Let A(l, 4) and B(3, 8) be any two points. Equation of the line joining A(l, 4) and B(3, 8) is
\(y-4=(\frac{8-4}{3-1})(x-1)\)
\(\Rightarrow\) y - 4 = 2(x - 1)  \(\Rightarrow\)  2x - y + 2 = 0
Length of perpendicular from point (2, -5) to line 2x - y + 2 = 0 is
\(=|\frac{2\times 2-(-5)+2}{\sqrt{(2)^2+(-1)^2}}|=|\frac{4+5+2}{\sqrt{5}}|=\frac{11}{\sqrt{5}}\)

Here m = - 4 and c = -4.
Putting these values in y = mx + c, we have
y = - 4x - 4 \(\Rightarrow\) 4x + y + 4 = 0.

Slope of AB = Slope o fCD
Slope of BC = Slope of DA
Also, slope of AB × slope of BC = −1

Let equation of line is \(\frac { x }{ a } +\frac { y }{ 2a } =1\) . Here, we have \(1=\frac { a+0 }{ 2 } \)  and \(2=\frac { b+0 }{ 2 } \)     
Ans. 2x + y - 4 = 0 

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then
Slope of AB = Slope of BC
\(\frac{b-0}{a-h}=\frac{k-b}{0-a}\)
\(\Rightarrow \frac{b}{a-h}=\frac{k-b}{-a}\)
\(\Rightarrow-a b=(k-b)(a-h)\)
\(\Rightarrow-a b=k a-k h-a b+b h\)
\(\Rightarrow k a+b h=k h\)
\(\text { On dividing both sides by } k h, \text { we obtain }\)
\(\frac{k a}{k h}+\frac{b h}{k h}=\frac{k h}{k h}\)
\(\Rightarrow \frac{a}{h}+\frac{b}{k}=1\)
\(\text { Hence, } \frac{a}{h}+\frac{b}{k}=1\)

Let the given points are \(A({ x }_{ 1 },{ y }_{ 1 })\equiv A(-1,1)\) and,\(B({ x }_{ 2 },{ y }_{ 2 })\equiv B(2,-4)\) then equation of line AB is
\(y-1=\frac { -4-1 }{ 2+1 } (x+1)\)
\( \left[ \because y-{ y }_{ 1 }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } (x-{ x }_{ 1 }) \right]\)
\( \Rightarrow y-1=\frac { -5 }{ 3 } (x+1) \Rightarrow 3y-3=-5x-5\)
\(\Rightarrow 5x+3y+2=0\)

Let(h,k) be the coordinates of the foot of the perpendicular from the point(2,3) on the line x+y-11=0
Then, the slope of the perpendicular line is\(\cfrac { k-3 }{ h-2 } \)
Again the slope of the given line x + y - 11= 0 is -1
using the condition of perpendicularity of lines,we have
(\(\cfrac { k-3 }{ h-2 } \))(-1)=-1 0r k-h=1...(i)
since,(h,k) lies on the given line, we have
h + k - 11 = 0 or h + k = 11.....(ii)
On solving eqs. (i) and (ii) we get h=5 and k=6
hence,(5,6) are the required coordinates of the foot of the perpendicular.

We know that,slope of a line joining two points (x₁,y₁) and (x₂,y²) is m \(=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
Slope of line joining the points (3,−2) and (7,−2) is
\(m=\frac { -2-(-2) }{ 7-3 } =\frac { 0 }{ 4 } =0\)

The coordinates of the new origin are h=1, k=2 and the original coordinates are given point are x = 3,y = -4.
The transformation relation between the old coordinates (x,y) and the new coordinates (X,Y) are given by
x = X + h,i.e.X = x - h ...(i)
y = Y+ k,i.e.Y=y-k. ..(ii)
On substituting the values x=3, y=4,h=1 and k=2 ineqs (i) and (ii), we get
X = 3 = 2 and Y = 42 = 6
Hence the coordinates of point (3,-4) in the new system are (2,-6).

Slope of required lines = \(\frac { -1 }{ 7 } \)
Now, let the equation of required lines be y = \(\frac { -1 }{ 7 } x+c.\)
Then, we have
\(\frac { \left| -1+\frac { 1 }{ 7 } -c \right| }{ \sqrt { { 1 }^{ 2 }+{ \left( \frac { 1 }{ 7 } \right) }^{ 2 } } } =1\Rightarrow c=\frac { -6\pm 5\sqrt { 2 } }{ 7 } \)
Ans. x + 7y + 6 \(\pm 5\sqrt { 2 } \) = 0

\({ P }_{ 1 }=\frac { \left| 0+0+3 \right| }{ \sqrt { { \left( -2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } } =\frac { 3 }{ \sqrt { 5 } } \)
\({ P }_{ 2 }=\frac { \left| 0+0-7 \right| }{ \sqrt { { \left( 1 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } } =\frac { 7 }{ \sqrt { 17 } } \)
Again,   \({ P }_{ 1 }=\frac { 3 }{ \sqrt { 5 } } \times \frac { \sqrt { 17 } }{ \sqrt { 17 } } =\sqrt { \frac { 153 }{ 85 } } \)
and \({ P }_{ 2 }=\frac { 7 }{ \sqrt { 17 } } \times \frac { \sqrt { 5 } }{ \sqrt { 5 } } =\sqrt { \frac { 245 }{ 85 } } \)
\(\therefore \sqrt { \frac { 245 }{ 85 } } >\sqrt { \frac { 153 }{ 85 } } \Rightarrow { P }_{ 2 }>{ P }_{ 1 }\)
Ans. x - 4y - 7 = 0 is farther.

First, find out the values of AB,BC,CA by distance formula and prove that \(\triangle ABC\) is an isosceles triangle, then prove that this triangle is also right-angled triangle by Converse of Pythagoras theorem.

(i) Let (x', y') the coordinates of the given point (x, y0 in the new system of translation of axes.
\(\therefore\) x'=x-h \(\Rightarrow\) x=x'+h=x'+1 and y=y'+k=y'+1
Substituting these values of x and y in the given equation x²+xy-3y²-y+2 = 0, we get
(x'+1)²+(x'+1)(y'+1)-3(y'+1)²-(y'+1)+2=0
\(\Rightarrow\) x'²+1+2x'+x'y'+x'+y'+1-3y'²-3-6y'-y'-1+2=0
\(\Rightarrow\)  x'²+x'y'-3y'²+3x'-6y' = 0
Hence the equation of the given pair of straight lines in a new system is
x²+xy-3y²+3x-6y=0
(ii) The orgin is shifted to a point (1, 1) and Let (x',y') be the new coordinates of the given point (x, y) in the line.
\(\therefore\) x = x' + h
\(\Rightarrow\) x = x' + 1 and y = y' + 1
Substituting these values of x and y in the given equation xy - y² - x+y=0 we get,
(x'+1) (y'+1)-(y'+1)²-(x'+1)+(y'+1) = 0
\(\Rightarrow\)  x'y' +x' +y' +1-y'²-1-2y'-x'-1+y'+1 = 0
\(\Rightarrow\)  x'y'-y'² = 0
Hence the equation of the given pair of straight lines in a new system is xy-y² = 0
(iii) Here h = 1 and k = 1
Let (x', y' ) be the coordinates of the new point
\(\therefore\)  x = x'+h, y=y' +k
\(\Rightarrow\)  x = x'+1, \(\Rightarrow\) y = y' +1
Now substituting the values of x and y is given equation xy-x-y+1 = 0, we get
(x' +1) (y'+1) - (x'+1)-(y'+1)+1=0
\(\Rightarrow\)  x'y' +x' +y' +1-x'-1-y'-1+1 =0
\(\Rightarrow\) x' y' = 0
Hence the equation of the given pair of straight lines in a new system of translation of axes is xy = 0.

(i) the origin is shifted to (-3, -2)
\(\therefore\) h = -3 and k = -2
The coordinates of the given point are (1, 1)
\(\therefore\) x = 1     and   y = 1
\(\therefore\)  x' = x-h and  y' = y-k
=1+3 and   = 1+2
=4   = 3
Hence the new coordinates of the given point are (4, 3).
(ii) Here h=-3 and k=-2 sfter shifting the origint and the coordinates of the given point are
x=0  and  y=1
\(\therefore\) x'=x-h  and y'=y-k
= 0+3 and = 1+2
\(\therefore\) x' = 3  and y = 3
Hence the new coordinates of the point are (3, 3)
(iii) Here
h = -3  and k=-2
x=5     and y=0
\(\therefore\) x'=x-h=5+3=8
and y'=y-k=0+2=2
Hence the new coordinates of the point are (8, 2)
(iv) Here
h=-3 and k = -2
x=-1 and y=-2
\(\therefore\)  x' = x-h= -1 + 3 = 2
and y' = y-k=-2 + 2 = 0
Hence the new coordinates of the given point are (2, 0).
(v) Here
h=-3 and k = -2
x = 3  and y=-5
\(\therefore\)  x'=x-h= 3+3 = 6
and y' = y-k = -5 + 2 = -3
Hence the new coordintes of the point are (6, -3).

\(\text { | et } A B C \text { be the riaht analed trianale where } \angle \mathrm{C}=90^{0}\)
\(\text { There are infinitely many such lines. }\)
\(\text { Let } m \text { be the slope of } A C \text { . }\)
\(\text { Slope of } B C=-\frac{1}{m}\)
\(\text { Equation of } A C: \quad y-3=m(x-1)\)
\(\Rightarrow x-1=\frac{1}{m}(y-3)\)
\(\text { Equation of BC: } \quad y-1=-\frac{1}{m}(x+4)\)
\(\Rightarrow x+4=-m(y-1)\)
\(\text { For a given value of } m \text { , we can get theseequations }\)
\(\text { For } m=0, \quad y-3=0 ; x+4=0\)
\(\text { For } m \rightarrow \infty, \quad x-1=0 : y-1=0\)