**Question I:**

The point undergoes the following transformations, successively:

a) reflection about the line .

b) translation through a distance 2 units along the positive directions of the x-axis.

c) rotation through an angle of about the origin in the anticlockwise direction.

d) reflection about .

**Hint: draw the diagrams at very step!**

Ans:

**Question 2:**

are n points in a plane whose co-ordinates are , , , respectively. , is bisected at the point , is divided in the ratio 1:2 at , is divided in the ratio at , is divided in the ratio at and so on until all n points are exhausted. Show that the co-ordinates of the final point so obtained are

.

**Solution 2:**

The co-ordinates of are .

Now, divides in the ratio . Hence, the co-ordinates of are

, or .

Again, divides in the ratio . Therefore, the co-ordinates of are , or

.

Proceeding in this manner,we can show that the coordinates of the final point obtained will be

.

**Remark: For a rigorous proof, prove the above by mathematical induction.**

**Question 3:**

A line L intersects the three sides BC, CA, and AB of a triangle ABC at P, Q and R, respectively. Show that

**Solution 3:**

Let , , and be the vertices of , and let be equation of the line L. If P divides BC in the ratio , then the coordinates of P are .

Also, as P lies on L, we have

…..call this relation I.

Similarly, we can obtain ….call this relation II.

and so, also, we can prove that …call this III.

Multiplying, I, II and III, we get the desired result.

**The above is the famous Menelaus’s theorem of plane geometry proved with elementary tools of co-ordinate geometry. As a homework quiz, try proving the equally famous Ceva’s theorem of plane geometry with elementary tools of co-ordinate geometry.**

**Question 4:**

A triangle has the lines and as two of its sides, with and being roots of the equation . If is the orthocentre of the triangle, show that the equation of the third side is .

**Solution 4:**

Since the given lines intersect at the origin, one of the triangle lies at the origin O(0,0). Let OA and OB be the given lines and , respectively. Let the equation of AB be . Now, as OH is perpendicular to AB, we have

, , say…call this equation I

Also, the coordinates of A and B are respectively,

and

Therefore, the equation of AB is

or …call this II.

Similarly, the equation of BH is ….call this III.

Solving II and III, we get the coordinates of H. Subtracting III from II, we get

Since and are the roots of the equation , we have and .

because y=b for H.

.

Hence, the equation of AB is

More later,

*Nalin Pithwa.*