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

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.

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