Coordinate Geometry

$Positive\, \, vector\, \, are\, \, \, 2i+j+k,6\hat { i } +\hat { j } +2\hat { k }  \ and\, \, 4i-5j+pk\, \, are\, \, collinear\, \, then\, \, P=2 \ if\, \, three\, \, position\, \, vectors\, \, are\, \, collinear\, \, then,\, \, its\, \, { { determinantsis } }\left( 0 \right)  \ \Rightarrow \left| \begin{matrix} 2\, \, \, \, \, \, \, 1\, \, \, \, \, \, \, 1 \ 6\, \, \, \, -1\, \, \, \, \, \, 2 \ 14\, \, -5\, \, \, \, P \  \end{matrix} \right| =0 \ \Rightarrow 2\left( { -P+10 } \right) -1\left( { 6P-28 } \right) +\left( { -30+14 } \right) =0 \ \Rightarrow -2P+20-6P+28-16=0 \ \Rightarrow -8P+32=0 \ \therefore P=4\, $

Consider the given points $A(\bar{a}),B(\bar{b}),C(\bar{c})$.
These three points determine two vectors $\vec{AB}$ and $\vec{AC}$

We know that, "two vectors $a,b$ are collinear if and only if $\vec{a} \times \vec{b}=0$".

Therefore two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$\vec{AB}=\vec{OB}-\vec{OA}=\bar{b}-\bar{a}$ and $\vec{AC}=\vec{OC}-\vec{OA}=\bar{c}-\bar{a}$

We have, two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$ \Rightarrow$ two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$

Since the three points determine two vectors $\vec{AB}$ and $\vec{AC}$, we conclude that

Three points $A(\bar{a}),B(\bar{b}),C(\bar{c})$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$.

$a^{2}+a-42=0\Rightarrow a=-7$ or $a=6.$
Since b is a root of $x^{2}+ax+a^{2}-37=0$
For $a=-7,$ we have $x^{2}-7x+49-37=0$
$\Rightarrow x^{2}-7x+12=0\Rightarrow x=4, 3 , so, b=4$ or $3.$
So the coordinates of P can be $(-7, 4)$ or $(-7, 3)$, 

$ y = x-3$ $\Rightarrow m=1$ ; $tan \theta =1$
By parametrization we have $x=2 \pm r cos\theta $
$y=1 \pm r sin\theta $

When three points are collinear, Slope of line joining any two points is same as the slope of line joining any other two points


Slope of line joining two points $ ({x} _{1}, {y} _{1}) $ and $ ({x} _{2}, {y} _{2}) $ is $\dfrac { {y} _{2} - {y} _{1}}{ {x} _{2} - {x} _{1}} $

So, Slope of line joining $ (5,5) ;  (7,7) $ is $ \dfrac {7-5}{7-5} = \dfrac {2}{2} = 1 $

And Slope of line joining $ (a,8) ;  (7,7) $ is $ \dfrac {7-a}{7-8} = a - 7 $

As they are collinear $ a - 7 = 1 => a = 8 $

shifted by $\sqrt{2}$ units $\Rightarrow r=\sqrt{2}$

for the line is tan $\theta =1$

$\therefore$ by parametric equations, we have

$x=x^{1}+r\ cos \theta$ ;  $y=y^{1}+r\ sin \theta$

$\Rightarrow x=4+\sqrt{2}\frac{1}{\sqrt{2}}$ ; $y=3+\sqrt{2}\frac{1}{\sqrt{2}}$

$x=5$  ;  $y=4$

The image of (1, 2) in y=x is (2, 1). The image of (2, 1) in y=0 is (2, -1). Thus, alpha=2 and beta=-1.

The family of lines (2x-3y+4) + k(x-2y+3) = 0 passes through a fixed point (intersection of the two lines). The locus of the image of a point reflected across a family of lines passing through a fixed point is a circle centered at that fixed point.

The line equation can be rewritten as (2x+3y-5)cos(theta) + (3x-5y+2)sin(theta) = 0. This passes through the intersection of 2x+3y=5 and 3x-5y=-2. Solving this gives P(1, 1). Reflecting P across 4x+6y=23 gives Q, then calculate distance to origin.

Given transformation is 
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right )$
So, the point $(-2,-7)$ $\rightarrow (-2-2(-7),-3(-2)-7)$
So, the image of the point $(-2,-7)$ under the given transformation is $ (12,-1)$

Let points are $A(-2,6)$ & $B(4,2)$

Point A(1,2) reflected in y=x gives B(2,1) (coordinates swap). B(2,1) reflected in y=0 gives (2,-1) (y-coordinate negated). Therefore alpha=2, beta=-1. This is identical to question 463562, testing the same two-step reflection concept.

The point $P(2,1)$ if shifted by $3\sqrt2$ parallel to the line $x+y =1$, in the direction of increasing ordinate, to reach point $Q(\alpha, \beta)$.