## Co-ordinate Geometry problems for IITJEE : equations of median, area of a triangle, and circles

Problem I:

If $A(x_{1}, y_{1})$, $B(x_{1}, y_{1})$ and $C(x_{3}, y_{3})$ are the vertices of a triangle ABC, then prove that the equation of the median through A is given by:

$\left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right |=0$

Solution I:

If D is the mid-point of BC, its co-ordinates are $( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} )$

Therefore, equation of the median AD is $\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1\\ \frac{x_{2}+x_{3}}{2} & \frac{y_{2}+y_{3}}{2} & 1 \end{array} \right|=0$, which in turn, implies that,

$\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2}+x_{3} & y_{2}+y_{3} & 2 \end{array}\right |=0$

Now apply the row transformation $R_{3} \rightarrow 2R_{3}$ to the previous determinant. So, we get

$\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right |=0$, using the sum property of determinants.

Hence, the proof.

Problem 2:

If $\triangle_{1}$ is the area of the triangle with vertices $(0,0), (a\tan {\alpha},b\cot{\alpha}), (a\sin{\alpha}, b\cos {\alpha})$, and $\triangle_{2}$ is the area of the triangle with vertices $(a,b)$, $(a\sec^{2}{\alpha}, b\csc^{2}{\alpha})$, and $(a+a\sin^{2}{\alpha}, b+b\cos^{2}{\alpha})$, and $\triangle_{3}$ is the area of the triangle with vertices $( 0, 0)$, $( a\tan{\alpha}, -b\cos{\alpha})$, $(a\sin{\alpha},b\cos{\alpha})$. Then, prove that there is no value of $\alpha$ for which the areas of triangles, $\triangle_{1}$, $\triangle_{2}$ and $\triangle_{3}$ are in GP.

Solution 2:

We have $\triangle_{1}=\frac{1}{2}|\left | \begin{array}{ccc}0 & 0 & 1 \\ a\tan{\alpha} & b\cot {\alpha} & 1 \\ a \sin{\alpha} & b\cos{\alpha} & 1 \end{array}\right ||=\frac{1}{2}ab|\sin{\alpha}-\cos{\alpha}|$, and

$\triangle_{2}=\frac{1}{2}|\left | \begin{array}{ccc}a & b & 1 \\ a\sec^{2}{\alpha} & b\csc^{2}{\alpha} & 1 \\ a + a\sin^{2}{\alpha} & b + b\cos^{2}{\alpha} & 1 \end{array} \right | |$.

Applying the following column transformations to the above determinant, $C_{1} \rightarrow -aC_{3}$ and $C_{2}-bC_{3}$, we get

$\triangle_{2}=\frac{1}{2}ab\left | \begin{array}{ccc}0 & 0 & 1 \\ \tan^{2}{\alpha} & \cot^{2}{\alpha} & 1 \\ \sin^{2}{\alpha} & \cos^{2}{\alpha} & 1 \end{array}\right | = \frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})$ and $\triangle_{3}=\frac{1}{2}|\left | \begin{array}{ccc} 0 & 0 & 1 \\ a\tan{\alpha} & -b\cot{\alpha} & 1 \\ a\sin{\alpha} & b\cos{\alpha} & 1 \end{array} \right | |=\frac{1}{2}ab |\sin {\alpha}+\cos{\alpha}|$

so that $\triangle_{1}\triangle_{3}=\frac{1}{2}ab\triangle_{2}$.

Now, $\triangle_{1}$, $\triangle_{2}$ and $\triangle_{3}$ are in GP, if $\triangle_{1}\triangle_{3}=\triangle_{2}^{2} \Longrightarrow \frac{1}{2}ab\triangle_{2}=\triangle_{2}^{2} \Longrightarrow \triangle_{2}=\frac{1}{2}ab$

$\Longrightarrow \triangle_{2}=\frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})=\frac{1}{2}ab \Longrightarrow (\sin^{2}{\alpha}-\cos^{2}{alpha})=1$, that is,

$\alpha = (2m+1)\pi/2$, where $m \in Z$. But, for this value of $\alpha$, the vertices of the given triangles are not defined. Hence, $\triangle_{1}$, and $\triangle_{2}$ and $\triangle_{3}$ cannot be in GP for any value of $\alpha$.

Problem 3:

Two points P and Q are taken on the line joining the points $A(0,0)$ and $B(3a,0)$ such that $AP=PQ=QB$. Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the length of the tangents from which to the three circles is equal to $b^{2}$, is

(a) $x^{2}+y^{2}-3ax+2a^{2}-b^{2}=0$

(b) $3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0$

(c) $x^{2}+y^{2}-5ax+6a^{2}-b^{2}=0$

(d) $x^{2}+y^{2}-ax-b^{2}=0$.

Ans. b.

Solution 3:

Since $AP=PQ=QB$, the co-ordinates of P are $(a,0)$ and of Q are $(2a,0)$, equations of the circles on AP, PQ, and QB as diameters are respectively.

So, we get

$(x-0)(x-a)+y^{2}=0$

$(x-a)(x-2a)+y^{2}=0$

$(x-2a)(x-3a)+y^{2}=0$

So, if $(h,k)$ be any point of the locus, then $3(h^{2}+k^{2})-9ah+8a^{2}=b^{2}$.

So, the required locus of $(h,k)$ is $3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0$.

More later,

Nalin Pithwa.

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