Category Archives: KVPY

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<m<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.

IITJEE Foundation Math and PRMO (preRMO) practice: another random collection of questions

Problem 1: Find the value of \frac{x+2a}{2b--x} + \frac{x-2a}{2a+x} + \frac{4ab}{x^{2}-4b^{2}} when x=\frac{ab}{a+b}

Problem 2: Reduce the following fraction to its lowest terms:

(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \div (\frac{x+y+z}{x^{2}+y^{2}+z^{2}-xy-yz-zx} - \frac{1}{x+y+z})+1

Problem 3: Simplify: \sqrt[4]{97-56\sqrt{3}}

Problem 4: If a+b+c+d=2s, prove that 4(ab+cd)^{2}-(a^{2}+b^{2}-c^{2}-d^{2})^{2}=16(s-a)(s-b)(s-c)(s-d)

Problem 5: If a, b, c are in HP, show that (\frac{3}{a} + \frac{3}{b} - \frac{2}{c})(\frac{3}{c} + \frac{3}{b} - \frac{2}{a})+\frac{9}{b^{2}}=\frac{25}{ac}.

May u discover the joy of Math! 🙂 🙂 🙂

Nalin Pithwa.

Can anyone have fun with infinite series?

Below is list of finitely many puzzles on infinite series to keep you a bit busy !! 🙂 Note that these puzzles do have an academic flavour, especially concepts of convergence and divergence of an infinite series.

Puzzle 1: A grandmother’s vrat (fast) requires her to keep odd number of lamps of finite capacity lit in a temple at any time during 6pm to 6am the next morning. Each oil-filled lamp lasts 1 hour and it burns oil at a constant rate. She is not allowed to light any lamp after 6pm but she can light any number of lamps before 6pm and transfer oil from some to the others throughout the night while keeping odd number of lamps lit all the time. How many fully-filled oil lamps does she need to complete her vrat?

Puzzle 2: Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 liters of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that p+2 is also a prime number. Then, the first number theorist would pipette out \frac{1}{p} litres of chemical and the second \frac{1}{(p+2)} litres. How many times do they have to play this game to empty the flask completely?

Puzzle 3: How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Puzzle 4: Imagine a tank that can be filled with infinite taps and can be emptied with infinite drains. The taps, turned on alone, can fill the empty tank to its full capacity in 1 hour, 3 hours, 5 hours, 7 hours and so on. Likewise, the drains opened alone, can drain a full tank in 2 hours, 4 hours, 6 hours, and so on. Assume that the taps and drains are sequentially arranged in the ascending order of their filling and emptying durations.

Now, starting with an empty tank, plumber A alternately turns on a tap for 1 hour and opens the drain for 1 hour, all operations done one at a time in a sequence. His sequence, by using t_{i} for i^{th} tap and d_{j} for j^{th} drain, can be written as follows: \{ t_{1}, d_{1}, t_{2}, d_{2}, \ldots\}_{A}.

When he finishes his operation, mathematically, after using all the infinite taps and drains, he notes that the tank is filled to a certain fraction, say, n_{A}<1.

Then, plumber B turns one tap on for 1 hour and then opens two drains for 1 hour each and repeats his sequence: \{ (t_{1},d_{1},d_{2}), (t_{2},d_{3},d_{4}), (t_{3},d_{4},d_{5}) \ldots \}_{B}.

At the end of his (B’s) operation, he finds that the tank is filled to a fraction that is exactly half of what plumber A had filled, that is, 0.5n_{A}.

How is this possible even though both have turned on all taps for 1 hour and opened all drains for 1 hour, although in different sequences?

I hope u do have fun!!

-Nalin Pithwa.

Pair of straight lines — quick review for IITJEE Math

Pair of Straight Lines:

I: Equation of Family of Lines:

A first degree equation ax+by+c=0 represents a straight line involving three constants a, b and c, which can be reduced to two by dividing both the sides of the equation by a non-zero constant. For instance, if a \neq 0 we can write the equation as

x+\frac{b}{a}y+\frac{c}{a}=0 or x+By+C=0

So, in order to determine an equation of a line, we need two conditions on the line to determine these constants. For instance, if we know two points on the line or a point on the line and its slope etc., we know the line. But, if we know just one condition, we have infinite number of lines satisfying the given condition. In this case, the equation of the line contains an arbitrary constant and for different values of the constant we have different lines satisfying the given condition and the constant is called a parameter.

Some Equations of Family of Lines:

  1. Family of lines with given slope. (Family of given lines) y = mx +k, where k is a parameter represents a family of lines in which each line has slope m.
  2. Family of lines through a point: y-y_{0}=k(x-x_{0}), where k is a parameter represents a family of lines in which each line passes through the point (x_{0},y_{0}).
  3. Family of lines parallel to a given line: ax + by + k=0, where k is a parameter represents a family of lines which are parallel to the line ax+by+c=0.
  4. Family of lines through intersection of two given lines:
    a_{1}x+b_{1}y+c_{1}+k(a_{2}x+b_{2}y+c_{2})=0, where k is a parameter represents a family of lines, in which each line passes through the point of intersection of two intersecting llines a_{1}x+b_{1}y+c_{1}=0 and a_{2}x+b_{2}y+c_{2}=0.
  5. Family of lines perpendicular to a given line:
    bx-ay+c=0 where k is a parameter represents a family of lines, in which each line is perpendicular to the line ax+by+c=0.
  6. Family of lines making a given intercept on axes:
    (a) \frac{x}{a}+\frac{y}{k}=1, where k is a parameter, represents a family of lines, in which each line makes an intercept a on the axis of x. and,

6(b) \frac{x}{k} + \frac{y}{b}=1, where k is a parameter, represents a family of lines, in which each line makes an intercept b on the axis of y.

7. Family of lines at a constant distance from the origin:

x\cos {\alpha}+y\sin{\alpha}=p, where \alpha is a parameter, represents a family of lines, in which each line is at a distance p from the origin.

II: Pair of Straight Lines:

A) Pair of straight lines through the origin: A homogeneous equation of the second degree ax^{2}+2hxy+by^{2}=0…call this equation 1; represents a pair of straight lines passing through the origin if and only if h^{2}-ab \geq 0. If b \neq 0, the given equation can be written as

y^{2}+\frac{2h}{b}xy+\frac{a}{b}x^{2}=0, or (\frac{y}{x})^{2}+\frac{2h}{b}(\frac{y}{x})+\frac{a}{b}=0, which, being quadratic in y/x, gives two values of y/x, say m_{1} and m_{2} and hence, the equations y=m_{1}x and y=m_{2}x of two straight lines passing through the origin. The slopes m_{1} and m_{2} of these straight lines are given by the relations

m_{1}+m_{2}=-\frac{2h}{b}…call this equation (2).

m_{1}m_{2}=\frac{a}{b}…call this equation (3).

B) Angle between the lines: represented by ax^{2}+2hxy+by^{2}=0. If \theta is an angle, between the lines represented by (1), then

\tan{\theta} = \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\frac{\pm\sqrt{(m_{1}+m_{2})^{2}-4m_{1}m_{2}}}{1+m_{1}m_{2}} = \frac{\pm \sqrt{4h^{2}-4ab}}{a+b}

\Longrightarrow \theta = \arctan (\frac{\pm 2\sqrt{h^{2}-ab}}{a+b})

If (a+b)=0, the lines are perpendicular, and if h^{2}=ab, then the lines are coincident.

C) Equation of the bisectors of the angles between the lines ax^{2}+2hxy=by^{2}=0.

Let y=m_{1}x and y=m_{2}x be the equation of the straight lines represented by the given equation. Then, equations of the angle bisectors are:

\frac{y-m_{1}x}{\sqrt{1+m_{1}^{2}}} = \frac{y-m_{2}x}{\sqrt{1+m_{1}^{2}}}

The joint equation of  these bisectors can be written as

(\frac{y-m_{1}x}{\sqrt{1+m_{1}^{2}}} + \frac{y-m_{2}x}{1+m_{2}^{2}})(\frac{y-m_{1}x}{1+m_{1}^{2}} - \frac{y-m_{2}x}{1+m_{2}^{2}})=0

or, (1+m_{2}^{2})(y-m_{1}x)^{2} - (1+m_{1}^{2})(y-m_{2}x)^{2}=0

or, x^{2}(m_{1}^{2}-m_{2}^{2}) - 2xy(m_{1}-m_{2})(1-m_{1}m_{2}) + y^{2}(m_{2}^{2}-m_{1}^{2})=0

or, \frac{x^{2}-y^{2}}{2xy} = \frac{1-m_{1}m_{2}}{m_{1}+m_{2}} = \frac{a-b}{2h}….this comes from (2) and (3).

or, \frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}

D) The general equation of second degree:

ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 represents a pair of straight lines if and only if

abc + 2fgh - af^{2} -bg^{2} - ch^{2} =0, that is, if

\left | \begin{array}{ccc}    a & h & g \\    h & b & f \\    g & f & c \end{array} \right |=0

If h^{2}=ab, the co-ordinates (x_{1},y_{1}) of the point of intersection of these lines are obtained by solving the equations

ax_{1}+hy_{1}+x=0 and hx_{1}+by_{1}+f=0

\Longrightarrow x_{1}=\frac{hf-bg}{ab-h^{2}} and y_{1}=\frac{gh-af}{ab-h^{2}}

E) Equation ax^{2}+2hxy +by^{2}=0 represents a pair of straight lines through the origin parallel to the lines given by the general equation of second degree given in (4) and hence, the angle between the lines given in (4) is same as in (2), that is,

\theta = \arctan {\frac{\pm {h^{2}-ab}}{a+b}}

so that if a+b=0, the lines are perpendicular and if h^{2}=ab, the lines are parallel.

F) Equation of the lines joining the origin to the points of intersection of a line and a conic:

Let L \equiv lx + my + n=0

and S \equiv ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c=0

be the equations of a line and a conic, respectively. Writing the equation of the line as

\frac{lx + my}{-n} = 1 and making S=0 homogeneous with its help, we get S=ax^{2}+2hxy+by^{2}+2(gx +fy)

(\frac{lx+my}{-n}) + c(\frac{lx+my}{-n})^{2}=0, which being a homogeneous equation of second degree, represents a pair of straight lines through the origin and passing through the points common to S=0 and L=0.

G) Some important results regarding pair of lines:

(i) Equation of the pair of lines through the origin perpendicular to the pair of lines ax^{2}+2hxy+by^{2}=0 is bx^{2}+2hxy +ay^{2}=0.

(ii) The product of the perpendicular lines drawn from the point (x_{1}, y_{1}) on the lines ax^{2}+2hxy+by^{2}=0 is |\frac{ax_{1}^{2}+2hx_{1}y_{1}+by_{1}^{2}}{\sqrt{(a-b)^{2}+4h^{2}}}|

(iii) The product of the perpendiculars drawn from the origin to the lines ax^{2}+2hxy + by^{2}+2gx+2fy+c=0 is \frac{c}{\sqrt{(a-b)^{2}+4h^{2}}}

(iv) Equations of the bisectors of the angles between the lines represented by ax^{2}+2hxy + by^{2} +2gx + 2fy + c=0 are given by

\frac{(x-x_{1})^{2}-(y-y_{1})^{2}}{a-b} = \frac{(x-x_{1})(y-y_{1})}{h} where (x_{1}, y_{1}) is the point of intersection of the lines represented by the given equation.

(v) If ax^{2}+2hxy + by^{2}+2gx + 2fy + c=0 represents two parallel straight lines, then the distance between them is 2\sqrt{\frac{g^{2}-ac}{a(a+b)}}

Every nth degree homogeneous equation a_{0}x^{n}+a_{1}x^{n-1}y + a_{2}x^{n-2}y + \ldots + a_{n-1}xy^{n-1}+a_{n}y^{n}=0, then

m_{1}+m_{2}+ \ldots + m_{n} = - \frac{a_{n-1}}{a_{n}}

and m_{1}m_{2}\ldots m_{n}=(-1)^{n}\frac{a_{0}}{a_{n}}

To be continued later,

-Nalin Pithwa.

Lagrange’s Mean Value Theorem and Cauchy’s Generalized Mean Value Theorem

Lagrange’s Mean Value Theorem:

If a function f(x) is continuous on the interval [a,b] and differentiable at all interior points of the interval, there will be, within [a,b], at least one point c, a<c<b, such that f(b)-f(a)=f^{'}(c)(b-a).

Cauchy’s Generalized Mean Value Theorem:

If f(x) and phi(x) are two functions continuous on an interval [a,b] and differentiable within it, and phi(x) does not vanish anywhere inside the interval, there will be, in [a,b], a point x=c, a<c<b, such that \frac{f(b)-f(a)}{phi(b)-phi(a)} = \frac{f^{'}(c)}{phi^{'}(c)}.

Some questions based on the above:

Problem 1:

Form Lagrange’s formula for the function y=\sin(x) on the interval [x_{1},x_{2}].

Problem 2:

Verify the truth of Lagrange’s formula for the function y=2x-x^{2} on the interval [0,1].

Problem 3:

Applying Lagrange’s theorem, prove the inequalities: (i) e^{x} \geq 1+x (ii) \ln (1+x) <x, for x>0. (iii) b^{n}-a^{n}<ab^{n-1}(b-a) for b>a. (iv) \arctan(x) <x.

Problem 4:

Write the Cauchy formula for the functions f(x)=x^{2}, phi(x)=x^{3} on the interval [1,2] and find c.

More churnings with calculus later!

Nalin Pithwa.

 

 

Some questions based on Rolle’s theorem

Problem 1:

Verify the truth of Rolle’s theorm for the following functions:

(a) y=x^{2}-3x+2 on the interval [1,2].

(b) y=x^{3}+5x^{2}-6x on the interval [0,1].

(c) y=(x-1)(x-2)(x-3) on the interval [1,3].

(d) y=\sin^{2}(x) on the interval [0,\pi].

Problem 2:

The function f(x)=4x^{3}+x^{2}-4x-1 has roots 1 and -1. Find the root of the derivative f^{'}(x) mentioned in Rolle’s theorem.

Problem 3:

Verify that between the roots of the function y=\sqrt[3]{x^{2}-5x+6} lies the root of its derivative.

Problem 4:

Verify the truth of Rolle’s theorem for the function y=\cos^{2}(x) on the interval [-\frac{\pi}{4},+\frac{\pi}{4}].

Problem 5:

The function y=1-\sqrt[5]{x^{4}} becomes zero at the end points of the interval [-1,1]. Make it clear that the derivative of the function does not vanish anywhere in the interval (-1,1). Explain why Rolle’s theorem is NOT applicable here.

Calculus is the fountainhead of many many ideas in mathematics and hence, technology. Expect more beautiful questions on Calculus !

-Nalin Pithwa

Some Applications of Derivatives — Part II

Derivatives in Economics.

Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals.

In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost (c) with respect to a level of production (x), so it is dc/dx.

For example, let c(x) represent the dollars needed needed to produce x tons of steel in one week. It costs more to produce x+h units, and the cost difference, divided by h, is the average increase in cost per ton per week:

\frac{c(x+h)-c(x)}{h}= average increase in cost/ton/wk to produce the next h tons of steel

The limit of this ratio as h \rightarrow 0 is the marginal cost of producing more steel when the current production level is x tons.

\frac{dc}{dx}=\lim_{h \rightarrow 0} \frac{c(x+h)-c(x)}{h}= marginal cost of production

Sometimes, the marginal cost of production is loosely defined to be the extra cost of producing one unit:

\frac{\triangle {c}}{\triangle {x}}=\frac{c(x+1)-c(x)}{1}

which is approximately the value of dc/dx at x. To see why this is an acceptable approximation, observe that if the slope  of c does not change quickly near x, then the difference quotient will be close to its limit, the derivative dc/dx, even if \triangle {x}=1. In practice, the approximation works best for large values of x.

Example: Marginal Cost

Suppose it costs c(x)=x^{3}-6x^{2}+15x  dollars to produce x radiators when 8 to 30 radiators are produced. Your shop currently produces 10 radiators a day. About how much extra cost will it cost to produce one more radiator a day?

Example : Marginal tax rate

To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by USD 1000, you can expect to have to pay an extra USD 280 in income taxes. This does not mean that you pay 28 percent of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes I with respect to income is dT/dI = 0.28. You will pay USD 0.28 out of every extra dollar you earn in taxes. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.

Example: Marginal revenue:

If r(x) = x^{3}-3x^{2}+12x gives the dollar revenue from selling x thousand candy bars, 5<= x<=20, the marginal revenue when x thousand are sold is

r^{'}(x) = \frac{d}{dx}(x^{3}-3x^{2}+12x)=3x^{2}-6x+12.

As with marginal cost, the marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 thousand candy bars a week, you can expect your revenue to increase by about r^{'}(10) = 3(100) -6(10) +12=252 USD, if you increase sales to 11 thousand bars a week.

Choosing functions to illustrate economics.

In case, you are wondering why economists use polynomials of low degree to illustrate complicated phenomena like cost and revenue, here is the rationale: while formulae for real phenomena are rarely available in any given instance, the theory of  economics can still provide valuable guidance. the functions about which theory speaks can often be illustrated with low degree polynomials on relevant intervals. Cubic polynomials provide a good balance between being easy to work with and being complicated enough to illustrate important points.

Ref: Calculus and Analytic Geometry by G B Thomas.

More later,

Nalin Pithwa

 

Could a one-sided limit not exist ?

Here is basic concept of limit :