## 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

$(y-\frac{m_{1}}{l+mm_{1}})=-\frac{1}{m_{2}}(x-\frac{1}{l+mm_{1}})$

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

$y=\frac{(1+m_{1}m_{2})m}{l^{2}+lm(m_{1}+m_{2})+m^{2}m_{1}m_{2}}$

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

$ax+by=\frac{1}{k}=\frac{ab(a+b-2h)}{a+b}$

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

More later,

Nalin Pithwa.

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