Tag Archives: Ceva’s theorem

Some random sample problems-solutions in Coordinate Geometry: I: IITJEE Mains Maths tutorials

Question I:

The point (4,1) undergoes the following transformations, successively:

a) reflection about the line y=x.

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

c) rotation through an angle of \pi/4 about the origin in the anticlockwise direction.

d) reflection about x=0.

Hint: draw the diagrams at very step!

Ans: (1/\sqrt{2}, 7/\sqrt{2})

Question 2:

A_{1}, A_{2}, A_{3}, \ldots, A_{n} are n points in a plane whose co-ordinates are (x_{1}, y_{1}), (x_{2},y_{2}), \ldots, (x_{n},y_{n}) respectively. A_{1}, A_{2} is bisected at the point G_{1}, G_{1}A_{3} is divided in the ratio 1:2 at G_{2}, G_{2}A_{4} is divided in the ratio 1:3 at G_{3}, G_{3}A_{3} is divided in the ratio 1:4 at G_{4} and so on until all n points are exhausted. Show that the co-ordinates of the final point so obtained are

(\frac{1}{n}(x_{1}+x_{2}+ \ldots + x_{n}) , \frac{1}{n}(y_{1}+y_{2}+ \ldots + y_{n}) ).

Solution 2:

The co-ordinates of G_{1} are (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}).

Now, G_{2} divides G_{1}A_{3} in the ratio 1:2. Hence, the co-ordinates of G_{2} are

( \frac{1}{3}(\frac{2(x_{1}+x_{2})}{2}+x_{3}), \frac{1}{3}(\frac{3(y_{1}+y_{2})}{2}+y_{3})), or (\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}).

Again, G_{3} divides G_{2}A_{4} in the ratio 1:4. Therefore, the co-ordinates of G_{3} are (\frac{1}{4}(\frac{3(x_{1}+x_{2}+x_{3})}{3}+x_{4}) ,\frac{1}{4}(\frac{3(y_{1}+y_{2}+y_{3})}{3}+y_{4}) ), or

( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4} ).

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

(\frac{1}{n}(x_{1}+x_{2}+x_{3}+\ldots + x_{n}),\frac{1}{n}(y_{1}+y_{2}+y_{3}+\ldots + y_{n})).

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 \frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1

Solution 3:

Let A(x_{1},y_{1}), B(x_{2},y_{2}), and C(x_{3},y_{3}) be the vertices of \triangle ABC, and let lx+my+n=0 be equation of the line L. If P divides BC in the ratio \lambda:1, then the coordinates of P are (\frac{\lambda x_{3}+x_{2}}{\lambda + 1} ,\frac{\lambda y_{3}+y_{2}}{\lambda + 1}).

Also, as P lies on L, we have l(\frac{\lambda x_{3}+x_{2}}{\lambda + 1})+m(\frac{\lambda y_{3}+y_{2}}{\lambda + 1})+n=0

\Longrightarrow \frac{lx_{2}+my_{2}+n}{lx_{3}+my_{3}+n}=\lambda=\frac{BP}{PC}…..call this relation I.

Similarly, we can obtain \frac{CQ}{QA}=-\frac{lx_{3}+my_{3}+n}{lx_{1}+my_{1}+n}….call this relation II.

and so, also, we can prove that \frac{AR}{RB}=-\frac{lx_{1}+my_{1}+n}{lx_{2}+my_{2}+n}…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 y=m_{1}x and y=m_{2}x as two of its sides, with m_{1} and m_{2} being roots of the equation bx^{2}+2hx+a=0. If H(a,b) is the orthocentre of the triangle, show that the equation of the third side is (a+b)(ax+by)=ab(a+b-2h).

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 y=m_{1}x and y=m_{2}x, respectively. Let the equation of AB be lx+my=1. Now, as OH is perpendicular to AB, we have

\frac{b}{a}=\frac{m}{l}, \Longrightarrow \frac{l}{a}=\frac{m}{b}=k, say…call this equation I

Also, the coordinates of A and B are respectively,

(\frac{1}{l+mm_{1}}, \frac{m_{1}}{l+mm_{1}}) and (\frac{1}{l+mm_{2}} , \frac{m_{2}}{l+mm_{2}})

Therefore, the equation of AB is


or x+m_{2}y=\frac{1+m_{1}m_{2}}{1+mm_{1}}…call this II.

Similarly, the equation of BH is x+m_{1}y=\frac{1+m_{1}m_{2}}{1+mm_{2}}….call this III.

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


Since m_{1} and m_{2} are the roots of the equation bx^{2}+2hx+a=0, we have m_{1}+m_{2}=-\frac{2h}{b} and m_{1}m_{2}=a/b.

\Longrightarrow y=\frac{(a+b)m}{bl^{2}-2hlm+am^{2}} \Longrightarrow \frac{m}{b}=\frac{bl^{2}-2hlm+am^{2}}{a+b} because y=b for H.

\Longrightarrow k=\frac{k^{2}(ba^{2}-2hab+ab^{2})}{a+b} \Longrightarrow k=\frac{a+b}{ab(a-2h+b)}.

Hence, the equation of AB is


\Longrightarrow (a+b)(ax+by)=ab(a+b-2h)

More later,

Nalin Pithwa.