Cartesian system and straight lines: IITJEE Mains: Problem solving skills

Problem 1:

The line joining A(b\cos{\alpha},b\sin{\alpha}) and B(a\cos{\beta},a\sin{\beta}) is produced to the point M(x,y) so that AM:MB=b:a, then find the value of x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}.

Solution 1:

As M divides AB externally in the ratio b:a, we have x=\frac{b(a\cos{\beta})-a(b\cos{\alpha})}{b-a} and y=\frac{b(a\sin{\beta})-a(b\sin{\alpha})}{b-a} which in turn

\Longrightarrow \frac{x}{y} = \frac{\cos{\beta}-cos{\alpha}}{\sin{\beta}-\sin{\alpha}}

= \frac{2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}}{2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\beta-\alpha}{2}}}

\Longrightarrow x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}=0

Problem 2:

If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points (a^{2}+1,a^{2}+1) and (2a,-2a), then where does the orthocentre lie?

Solution 2:

From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre (0,0) and the centroid (\frac{(a+1)^{2}}{2},\frac{(a-1)^{2}}{2}), that is, y.\frac{(a+1)^{2}}{2} = x.\frac{(a-1)^{2}}{2}, or (a-1)^{2}x-(a+1)^{2}y=0. That is, the orthocentre lies on this line.

Problem 3:

If a, b, c are unequal and different from 1 such that the points (\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1}), (\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1}) and (\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1}) are collinear, then which of the following option is true?

a: bc+ca+ab+abc=0

b: a+b+c=abc

c: bc+ca+ab=abc

d: bc+ca+ab-abc=3(a+b+c)

Solution 3:

Suppose the given points lie on the line lx+my+n=0 then a, b, c are the roots of the equation :

lt^{3}+m(t^{2}-3)+n(t-1)=0, or

lt^{3}+mt^{2}+nt-(3m+n)=0

\Longrightarrow a+b+c=-\frac{m}{l} and ab+bc+ca=\frac{n}{l}, that is, abc=(3m+n)/l

Eliminating l, m, n, we get abc=-3(a+b+c)+bc+ca+ab

\Longrightarrow bc+ca+ab-abc=3(a+b+c), that is, option (d) is the answer.

Problem 4:

If p, x_{1}, x_{2}, \ldots, x_{i}, \ldots and q, y_{1}, y_{2}, \ldots, y_{i}, \ldots are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points A_{i}(x_{i},y_{i}) with i=1,2,3 \ldots, n lie?

Solution 4:

Note: Centre of Mean Position is (\frac{\sum{xi}}{n},\frac{\sum {yi}}{n}).

Let the coordinates of the centre of mean position of the points A_{i}, i=1,2,3, \ldots,n be (x,y) then

x=\frac{x_{1}+x_{2}+x_{3}+\ldots + x_{n}}{n} and y=\frac{y_{1}+y_{2}+\ldots + y_{n}}{n}

\Longrightarrow x = \frac{np+a(1+2+\ldots+n)}{n}, y=\frac{nq+b(1+2+\ldots+n)}{n}

\Longrightarrow x=p+ \frac{n(n+1)}{2n}a and y=q+ \frac{n(n+1)}{2n}b

\Longrightarrow x=p+\frac{n+1}{2}a, and y=q+\frac{n+1}{2}b

\Longrightarrow 2\frac{(x-p)}{a}=2\frac{(y-q)}{b} \Longrightarrow bx-ay=bp-aq, that is, the CM lies on this line.

Problem 5:

The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}?

Solution 5:

Equation of the line L in the two coordinate systems is \frac{x}{a} + \frac{y}{b}=1, and \frac{X}{p} + \frac{Y}{q}=1 where (X,Y) are the new coordinate of a point (x,y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.

\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}}

\Longrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}

or \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}=0. So, the value is zero.

Problem 6:

Let O be the origin, A(1,0) and B(0,1) and P(x,y) are points such that xy>0 and x+y<1, then which of the following options is true:

a: P lies either inside the triangle OAB or in the third quadrant

b: P cannot lie inside the triangle OAB

c: P lies inside the triangle OAB

d: P lies in the first quadrant only.

Solution 6:

Since xy>0, P either lies in the first quadrant or in the third quadrant. The inequality x+y<1 represents all points below the line x+y=1. So that xy>0 and x+y<1 imply that either P lies inside the triangle OAB or in the third quadrant.

Problem 7:

An equation of a line through the point (1,2) whose distance from the point A(3,1) has the greatest value is :

option i: y=2x

option ii: y=x+1

option iii: x+2y=5

option iv: y=3x-1

Solution 7:

Let the equation of the line through (1,2) be y-2=m(x-1). If p denotes the length of the perpendicular from (3,1) on this line, then p=|\frac{2m+1}{\sqrt{m^{2}+1}}|

\Longrightarrow p^{2}=\sqrt{\frac{4m^{2}+4m+1}{m^{2}+1}}=4+ \frac{4m-3}{m^{2}+1}=s, say

then p^{2} is greatest if and only if s is greatest.

Now, \frac{ds}{dm} = \frac{(m^{2}+1)(4)-2m(4m-3)}{(m^{2}+1)^{2}} = \frac{-2(2m-1)(m-2)}{(m^{2}+1)^{2}}

\frac{ds}{dm} = 0 so that \Longrightarrow m = \frac{1}{2}, 2. Also, \frac{ds}{dm}<0, if m<\frac{1}{2}, and

\frac{ds}{dm} >0, if 1/2<m<2

and \frac{ds}{dm} <0, if m>2. So s is greatest for m=2. And, thus, the equation of the required line is y=2x.

Problem 8:

The points A(-4,-1), B(-2,-4), Slatex C(4,0)$ and D(2,3) are the vertices of a :

option a: parallelogram

option b: rectangle

option c: rhombus

option d: square.

Note: more than one option may be right. Please mark all that are right.

Solution 8:

Mid-point of AC = (\frac{-4+4}{2},\frac{-1+0}{2})=(0, \frac{-1}{2})

Mid-point of BD = (\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})

\Longrightarrow the diagonals AC and BD bisect each other.

\Longrightarrow ABCD is a parallelogram.

Next, AC= \sqrt{(-4-4)^{2}+(-1+0)^{2}}=\sqrt{64+1}=\sqrt{65} and BD=\sqrt{(-2-2)^{2}+(-4+3)^{2}}=\sqrt{16+49}=\sqrt{65} and since the diagonals are also equal, it is a rectangle.

As AB=\sqrt{(-4+2)^{2}+(-1+4)^{2}}=\sqrt{13} and BC=\sqrt{(-2-4)^{2}+(-4)^{2}}=\sqrt{36+16}=sqrt{52}, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.

Problem 9:

Equations (b-c)x+(c-a)y+(a-b)=0 and (b^{3}-c^{3})x+(c^{3}-a^{3})y+a^{3}-b^{3}=0 will represent the same line if

option i: b=c

option ii: c=a

option iii: a=b

option iv: a+b+c=0

Solution 9:

The two lines will be identical if there exists some real number k, such that

b^{3}-c^{3}=k(b-c), and c^{3}-a^{3}=k(c-a), and a^{3}-b^{3}=k(a-b).

\Longrightarrow b-c=0 or b^{2}+c^{2}+bc=k

\Longrightarrow c-a=0 or c^{2}+a^{2}+ac=k, and

\Longrightarrow a-b=0 or a^{2}+b^{2}+ab=k

That is, b=c or c=a, or a=b.

Next, b^{2}+c^{2}+bc=c^{2}+a^{2}+ca \Longrightarrow b^{2}-a^{2}=c(a-b). Hence, a=b, or a+b+c=0.

Problem 10:

The circumcentre of a triangle with vertices A(a,a\tan{\alpha}), B(b, b\tan{\beta}) and C(c, c\tan{\gamma}) lies at the origin, where 0<\alpha, \beta, \gamma < \frac{\pi}{2} and \alpha + \beta + \gamma = \pi. Show that it’s orthocentre lies on the line 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}x-4\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}y=y

Solution 10:

As the circumcentre of the triangle is at the origin O, we have OA=OB=OC=r, where r is the radius of the circumcircle.

Hence, OA^{2}=r^{2} \Longrightarrow a^{2}+a^{2}\tan^{2}{\alpha}=r^{2} \Longrightarrow a = r\cos{\alpha}

Therefore, the coordinates of A are (r\cos{\alpha},r\sin{\alpha}). Similarly, the coordinates of B are (r\cos{\beta},r\sin{\beta}) and those of C are (r\cos{\gamma},r\sin{\gamma}). Thus, the coordinates of the centroid G of \triangle ABC are

(\frac{1}{3}r(\cos{\alpha}+\cos{\beta}+\cos{\gamma}),\frac{1}{3}r(\sin{\alpha}+\sin{\beta}+\sin{\gamma})).

Now, if P(h,k) is the orthocentre of \triangle ABC, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.

Hence, \frac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{\cos{\alpha}+\cos{\beta}+\cos{\gamma}}=\frac{k}{h}

\Longrightarrow \frac{4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}}{1+4\sin{(\frac{\alpha}{2})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}}= \frac{k}{h}

because \alpha+\beta+\gamma=\pi.

Hence, the orthocentre P(h,k) lies on the line

4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}x-4\sin{(\frac{}{})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}y=y.

Hope this gives an assorted flavour. More stuff later,

Nalin Pithwa.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s

%d bloggers like this: