Co-ordinate Geometry : IITJEE Mains practice: some random problems again

Problem 1:

The line Ax+By+C=0 cuts the circle x^{2}+y^{2}+ax+by+c=0 at P and Q. The line A^{'}x+B^{'}y+C^{'}=0 cuts the circle x^{2}+y^{2}+a^{'}x+b^{'}y+c^{'}=0 at R and S. If P, Q, R and S are concyclic, prove that

\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0.

Solution I;

An equation of a circle through P and Q is x^{2}+y^{2}+ax+by+c +\lambda (Ax+By+C)=0…call this equation I.

And, an equation of a circle through R and S is x^{2}+y^{2}+a^{'}x+b^{'}y + c^{'}+\mu (A^{'}x+B^{'}y+C^{'})=0…call this equation II.

If P, Q, R and S are concyclic, then I and II represent the same circle for same values of \lambda and \mu.

\Longrightarrow a+ \lambda A=a^{'}+\mu A^{'} or a-a^{'} + \lambda A - \mu A^{'}=0

so also,

b + \lambda B = b^{'} + \mu B^{'} or b-b^{'}+\lambda B - \mu B^{'}=0

c + \lambda C = c^{'} + \mu C^{'} or c-c^{'} + \lambda C - \mu C^{'}=0.

Eliminating \lambda and \mu, we get the following:

\left | \begin{array}{ccc} a-a^{'} & A & -A^{'}\\ b-b^{'} & B & -B^{'}\\ c-c^{'} & C & -C^{'} \end{array} \right |=0, that is,

\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0

Problem II:

A straight line is such that the algebraic sum of the perpendiculars falling upon it from any number of fixed points is zero. Show that it always passes through a fixed point.

Solution II:

Let (x_{i},y_{i}) where i=1,2,3,\ldots be n fixed points. Let ax+by+c=0 be the given line. Thus, as per given hypthesis, we have

\sum_{i=1}^{n}\frac{ax{i}+by_{i}+c}{\sqrt{(a^{2}+b^{2})}}=0 \Longrightarrow a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}y_{i}+nc=0 \Longrightarrow a\overline{x}+b\overline{y}+c=0 where \overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i} and \overline{y}=\frac{}{}\sum_{i=1}^{n}y_{i}

which shows that the given line passes through the fixed point (\frac{1}{n}\sum_{i=1}^{n}x_{i}, \frac{1}{n}\sum_{i=1}^{n}y_{i}).

Problem III:

The straight lines L \equiv ax+by+c=0 and L_{1} \equiv a_{1}x+b_{1}y+c_{1}=0 are intersecting. Find the straight line L_{2} such that L is the bisector of the angle between L_{1} and L_{2}.

Solution III:

Let the equation of the line L_{2} be L_{1}+ \lambda L=0 \Longrightarrow (a_{1}+\lambda a)x+(b_{1}+\lambda b)y+\lambda c=0 where the slopes of L_{2}, L, L_{1} are respectively

-\frac{a_{1}+\lambda a}{b_{1}+\lambda b}, \frac{-a}{b}, -\frac{a_{1}}{b_{1}}.

Since L is the bisector of the angle between L_{2} and L_{1} we have

\frac{-(\frac{a_{1}+\lambda a}{b_{1}+\lambda b})+\frac{a}{b}}{1+\frac{a(a_{1}+\lambda a)}{b(b_{1}+\lambda b)}}=\frac{-\frac{a}{b}+\frac{a_{1}}{b_{1}}}{1+\frac{aa_{1}}{bb_{1}}}

\Longrightarrow \frac{-b(a_{1}+\lambda a)+a(a_{1}+\lambda b)}{b(b_{1}+\lambda b)+a(a_{1}+\lambda a)}=-\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}

\Longrightarrow \frac{ab_{1}-a_{1}b}{\lambda (a^{2}+b^{2})+aa_{1}+bb_{1}} = -\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}

\Longrightarrow \lambda = - \frac{2(aa_{1}+bb_{1})}{a^{2}+b^{2}}

Hence, the equation of the required line L_{1} is (a^{2}+b^{2})(a_{1}x+b_{1}y+c_{1})=2(aa_{1}+bb_{1})(ax+by+c).

Problem IV:

If a, b are real numbers and c>0, find the locus represented by |ay-bx|=c\sqrt{(x-a)^{2}+(y-b)^{2}}.

PS: Please draw a right angled triangle PMA, with right angle at M, and P being (x,y) and A being (a,b).

Solution IV:

Let x=a+r\cos {\theta} and y=b+r\sin {\theta}, then the given equation becomes a\sin {\theta}-b\cos {\theta}=c.

\Longrightarrow r\sin{(\theta-\alpha)}=c where r=\sqrt{a^{2}+b^{2}} and \tan {(\alpha)}=\frac{b}{a} which is the slope of ay-bx, which in turn implies \frac{c}{r}=\sin (\theta -\alpha) \leq 1

\Longrightarrow c \leq r, or c \leq \sqrt{a^{2}+b^{2}}. The given equation now becomes

\frac{|ay-bx|}{\sqrt{a^{2}+b^{2}}}=\frac{c}{\sqrt{a^{2}+b^{2}}}\sqrt{(x-a)^{2}+(y-b)^{2}}….call this as relation I.

If M is the foot of the perpendicular from a point P(x,y) on the line ay-bx=0 and A is the point (a,b) which clearly lies on this line, then from relation I, we have

\frac{PM}{PA}=\frac{c}{\sqrt{a^{2}+b^{2}}}=\sin {(\theta - \alpha)}. Hence, the locus of P is a straight line through the point (a,b) inclined at an angle \arcsin {\frac{c}{\sqrt{a^{2}+b^{2}}}} with the line ay-bx=0.

Problem V:

Find the co-ordinates of the orthocentre of the triangle formed by the lines y=0 and (1+t)x-ty+t(1+t)=0 and (1+u)x-uy+u(1+u)=0, where t \neq u, and show that for all values of t and u, the orthocentre lies on the line x+y=0.

Solution V:

Let the equation of the side BC be y=0. Then, the coordinates of B and C are (-t,0) and (-u,0), respectively, where (1+t)x-ty+t(1+t)=0 and (1+u)x-uy+u(1+u)=0 are equations of AB and AC, respectively.

PS: Please draw the diagram on your own for a better understanding of the solution presented.

Now, equation of BE is y={\frac{-u}{1+u}}(x+t)…let us call this equaiton I.

And, equation of CF is y=\frac{-t}{1+t}(x+u)…let us call this equation II.

Solving I and II, we get the following:

x(\frac{u}{1+u}-\frac{t}{1+t})=\frac{tu}{1+t}-\frac{tu}{1+u}, which in turn implies that

x=tu and y=-tu, so that the orthocentre is the point (tu,-tu) which lies on the line x+y=0.

Cheers,

Nalin Pithwa

One Comment

  1. Posted August 13, 2018 at 5:37 am | Permalink | Reply

    Quite resourceful blog for JEE Aspirants !

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