radical centre of the three circles, if the centres of these circles are non-collinear. 4: Special Forms of Equation of a Circle: 1. An equation of a circle with centre $(r,r)$ and radius $|r|$ is $(x-r)^{2}+(y-r)^{2}=r^{2}$. This touches the co-ordinate axes at the points $(r,0)$ and $(0,r)$. 2. An equation of a circle with centre $(x_{1},r)$, radius $|r|$ is $(x-x_{1})^{2}+(y-r)^{2}=r^{2}$. This touches the x-axis at $(x_{1},0)$. 3. An equation of a circle with centre $(\frac{a}{2},\frac{b}{2})$ and radius $\sqrt{\frac{(a^{2}+b^{2})}{4}}$ is $x^{2}+y^{2}-ax-by=0$. This circle passes through the origin $(0,0)$, and has intercepts a and b on the x and y axes, respectively. 5: Systems of Circles: Let $S \equiv x^{2}+y^{2}+2gx+2fy+c$; and $S^{'} \equiv x^{2}+y^{2}+2g^{'}x+2f^{'}y+c^{'}$ and $L \equiv ax + by + k^{'}$. 1. If two circles $S=0$ and $S^{'}=0$ intersect at real and distinct points, then $S+\lambda S^{'}=0$ where $\lambda \neq -1$ represents a family of circles passing through these points, where $\lambda$ is a parameter, and $S-S^{'}=0$ when $\lambda=-1$ represents the chord of the circles. 2. If two circles $S =0$ and $S^{'}=0$ touch each other, then $S-S^{'}=0$ represents equation of the common tangent to the two circles at their point of contact. 3. If two circles $S=0$ and $S^{'}=0$ intersect each other orthogonally (the tangents at the point of intersection of the two circles are at right angles), then $2gg^{'}+2ff^{'}=c+c^{'}$. 4. If the circle $S=0$ intersects the line $L=0$ at two real and distinct points, then $S+\lambda L=0$ represents a family of circles passing through these points. 5. If $L=0$ is a tangent to the circle $S=0$ at P, then $S+\lambda L=0$ represents a family of circles touching $S=0$ at P, and having $L=0$ as the common tangent at P. 6. Coaxial Circles: A system of circles is said to be coaxial if every pair of circles of the system have the same radical axis. The simplest form of the equation of a coaxial system of circles is : $x^{2}+y^{2}+2gx+c=0$, where g is a variable and c is constant, the common radical axis of the system being y-axis and the line of centres being x-axis. The Limiting points of the coaxial system of circles are the members of the system which are of zero radius. Thus, the limiting points of the coaxial system of circles $x^{2}+y^{2}+2gx+c=0$ are $(\pm \sqrt{c},0)$ if $c>0$. The equation $S+\lambda S^{'}=0$ ($\lambda \neq -1$) represents a family of coaxial circles, two of whose members are given to be $S=0$ and $S^{'}=0$. 7. Conjugate systems (or orthogonal systems) of circles : Two system of circles such that every circle of one system cuts every circle of the other system orthogonally are said to be conjugate system of circles. For instance, $x^{2}+y^{2}+2gx+c=0$ and $x^{2}+y^{2}+2fy-c=0$, where g and f are variables and c is constant, represent two systems of coaxial circles which are conjugate. 6: Common tangents to two circles: If $(x-g_{1})^{2}+(y-f_{1})^{2}=a_{1}^{2}$ and $(x-g_{2})^{2}+(y-f_{2})^{2}=a_{2}^{2}$ are two circles with centres $C_{1}(g_{1},f_{1})$ and $C_{2}(g_{2},f_{2})$ and radii $a_{1}$ and $a_{2}$ respectively, then we have the following results regarding their common tangents: 1. When $C_{1}C_{2}>a_{1}+a_{2}$, that is, distance between the centres is greater than the sum of their radii, the two circles do not intersect with each other, and four common tangents can be drawn to circles. Two of them are direct common tangents and other two are transverse common tangents. The points $T_{1},T_{2}$ of intersection of direct common tangents and transverse common tangents respectively, always lie on the line joining the centres of the two circles and divide it externally and internally respectively in the ratio of their radii. 2. When $C_{1}C_{2}=a_{1}+a_{2}$, that is, the distance between the centres is equal to the sum of their radii, the two circles touch each other externally, two direct tangents are real and distinct and the transverse tangents coincide. 3. When $C_{1}C_{2}, that is, the distance between the centres is less than the sum of the radii, the circles intersect at two real and distinct points, the two direct common tangents are real and distinct while the transverse common tangents are imaginary. 4. When $C_{1}C_{2}=|a_{1}-a_{2}|$ with $a_{1} \neq a_{2}$, that is, the distance between the centres is equal to the difference of their radii, the circles touch each other internally, two direct common tangents are real and coincident, while the transeverse common tangents are imaginary. 5. When $C_{1}C_{2}<|a_{1}-a_{2}|$, with $a_{1} \neq a_{2}$, that is, the distance between the centres is less than the difference of the radii, one circle with smaller radius lies inside the other and the four common tangents are all imaginary. To be continued, Nalin Pithwa. ### Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them. Problem 1: Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points $R_{1}$, $R_{2}$, $R_{3}$, $\ldots$, $R_{n}$ and on it a point R is taken such that $\frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}$ Show that the locus of R is a straight line. Solution 1: Let equations of the given lines be $a_{i}x+b_{i}y+c_{i}=0$, $i=1,2,\ldots, n$, and the point O be the origin $(0,0)$. Then, the equation of the line through O can be written as $\frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r$ where $\theta$ is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O. Let $r, r_{1}, r_{2}, \ldots, r_{n}$ be the distances of the points $R, R_{1}, R_{2}, \ldots, R_{n}$ from O which in turn $\Longrightarrow OR=r$ and $OR_{i}=r_{i}$, where $i=1,2,3 \ldots n$. Then, coordinates of R are $(r\cos{\theta}, r\sin{\theta})$ and of $R_{i}$ are $(r_{i}\cos{\theta},r_{i}\sin{\theta})$ where $i=1,2,3, \ldots, n$. Since $R_{i}$ lies on $a_{i}x+b_{i}y+c_{i}=0$, we can say $a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0$ for $i=1,2,3, \ldots, n$ $\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}$, for $i=1,2,3, \ldots, n$ $\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$ $\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$ …as given… $\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0$ Hence, the locus of R is $(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0$ which is a straight line. Problem 2: Determine all values of $\alpha$ for which the point $(\alpha,\alpha^{2})$ lies inside the triangle formed by the lines $2x+3y-1=0$, $x+2y-3=0$, $5x-6y-1=0$. Solution 2: Solving equations of the lines two at a time, we get the vertices of the given triangle as: $A(-7,5)$, $B(1/3,1/9)$ and $C(5/4, 7/8)$. So, AB is the line $2x+3y-1=0$, AC is the line $x+2y-3=0$ and BC is the line $5x-6y-1=0$ Let $P(\alpha,\alpha^{2})$ be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line $5x-6y-1=0$, both $5(-7)-6(5)-1$ and $5\alpha-6\alpha^{2}-1$ must have the same sign. $\Longrightarrow 5\alpha-6\alpha^{2}-1<0$ or $6\alpha^{2}-5\alpha+1>0$ which in turn $\Longrightarrow (3\alpha-1)(2\alpha-1)>0$ which in turn $\Longrightarrow$ either $\alpha<1/3$ or $\alpha>1/2$….call this relation I. Again, since B and P lie on the same side of the line $x+2y-3=0$, $(1/3)+(2/9)-3$ and $\alpha+2\alpha^{2}-3$ have the same sign. $\Longrightarrow 2\alpha^{2}+\alpha-3<0$ and $\Longrightarrow (2\alpha+3)(\alpha-1)<0$, that is, $-3/2 <\alpha <1$…call this relation II. Lastly, since C and P lie on the same side of the line $2x+3y-1=0$, we have $2 \times (5/4) + 3 \times (7/8) -1$ and $2\alpha+3\alpha^{2}-1$ have the same sign. $\Longrightarrow 3\alpha^{2}+2\alpha-1>0$ that is $(3\alpha-1)(\alpha+1)>0$ $\alpha<-1$ or $\alpha>1/3$….call this relation III. Now, relations I, II and III hold simultaneously if $-3/2 < \alpha <-1$ or $1/2<\alpha<1$. Problem 3: A variable straight line of slope 4 intersects the hyperbola $xy=1$ at two points. Find the locus of the point which divides the line segment between these two points in the ratio $1:2$. Solution 3: Let equation of the line be $y=4x+c$ where c is a parameter. It intersects the hyperbola $xy=1$ at two points, for which $x(4x+c)=1$, that is, $\Longrightarrow 4x^{2}+cx-1=0$. Let $x_{1}$ and $x_{2}$ be the roots of the equation. Then, $x_{1}+x_{2}=-c/4$ and $x_{1}x_{2}=-1/4$. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are $(x_{1}, \frac{1}{x_{1}})$ and that of B are $(x_{2}, \frac{1}{x_{2}})$. Let $R(h,k)$ be the point which divides AB in the ratio $1:2$, then $h=\frac{2x_{1}+x_{2}}{3}$ and $k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}$, that is, $\Longrightarrow 2x_{1}+x_{2}=3h$…call this equation I. and $x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k$….call this equation II. Adding I and II, we get $3(x_{1}+x_{2})=3(h-\frac{k}{4})$, that is, $3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}$….call this equation III. Subtracting II from I, we get $x_{1}-x_{2}=3(h+\frac{k}{4})$ $\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}$ $\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}$ $\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}$ $\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})$ $\Longrightarrow 16h^{2}+10hk+k^{2}-2=0$ so that the locus of $R(h,k)$ is $16x^{2}+10xy+y^{2}-2=0$ More later, Nalin Pithwa. ### Cartesian system and straight lines: IITJEE Mains: Problem solving skills Problem 1: The line joining $A(b\cos{\alpha},b\sin{\alpha})$ and $B(a\cos{\beta},a\sin{\beta})$ is produced to the point $M(x,y)$ so that $AM:MB=b:a$, then find the value of $x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}$. Solution 1: As M divides AB externally in the ratio $b:a$, we have $x=\frac{b(a\cos{\beta})-a(b\cos{\alpha})}{b-a}$ and $y=\frac{b(a\sin{\beta})-a(b\sin{\alpha})}{b-a}$ which in turn $\Longrightarrow \frac{x}{y} = \frac{\cos{\beta}-cos{\alpha}}{\sin{\beta}-\sin{\alpha}}$ $= \frac{2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}}{2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\beta-\alpha}{2}}}$ $\Longrightarrow x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}=0$ Problem 2: If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points $(a^{2}+1,a^{2}+1)$ and $(2a,-2a)$, then where does the orthocentre lie? Solution 2: From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre $(0,0)$ and the centroid $(\frac{(a+1)^{2}}{2},\frac{(a-1)^{2}}{2})$, that is, $y.\frac{(a+1)^{2}}{2} = x.\frac{(a-1)^{2}}{2}$, or $(a-1)^{2}x-(a+1)^{2}y=0$. That is, the orthocentre lies on this line. Problem 3: If a, b, c are unequal and different from 1 such that the points $(\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1})$, $(\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1})$ and $(\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1})$ are collinear, then which of the following option is true? a: $bc+ca+ab+abc=0$ b: $a+b+c=abc$ c: $bc+ca+ab=abc$ d: $bc+ca+ab-abc=3(a+b+c)$ Solution 3: Suppose the given points lie on the line $lx+my+n=0$ then a, b, c are the roots of the equation : $lt^{3}+m(t^{2}-3)+n(t-1)=0$, or $lt^{3}+mt^{2}+nt-(3m+n)=0$ $\Longrightarrow a+b+c=-\frac{m}{l}$ and $ab+bc+ca=\frac{n}{l}$, that is, $abc=(3m+n)/l$ Eliminating l, m, n, we get $abc=-3(a+b+c)+bc+ca+ab$ $\Longrightarrow bc+ca+ab-abc=3(a+b+c)$, that is, option (d) is the answer. Problem 4: If $p, x_{1}, x_{2}, \ldots, x_{i}, \ldots$ and $q, y_{1}, y_{2}, \ldots, y_{i}, \ldots$ are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points $A_{i}(x_{i},y_{i})$ with $i=1,2,3 \ldots, n$ lie? Solution 4: Note: Centre of Mean Position is $(\frac{\sum{xi}}{n},\frac{\sum {yi}}{n})$. Let the coordinates of the centre of mean position of the points $A_{i}$, $i=1,2,3, \ldots,n$ be $(x,y)$ then $x=\frac{x_{1}+x_{2}+x_{3}+\ldots + x_{n}}{n}$ and $y=\frac{y_{1}+y_{2}+\ldots + y_{n}}{n}$ $\Longrightarrow x = \frac{np+a(1+2+\ldots+n)}{n}$, $y=\frac{nq+b(1+2+\ldots+n)}{n}$ $\Longrightarrow x=p+ \frac{n(n+1)}{2n}a$ and $y=q+ \frac{n(n+1)}{2n}b$ $\Longrightarrow x=p+\frac{n+1}{2}a$, and $y=q+\frac{n+1}{2}b$ $\Longrightarrow 2\frac{(x-p)}{a}=2\frac{(y-q)}{b} \Longrightarrow bx-ay=bp-aq$, that is, the CM lies on this line. Problem 5: The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}$? Solution 5: Equation of the line L in the two coordinate systems is $\frac{x}{a} + \frac{y}{b}=1$, and $\frac{X}{p} + \frac{Y}{q}=1$ where $(X,Y)$ are the new coordinate of a point $(x,y)$ when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed. $\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}}$ $\Longrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}$ or $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}=0$. So, the value is zero. Problem 6: Let O be the origin, $A(1,0)$ and $B(0,1)$ and $P(x,y)$ are points such that $xy>0$ and $x+y<1$, then which of the following options is true: a: P lies either inside the triangle OAB or in the third quadrant b: P cannot lie inside the triangle OAB c: P lies inside the triangle OAB d: P lies in the first quadrant only. Solution 6: Since $xy>0$, P either lies in the first quadrant or in the third quadrant. The inequality $x+y<1$ represents all points below the line $x+y=1$. So that $xy>0$ and $x+y<1$ imply that either P lies inside the triangle OAB or in the third quadrant. Problem 7: An equation of a line through the point $(1,2)$ whose distance from the point $A(3,1)$ has the greatest value is : option i: $y=2x$ option ii: $y=x+1$ option iii: $x+2y=5$ option iv: $y=3x-1$ Solution 7: Let the equation of the line through $(1,2)$ be $y-2=m(x-1)$. If p denotes the length of the perpendicular from $(3,1)$ on this line, then $p=|\frac{2m+1}{\sqrt{m^{2}+1}}|$ $\Longrightarrow p^{2}=\sqrt{\frac{4m^{2}+4m+1}{m^{2}+1}}=4+ \frac{4m-3}{m^{2}+1}=s$, say then $p^{2}$ is greatest if and only if s is greatest. Now, $\frac{ds}{dm} = \frac{(m^{2}+1)(4)-2m(4m-3)}{(m^{2}+1)^{2}} = \frac{-2(2m-1)(m-2)}{(m^{2}+1)^{2}}$ $\frac{ds}{dm} = 0$ so that $\Longrightarrow m = \frac{1}{2}, 2$. Also, $\frac{ds}{dm}<0$, if $m<\frac{1}{2}$, and $\frac{ds}{dm} >0$, if $1/2 and $\frac{ds}{dm} <0$, if $m>2$. So s is greatest for $m=2$. And, thus, the equation of the required line is $y=2x$. Problem 8: The points $A(-4,-1)$, $B(-2,-4)$, Slatex C(4,0)$ and $D(2,3)$ are the vertices of a :

option a: parallelogram

option b: rectangle

option c: rhombus

option d: square.

Note: more than one option may be right. Please mark all that are right.

Solution 8:

Mid-point of AC = $(\frac{-4+4}{2},\frac{-1+0}{2})=(0, \frac{-1}{2})$

Mid-point of BD = $(\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})$

$\Longrightarrow$ the diagonals AC and BD bisect each other.

$\Longrightarrow$ ABCD is a parallelogram.

Next, $AC= \sqrt{(-4-4)^{2}+(-1+0)^{2}}=\sqrt{64+1}=\sqrt{65}$ and $BD=\sqrt{(-2-2)^{2}+(-4+3)^{2}}=\sqrt{16+49}=\sqrt{65}$ and since the diagonals are also equal, it is a rectangle.

As $AB=\sqrt{(-4+2)^{2}+(-1+4)^{2}}=\sqrt{13}$ and $BC=\sqrt{(-2-4)^{2}+(-4)^{2}}=\sqrt{36+16}=sqrt{52}$, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.

Problem 9:

Equations $(b-c)x+(c-a)y+(a-b)=0$ and $(b^{3}-c^{3})x+(c^{3}-a^{3})y+a^{3}-b^{3}=0$ will represent the same line if

option i: $b=c$

option ii: $c=a$

option iii: $a=b$

option iv: $a+b+c=0$

Solution 9:

The two lines will be identical if there exists some real number k, such that

$b^{3}-c^{3}=k(b-c)$, and $c^{3}-a^{3}=k(c-a)$, and $a^{3}-b^{3}=k(a-b)$.

$\Longrightarrow b-c=0$ or $b^{2}+c^{2}+bc=k$

$\Longrightarrow c-a=0$ or $c^{2}+a^{2}+ac=k$, and

$\Longrightarrow a-b=0$ or $a^{2}+b^{2}+ab=k$

That is, $b=c$ or $c=a$, or $a=b$.

Next, $b^{2}+c^{2}+bc=c^{2}+a^{2}+ca \Longrightarrow b^{2}-a^{2}=c(a-b)$. Hence, $a=b$, or $a+b+c=0$.

Problem 10:

The circumcentre of a triangle with vertices $A(a,a\tan{\alpha})$, $B(b, b\tan{\beta})$ and $C(c, c\tan{\gamma})$ lies at the origin, where $0<\alpha, \beta, \gamma < \frac{\pi}{2}$ and $\alpha + \beta + \gamma = \pi$. Show that it’s orthocentre lies on the line $4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}x-4\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}y=y$

Solution 10:

As the circumcentre of the triangle is at the origin O, we have $OA=OB=OC=r$, where r is the radius of the circumcircle.

Hence, $OA^{2}=r^{2} \Longrightarrow a^{2}+a^{2}\tan^{2}{\alpha}=r^{2} \Longrightarrow a = r\cos{\alpha}$

Therefore, the coordinates of A are $(r\cos{\alpha},r\sin{\alpha})$. Similarly, the coordinates of B are $(r\cos{\beta},r\sin{\beta})$ and those of C are $(r\cos{\gamma},r\sin{\gamma})$. Thus, the coordinates of the centroid G of $\triangle ABC$ are

$(\frac{1}{3}r(\cos{\alpha}+\cos{\beta}+\cos{\gamma}),\frac{1}{3}r(\sin{\alpha}+\sin{\beta}+\sin{\gamma}))$.

Now, if $P(h,k)$ is the orthocentre of $\triangle ABC$, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.

Hence, $\frac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{\cos{\alpha}+\cos{\beta}+\cos{\gamma}}=\frac{k}{h}$

$\Longrightarrow \frac{4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}}{1+4\sin{(\frac{\alpha}{2})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}}= \frac{k}{h}$

because $\alpha+\beta+\gamma=\pi$.

Hence, the orthocentre $P(h,k)$ lies on the line

$4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}x-4\sin{(\frac{}{})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}y=y$.

Hope this gives an assorted flavour. More stuff later,

Nalin Pithwa.

### 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:

1. An equation of a line parallel to the x-axis is $y=k$ and that of the x-axis itself is $y=0$.
2. An equation of a line parallel to the y-axis is $x=h$ and that of the y-axis itself is $x=0$.
3. 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 $x_{1}y=y_{1}x$.
4. Slope-intercept form: An equation of a line with slope m and making an intercept c on the y-axis is $y=mx+c$.
5. Point-slope form: An equation of a line with slope m and passing through $(x_{1},y_{1})$ is $y-y_{1}=x-x_{1}$.
6. Two-point form: An equation of a line passing through the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $\frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}$.
7. Intercept form: An equation of a line making intercepts a and b on the x-axis and y-axis respectively, is $\frac{x}{a} + \frac{y}{b}=1$.
8. 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 $y=y_{1}+r\sin{\theta}$.
9. 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 $x\cos{\alpha}+y\sin{\alpha}=p$.
10. 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:

1. Two lines given by the equations $ax+by+c=0$ and $a^{'}x+b^{'}y+c^{'}=0$ 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.
2. 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 $\frac{|c-c^{'}|}{\sqrt{a^{2}+b^{2}}}$
3. 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 $\left | \begin{array}{ccc} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} \right|=0$
4. 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.
5. 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

1. parallelogram: show that the diagonals AC and BD bisect each other.
2. rhombus: show that the diagonals AC and BD bisect each other and a pair of adjacent sides, say, AB and BC are equal.
3. 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.
4. 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:

1. 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 $Y=y\cos{\theta}-x\sin{\theta}$.
2. 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.