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Awesome inspiration for Mathematics !
The Early History of Calculus Problems, II: AMS feature column
The Early History of Calculus Problems: AMS Feature column
Vladimir Voevodsky: Obit: The Washington Post
Fun with Math websites: MAA
Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II
I have a collection of some “random”, yet what I call ‘beautiful” questions in Coordinate 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 , , , , and on it a point R is taken such that
Show that the locus of R is a straight line.
Solution 1:
Let equations of the given lines be , , and the point O be the origin .
Then, the equation of the line through O can be written as where is the angle made by the line with the positive direction of xaxis and r is the distance of any point on the line from the origin O.
Let be the distances of the points from O which in turn and , where .
Then, coordinates of R are and of are where .
Since lies on , we can say for
, for
…as given…
Hence, the locus of R is which is a straight line.
Problem 2:
Determine all values of for which the point lies inside the triangle formed by the lines , , .
Solution 2:
Solving equations of the lines two at a time, we get the vertices of the given triangle as: , and .
So, AB is the line , AC is the line and BC is the line
Let 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 , both and must have the same sign.
or which in turn which in turn either or ….call this relation I.
Again, since B and P lie on the same side of the line , and have the same sign.
and , that is, …call this relation II.
Lastly, since C and P lie on the same side of the line , we have and have the same sign.
that is
or ….call this relation III.
Now, relations I, II and III hold simultaneously if or .
Problem 3:
A variable straight line of slope 4 intersects the hyperbola at two points. Find the locus of the point which divides the line segment between these two points in the ratio .
Solution 3:
Let equation of the line be where c is a parameter. It intersects the hyperbola at two points, for which , that is, .
Let and be the roots of the equation. Then, and . If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are and that of B are .
Let be the point which divides AB in the ratio , then and , that is, …call this equation I.
and ….call this equation II.
Adding I and II, we get , that is,
….call this equation III.
Subtracting II from I, we get
so that the locus of is
More later,
Nalin Pithwa.
Prioritize your passions and commitments, says Dr. Shawna Pandya
Kids nowadays need counselling for a choice of career. In my humble opinion, excellence in any field of knowledge/human endeavour gives deep satisfaction as well as a means of livelihood. But, I am a mere mortal, most of my students are bright, ambitious, multitalented and hardworking. I would like to present to them the views of one of my “idols”, though not a mathematician…
Dr. Shawna Pandya:
Hats off to Dr. Shawna Pandya, belated though from me…:)
— Nalin Pithwa.
Cartesian system and straight lines: IITJEE Mains: Problem solving skills
Problem 1:
The line joining and is produced to the point so that , then find the value of .
Solution 1:
As M divides AB externally in the ratio , we have and which in turn
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 and , 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 and the centroid , that is, , or . That is, the orthocentre lies on this line.
Problem 3:
If a, b, c are unequal and different from 1 such that the points , and are collinear, then which of the following option is true?
a:
b:
c:
d:
Solution 3:
Suppose the given points lie on the line then a, b, c are the roots of the equation :
, or
and , that is,
Eliminating l, m, n, we get
, that is, option (d) is the answer.
Problem 4:
If and are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points with lie?
Solution 4:
Note: Centre of Mean Position is .
Let the coordinates of the centre of mean position of the points , be then
and
,
and
, and
, 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 ?
Solution 5:
Equation of the line L in the two coordinate systems is , and where are the new coordinate of a point 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.
or . So, the value is zero.
Problem 6:
Let O be the origin, and and are points such that and , 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 , P either lies in the first quadrant or in the third quadrant. The inequality represents all points below the line . So that and imply that either P lies inside the triangle OAB or in the third quadrant.
Problem 7:
An equation of a line through the point whose distance from the point has the greatest value is :
option i:
option ii:
option iii:
option iv:
Solution 7:
Let the equation of the line through be . If p denotes the length of the perpendicular from on this line, then
, say
then is greatest if and only if s is greatest.
Now,
so that . Also, , if , and
, if
and , if . So s is greatest for . And, thus, the equation of the required line is .
Problem 8:
The points , , Slatex C(4,0)$ and 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:
Midpoint of AC =
Midpoint of BD =
the diagonals AC and BD bisect each other.
ABCD is a parallelogram.
Next, and and since the diagonals are also equal, it is a rectangle.
As and , the adjacent sides are not equal and hence, it is neither a rhombus nor a square.
Problem 9:
Equations and will represent the same line if
option i:
option ii:
option iii:
option iv:
Solution 9:
The two lines will be identical if there exists some real number k, such that
, and , and .
or
or , and
or
That is, or , or .
Next, . Hence, , or .
Problem 10:
The circumcentre of a triangle with vertices , and lies at the origin, where and . Show that it’s orthocentre lies on the line
Solution 10:
As the circumcentre of the triangle is at the origin O, we have , where r is the radius of the circumcircle.
Hence,
Therefore, the coordinates of A are . Similarly, the coordinates of B are and those of C are . Thus, the coordinates of the centroid G of are
.
Now, if is the orthocentre of , 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,
because .
Hence, the orthocentre lies on the line
.
Hope this gives an assorted flavour. More stuff later,
Nalin Pithwa.
Frenchmen and mathematics ! :) :) :)
Mathematicians are like Frenchmen: whatever you tell to them they translate in their own language and forthwith it is something entirely different. — GOETHE.
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