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