Geometry

Given pair of lines: $x^2+2xy\sec\alpha+y^2=0$

This is a homogeneous second-degree equation. Factorising it,

$x^2+2xy\sec\alpha+y^2=0$

Multiply by $\cos^2\alpha$: $x^2\cos^2\alpha+2xy\cos\alpha+y^2\cos^2\alpha=0$

Now, $(x\cos\alpha+y(1+\sin\alpha))(x\cos\alpha+y(1-\sin\alpha))=0$

Hence the separate equations of the pair of lines are: $\boxed{x\cos\alpha+y(1+\sin\alpha)=0}$ and $\boxed{x\cos\alpha+y(1-\sin\alpha)=0}$

Now, for the pair of lines $ax^2+2hxy+by^2=0,$ the angle $\theta$ between them is given by $\tan\theta=\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Here, $a=1,\quad b=1,\quad h=\sec\alpha$

So, $\tan\theta=\left|\frac{2\sqrt{\sec^2\alpha-1}}{2}\right|$ $=\left|\tan\alpha\right|$

Therefore, $\boxed{\theta=\tan^{-1}|\tan\alpha|}$

For an acute angle $\alpha$,

So the angle between the two lines is $\alpha$.

$r^2$

In inversion, the product of the distance of a point from the centre of inversion and the distance of its inverse point from the same centre is equal to the square of the radius of the inversion circle. So $OP \times OP' = r^2$.

Factor the given equation:

$6x^2+5xy-6y^2-3x+2y=0$

$(3x-2y)(2x+3y-1)=0$

Hence, the separate equations of the pair of lines are:

$\boxed{3x-2y=0}$

and

$\boxed{2x+3y-1=0}$

$r^2$

In an inversion circle of radius $r$ with centre $O$, the product of the distance of a point from the centre and the distance of its inverse point from the centre is always equal to the square of the radius. So $OA \times OA' = r^2$.

For the pair of lines
$x^2 - 4xy + y^2 = 0,$ compare with
$ax^2 + 2hxy + by^2 = 0.$

Here, $a = 1,\quad 2h = -4 \Rightarrow h = -2,\quad b = 1.$

The angle $\theta$ between the pair of lines is given by
$\tan\theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|.$

Substituting the values, $\tan\theta = \left|\frac{2\sqrt{(-2)^2 - 1\cdot 1}}{1+1}\right|$ $= \left|\frac{2\sqrt{4-1}}{2}\right|$ $= \sqrt{3}.$

So, $\tan\theta = \tan 60^\circ$

$\therefore \theta = 60^\circ.$

Hence, the angle between the pair of lines is $60^\circ$.

$m_1 \times m_2 = -1$

Two lines are perpendicular when the product of their slopes equals $-1$, i.e. $m_1 m_2 = -1$ (equivalently $m_2 = -\dfrac{1}{m_1}$).

Circle

A plane cutting a cone parallel to its base gives a closed curve at a constant distance from the axis — a circle.

Matrices.

$R_1$ (reflection on $y=x$) $=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $R_2$ (rotation $+90^\circ$ about the origin) $=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

$R_1\circ R_2$ means $R_2$ is applied first, then $R_1$:

$R_1R_2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$

This is the matrix of reflection on the X-axis, so the single transformation is a reflection in the X-axis, $(x,y)\to(x,-y)$.

Image of $\triangle ABC$.

$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}5&3&6\\6&4&1\end{pmatrix}=\begin{pmatrix}5&3&6\\-6&-4&-1\end{pmatrix}$

So $A'(5,-6),\ B'(3,-4),\ C'(6,-1)$.

Graph. On the same graph paper plot $A(5,6),\ B(3,4),\ C(6,1)$ and join them to form $\triangle ABC$. Plot $A'(5,-6),\ B'(3,-4),\ C'(6,-1)$ and join them to form $\triangle A'B'C'$. The two triangles are mirror images of each other in the X-axis.

Compare $2x^2+kxy+3y^2=0$ with $ax^2+2hxy+by^2=0$: $a=2$, $2h=k$ so $h=\dfrac k2$, $b=3$.

The angle between the lines satisfies

$\tan\theta=\frac{2\sqrt{h^2-ab}}{a+b}$

With $\theta=45^\circ$, $\tan\theta=1$:

$1=\frac{2\sqrt{\frac{k^2}{4}-6}}{5}\ \Rightarrow\ \sqrt{\frac{k^2}{4}-6}=\frac52$

Squaring,

$\frac{k^2}{4}-6=\frac{25}{4}\ \Rightarrow\ k^2=25+24=49\ \Rightarrow\ k=7\ (\text{positive value})$

Separate equations. The equation becomes

$2x^2+7xy+3y^2=0$ $2x^2+6xy+xy+3y^2=0$ $(2x+y)(x+3y)=0$

Hence the lines are $2x+y=0$ and $x+3y=0$.

Check: their slopes are $-2$ and $-\frac13$, so $\tan\theta=\left|\dfrac{-2+\frac13}{1+\frac23}\right|=1$, giving $\theta=45^\circ$.

tan θ = $\left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$

The tangent of the angle between the two lines is the absolute value of the difference of the slopes divided by 1 plus the product of the slopes.

The unit square has vertices $(0,0),(1,0),(1,1),(0,1)$, i.e. the matrix $\begin{pmatrix}0&1&1&0\\0&0&1&1\end{pmatrix}$.

$\begin{pmatrix}a&2\\b&2\end{pmatrix}\begin{pmatrix}0&1&1&0\\0&0&1&1\end{pmatrix}=\begin{pmatrix}0&a&a+2&2\\0&b&b+2&2\end{pmatrix}$

Equating with $\begin{pmatrix}0&4&c&2\\0&1&3&d\end{pmatrix}$:

$a=4,\quad b=1,\quad c=a+2=6,\quad b+2=3\ (\text{checks}),\quad d=2$

$\therefore\ a=4,\ b=1,\ c=6,\ d=2$

Circle B: $x^2+y^2-4x+6y-12=0$ gives $g=-2,\ f=3$, so its centre is $Y(-g,-f)=(2,-3)$.

Circle A has centre $X(-4,5)$ and passes through $Y(2,-3)$, so its radius is

$r=XY=\sqrt{(2+4)^2+(-3-5)^2}=\sqrt{36+64}=10.$

Equation of circle A:

$(x+4)^2+(y-5)^2=100$

$x^2+8x+16+y^2-10y+25-100=0$

$\therefore\ x^2+y^2+8x-10y-59=0$

Let
$R$ = reflection on the line $y=-x$
$S$ = rotation about origin through $180^\circ$

Matrix of $R$:

Matrix of $S$:

So,

This matrix represents reflection on the line $y=x$.

Hence, the combined transformation $R\circ S$ is the reflection about the line $y=x$.

Now, under reflection on $y=x$,

Therefore,

So, the vertices of the image triangle are

Graph:
Plot the points $A(1,2)$, $B(-4,3)$, $C(3,5)$ and join them to form $\Delta ABC$.
Plot the points $A'(2,1)$, $B'(3,-4)$, $C'(5,3)$ and join them to form $\Delta A'B'C'$.
The line $y=x$ is the mirror line of the combined transformation.

Let $d^2=(x-h)^2+(y-k)^2$. The inverse point $P'$ lies on ray $CP$ with $CP\cdot CP'=r^2$, so $P'(x',y')=\left(h+\frac{r^2(x-h)}{(x-h)^2+(y-k)^2},\ \ k+\frac{r^2(y-k)}{(x-h)^2+(y-k)^2}\right)$

Let the matrix be $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$.

From $A(2,0)\to A'(0,2)$: $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}2\\0\end{pmatrix}=\begin{pmatrix}2a\\2c\end{pmatrix}=\begin{pmatrix}0\\2\end{pmatrix}\Rightarrow a=0,\ c=1$

From $B(2,9)\to B'(9,2)$: $2a+9b=9\Rightarrow b=1,\qquad 2c+9d=2\Rightarrow d=0$

$M=\begin{pmatrix}0&1\\1&0\end{pmatrix}$

Check with $C(6,7)$: $M\begin{pmatrix}6\\7\end{pmatrix}=\begin{pmatrix}7\\6\end{pmatrix}=C'$. This matrix is the reflection in the line $y=x$.

Solution:

Let
$R_1(x,y)=(-x,y)$
be reflection on the $y$-axis, and
$R_2(x,y)=(y,-x)$
be rotation about the origin through $-90^\circ$ (clockwise $90^\circ$).

Now, $R_1\circ R_2(x,y)=R_1(R_2(x,y))$ $=R_1(y,-x)$ $=(-y,-x).$

The transformation $(x,y)\to(-y,-x)$ is exactly the reflection in the line
$y=-x \quad \text{or} \quad x+y=0.$

Hence, the combined transformation $R_1\circ R_2$ is a single reflection about the line $y=-x$.

Now applying this to the vertices of $\triangle ABC$:

$A(2,0)\to A'(-0,-2)=(0,-2)$ $B(3,1)\to B'(-1,-3)$ $C(1,1)\to C'(-1,-1)$

So the image triangle $A'B'C'$ has vertices
$A'(0,-2),\quad B'(-1,-3),\quad C'(-1,-1).$

Graph:
Plot the points $A(2,0)$, $B(3,1)$, $C(1,1)$ and join them to form $\triangle ABC$.
Plot the points $A'(0,-2)$, $B'(-1,-3)$, $C'(-1,-1)$ and join them to form $\triangle A'B'C'$ on the same graph paper.
The line $y=-x$ is the mirror line of the transformation.

The given circle $x^2+y^2+4x-6y-36=0$ has $g=2,\ f=-3$, so its centre is $(-g,-f)=(-2,3)$.

The required circle has centre $(3,5)$ and passes through $(-2,3)$, so its radius is

$r=\sqrt{(3+2)^2+(5-3)^2}=\sqrt{25+4}=\sqrt{29}.$

Equation of the circle:

$(x-3)^2+(y-5)^2=29$

$x^2-6x+9+y^2-10y+25-29=0$

$\therefore\ x^2+y^2-6x-10y+5=0$

Given lines:

$cx + dy + e = 0 \quad \text{and} \quad fx + gy + h = 0$

Their slopes are

$m_1 = -\frac{c}{d}, \qquad m_2 = -\frac{f}{g}$

Since the lines are parallel, their slopes are equal:

$-\frac{c}{d} = -\frac{f}{g}$

$\frac{c}{d} = \frac{f}{g}$

Cross-multiplying,

$cg = df$

$\therefore cg - df = 0$

Hence proved.

Given lines:

$2x + 3y - 8 = 0 \quad \text{... (i)}$

$4x + 6y = 6 \Rightarrow 4x + 6y - 6 = 0 \quad \text{... (ii)}$

Slope of line (i):

$m_1 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{2}{3}$

Slope of line (ii):

$m_2 = -\frac{4}{6} = -\frac{2}{3}$

Since $m_1 = m_2 = -\dfrac{2}{3}$, the two lines have equal slopes.

Therefore, the given lines are parallel to each other.

Let the matrix be $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Then $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&2&5\\0&4&-1\end{pmatrix}=\begin{pmatrix}2&8&9\\0&12&-3\end{pmatrix}$

The product is $\begin{pmatrix}a&2a+4b&5a-b\\c&2c+4d&5c-d\end{pmatrix}$.

Column 1: $a=2,\ c=0$.

Column 2: $2a+4b=8\Rightarrow4+4b=8\Rightarrow b=1$; $2c+4d=12\Rightarrow d=3$.

Check column 3: $5a-b=10-1=9$ and $5c-d=-3$. ✓

$\boxed{M=\begin{pmatrix}2&1\\0&3\end{pmatrix}}$

The inversion point $M'$ lies on ray $OM$ with $OM\cdot OM' = r^2$. Hence

$M'\left(\frac{r^2x}{x^2+y^2},\ \frac{r^2y}{x^2+y^2}\right)$

Factorise the given equation:

$2x^2+3xy-2y^2=(2x-y)(x+2y)=0$

So the given lines are $y=2x$ (slope $2$) and $x+2y=0$ (slope $-\frac12$).

The line through the origin perpendicular to $y=2x$ has slope $-\frac12$, that is $x+2y=0$.

The line through the origin perpendicular to $x+2y=0$ has slope $2$, that is $2x-y=0$.

The required pair is $(x+2y)(2x-y)=0$, so the single equation is

$2x^2+3xy-2y^2=0$

It is the same as the given equation, because the given lines are already perpendicular to each other (coefficient of $x^2$ plus coefficient of $y^2$ is $2-2=0$).

Let the circle be

$x^2+y^2+2gx+2fy+c=0,$

with centre $(-g,-f)$.

Centre on the line. Putting $(-g,-f)$ in $4x+y-16=0$:

$-4g-f-16=0 \;\Rightarrow\; 4g+f=-16 \qquad (i)$

Passes through $(4,1)$:

$16+1+8g+2f+c=0 \;\Rightarrow\; 8g+2f+c=-17 \qquad (ii)$

Passes through $(6,5)$:

$36+25+12g+10f+c=0 \;\Rightarrow\; 12g+10f+c=-61 \qquad (iii)$

(iii) $-$ (ii): $4g+8f=-44$, so $g+2f=-11$ $\;(iv)$.

From (i), $f=-16-4g$. Putting in (iv): $g-32-8g=-11$, so $-7g=21$ and $g=-3$. Then $f=-16+12=-4$.

From (ii): $c=-17-8(-3)-2(-4)=-17+24+8=15$.

$\boxed{x^2+y^2-6x-8y+15=0}$

Check: centre $(3,4)$ gives $4(3)+4-16=0$, and $(6,5)$ gives $36+25-36-40+15=0$.

Factorise the given equation:

$x^2-xy-2y^2=(x-2y)(x+y)=0,$

so the lines are $x-2y=0$ (slope $\tfrac12$) and $x+y=0$ (slope $-1$).

Lines through the origin perpendicular to these have slopes $-2$ and $1$:

$y=-2x \Rightarrow 2x+y=0,\qquad y=x \Rightarrow x-y=0.$

Their combined equation is

$(2x+y)(x-y)=0 \Rightarrow 2x^2-2xy+xy-y^2=0$

$\boxed{2x^2-xy-y^2=0}$

$h^2 = ab$

The pair of straight lines $ax^2 + 2hxy + by^2 = 0$ is coincident when the two lines have the same slope, which happens when the discriminant of the quadratic in $y/x$ is zero, i.e. $h^2 = ab$.

A circle is formed when the cutting plane is parallel to the base of the cone, i.e. perpendicular to the axis of the cone (and does not pass through the vertex).

At infinity

In an inversion, the distance of the image from the centre is given by $OP' = \dfrac{r^2}{OP}$. When P lies at the centre, $OP = 0$, so $OP' \to \infty$; hence the image P' lies at infinity.

Completing the squares: $(x-1)^2+(y-5)^2=25$ so the centre is $C(1,5)$.

The centre is the mid-point of every diameter. Let the other end be $B(x_2,y_2)$ and $A(1,0)$: $\frac{1+x_2}{2}=1\Rightarrow x_2=1,\qquad\frac{0+y_2}{2}=5\Rightarrow y_2=10$

The other end of the diameter is $(1,10)$.

For reflection on the (x)-axis, the transformation matrix is

Let the vertices be written as column matrices:

Now,

Hence, the vertices of the image triangle are

Given: P(3, 7) and R(−1, 3) are opposite vertices of square PQRS, so PR is one diagonal. QS is the other diagonal.

Key facts used: The diagonals of a square bisect each other at right angles. So QS passes through the mid-point of PR and is perpendicular to PR.

Step 1 — Mid-point of PR (which lies on QS):

$M = \left(\frac{3+(-1)}{2},\ \frac{7+3}{2}\right) = (1,\ 5)$

Step 2 — Slope of PR:

$m_{PR} = \frac{3-7}{-1-3} = \frac{-4}{-4} = 1$

Step 3 — Slope of QS (perpendicular to PR, so $m_1 m_2 = -1$):

$m_{QS} = -\frac{1}{1} = -1$

Step 4 — Equation of QS through (1, 5) with slope −1:

$y - 5 = -1(x - 1)$ $y - 5 = -x + 1$ $x + y = 6$

Verification: The other two vertices are Q(3, 3) and S(−1, 7); the line through them is indeed $x + y = 6$.

$\boxed{x + y = 6}$