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*cheers to MAA! *

Nalin Pithwa.

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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 x-axis 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.

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Dr. Shawna Pandya:

Hats off to Dr. Shawna Pandya, belated though from me…:-)

— Nalin Pithwa.

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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:

Mid-point of AC =

Mid-point 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.

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