Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II

I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.

Problem 1:

Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points R_{1}, R_{2}, R_{3}, \ldots, R_{n} and on it a point R is taken such that \frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}

Show that the locus of R is a straight line.

Solution 1:

Let equations of the given lines be a_{i}x+b_{i}y+c_{i}=0, i=1,2,\ldots, n, and the point O be the origin (0,0).

Then, the equation of the line through O can be written as \frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r where \theta is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O.

Let r, r_{1}, r_{2}, \ldots, r_{n} be the distances of the points R, R_{1}, R_{2}, \ldots, R_{n} from O which in turn \Longrightarrow OR=r and OR_{i}=r_{i}, where i=1,2,3 \ldots n.

Then, coordinates of R are (r\cos{\theta}, r\sin{\theta}) and of R_{i} are (r_{i}\cos{\theta},r_{i}\sin{\theta}) where i=1,2,3, \ldots, n.

Since R_{i} lies on a_{i}x+b_{i}y+c_{i}=0, we can say a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0 for i=1,2,3, \ldots, n

\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}, for i=1,2,3, \ldots, n

\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}

\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta} …as given…

\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0

Hence, the locus of R is (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0 which is a straight line.

Problem 2:

Determine all values of \alpha for which the point (\alpha,\alpha^{2}) lies inside the triangle formed by the lines 2x+3y-1=0, x+2y-3=0, 5x-6y-1=0.

Solution 2:

Solving equations of the lines two at a time, we get the vertices of the given triangle as: A(-7,5), B(1/3,1/9) and C(5/4, 7/8).

So, AB is the line 2x+3y-1=0, AC is the line x+2y-3=0 and BC is the line 5x-6y-1=0

Let P(\alpha,\alpha^{2}) be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line 5x-6y-1=0, both 5(-7)-6(5)-1 and 5\alpha-6\alpha^{2}-1 must have the same sign.

\Longrightarrow 5\alpha-6\alpha^{2}-1<0 or 6\alpha^{2}-5\alpha+1>0 which in turn \Longrightarrow (3\alpha-1)(2\alpha-1)>0 which in turn \Longrightarrow either \alpha<1/3 or \alpha>1/2….call this relation I.

Again, since B and P lie on the same side of the line x+2y-3=0, (1/3)+(2/9)-3 and \alpha+2\alpha^{2}-3 have the same sign.

\Longrightarrow 2\alpha^{2}+\alpha-3<0 and \Longrightarrow (2\alpha+3)(\alpha-1)<0, that is, -3/2 <\alpha <1…call this relation II.

Lastly, since C and P lie on the same side of the line 2x+3y-1=0, we have 2 \times (5/4) + 3 \times (7/8) -1 and 2\alpha+3\alpha^{2}-1 have the same sign.

\Longrightarrow 3\alpha^{2}+2\alpha-1>0 that is (3\alpha-1)(\alpha+1)>0

\alpha<-1 or \alpha>1/3….call this relation III.

Now, relations I, II and III hold simultaneously if -3/2 < \alpha <-1 or 1/2<\alpha<1.

Problem 3:

A variable straight line of slope 4 intersects the hyperbola xy=1 at two points. Find the locus of the point which divides the line segment between these two points in the ratio 1:2.

Solution 3:

Let equation of the line be y=4x+c where c is a parameter. It intersects the hyperbola xy=1 at two points, for which x(4x+c)=1, that is, \Longrightarrow 4x^{2}+cx-1=0.

Let x_{1} and x_{2} be the roots of the equation. Then, x_{1}+x_{2}=-c/4 and x_{1}x_{2}=-1/4. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are (x_{1}, \frac{1}{x_{1}}) and that of B are (x_{2}, \frac{1}{x_{2}}).

Let R(h,k) be the point which divides AB in the ratio 1:2, then h=\frac{2x_{1}+x_{2}}{3} and k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}, that is, \Longrightarrow 2x_{1}+x_{2}=3h…call this equation I.

and x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k….call this equation II.

Adding I and II, we get 3(x_{1}+x_{2})=3(h-\frac{k}{4}), that is,

3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}….call this equation III.

Subtracting II from I, we get x_{1}-x_{2}=3(h+\frac{k}{4})

\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}

\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}

\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}

\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})

\Longrightarrow 16h^{2}+10hk+k^{2}-2=0

so that the locus of R(h,k) is 16x^{2}+10xy+y^{2}-2=0

More later,

Nalin Pithwa.

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