Note: Equation of a like is also defined as \displaystyle y - y_1 = m ( x-x_1)

Question 1: Find the equation of the straight through the following pair of points:

i) \displaystyle (0, 0)  \text{ and }  (2,-2)           ii) \displaystyle (a, b) \text{ and } (a + c \sin \alpha, b + c \cos \alpha)

iii) \displaystyle (0, -a) \text{ and } ( b, 0)           iv) \displaystyle (a, b) \text{ and } (a + b , a -b)

v) \displaystyle ( a t_1, a/t_1)  \text{ and } ( a t_2, a/t_2)           vi) \displaystyle (a \cos \alpha, a\sin \alpha) \text{ and } (a \cos \beta, a\sin \beta)

Answer:

i)       Given points \displaystyle A(0, 0) and \displaystyle B(2,-2)

\displaystyle  \text{ Slope of } AB = m =   \frac{-2-0}{2-0}   =   \frac{-2}{2}   = 1

Substituting in \displaystyle y - y_1 = m ( x-x_1) we get the equation of \displaystyle AB   as:

\displaystyle y - 0 = -1(x-0) \hspace{0.5cm} \Rightarrow y + x = 0

ii)      Given points \displaystyle A(a, b) and \displaystyle B(a + c \sin \alpha, b + c \cos \alpha)

\displaystyle \text{ Slope of } AB = m =   \frac{b+c \cos \alpha - b}{a+c \sin \alpha - a}   =   \frac{c \cos \alpha}{c \sin \alpha}   = \cot \alpha

Substituting in \displaystyle y - y_1 = m ( x-x_1) we get the equation of \displaystyle AB   as:

\displaystyle y - b = \cot \alpha (x-a) \hspace{0.5cm} \Rightarrow y  - x \cot \alpha + a - b  = 0

iii)     Given points \displaystyle A(0, -a) and \displaystyle B( b, 0)

\displaystyle \text{ Slope of } AB = m =   \frac{0-(-a)}{b-0}   =   \frac{a}{b}  

Substituting in \displaystyle y - y_1 = m ( x-x_1) we get the equation of \displaystyle AB   as:

\displaystyle y - (-a) =   \frac{a}{b}   (x-0)

\displaystyle \Rightarrow by + ab = ax

\displaystyle \Rightarrow ax-by-ab=0

iv)    Given points \displaystyle A(a, b) and \displaystyle B(a + b , a -b)

\displaystyle \text{ Slope of } AB = m =   \frac{a-b-b}{a+b-a}   =   \frac{a-2b}{b}  

Substituting in \displaystyle y - y_1 = m ( x-x_1) we get the equation of \displaystyle AB   as:

\displaystyle y - b =   \frac{a-2b}{b}   (x-a)

\displaystyle \Rightarrow yb-b^2=(a-2b)x-a(a-2b)

\displaystyle \Rightarrow yb-b^2 =  (a-2b)x - a^2+2ab

\displaystyle \Rightarrow (a-2b)x - yb + b^2 + 2ab - a^2=0

v)      Given points \displaystyle A( a t_1, a/t_1) and \displaystyle B( a t_2, a/t_2)

\displaystyle \text{ Slope of } AB = m =   \frac{\frac{a}{t_2} - \frac{a}{t_1}}{at_2 - at_1}   =   \frac{t_1-t_2}{t_1t_2(t_2-t_1)}   = -   \frac{1}{t_1t_2}  

Substituting in \displaystyle y - y_1 = m ( x-x_1) we get the equation of \displaystyle AB   as:

\displaystyle y -   \frac{a}{t_1}   =   \frac{-1}{t_1t_2}   (x-at_1)

\displaystyle \Rightarrow t_1t_2y- at_2 = -x + at_1

\displaystyle \Rightarrow x + t_1t_2y-a(t_1+t_2)=0

vi)     Given points \displaystyle A(a \cos \alpha, a\sin \alpha) and \displaystyle B(a \cos \beta, a\sin \beta)

\displaystyle \text{ Slope of } AB = m =   \frac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha}   =   \frac{\sin \beta - \sin \alpha}{ \cos \beta -  \cos \alpha}  

Substituting in \displaystyle y - y_1 = m ( x-x_1) we get the equation of \displaystyle AB   as:

\displaystyle y - a \sin \alpha =   \frac{\sin \beta - \sin \alpha}{ \cos \beta -  \cos \alpha}   (x-a \cos \alpha)

\displaystyle \Rightarrow y ( \cos \beta -  \cos \alpha) - a \sin \alpha ( \cos \beta -  \cos \alpha) = x ( \sin \beta - \sin \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)

\displaystyle \Rightarrow y ( \cos \beta -  \cos \alpha) - x ( \sin \beta - \sin \alpha) = a \sin \alpha ( \cos \beta -  \cos \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)

\displaystyle \Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{  \beta + \alpha }{2} \Big) = a ( \sin \alpha \cos \beta - \cos \alpha \sin \beta)

\displaystyle \Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{  \beta + \alpha }{2} \Big) = 2a \sin \Big(\frac{\alpha-\beta}{2} \Big) \cos \Big(\frac{ \alpha - \beta}{2} \Big)

\displaystyle \Rightarrow y \sin \Big(\frac{\beta + \alpha}{2} \Big)+ x  \cos \Big(\frac{  \beta + \alpha }{2} \Big) = a  \cos \Big(\frac{ \alpha - \beta}{2} \Big)

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Question 2: Find the equation to the sides of the triangles the coordinates of whose angular points are respectively: (i) \displaystyle (1, 4),(2, -3) and \displaystyle (-1, -2)       ii) \displaystyle (0, 1), (2, 0) and \displaystyle ( -1, -2)

Answer:

i)       Given points \displaystyle A(1, 4), B(2, -3) and \displaystyle C(-1, -2)

\displaystyle \text{ Slope of } AB = m_1 =   \frac{-3-4}{2-1}   =   \frac{-7}{1}   = -7

Therefore equation of \displaystyle AB :

\displaystyle y - 4 = - 7 ( x-1) \hspace{0.5cm} \Rightarrow 7x + y - 11 = 0

\displaystyle \text{ Slope of } BC = m_2 =   \frac{-2-(-3)}{-1-2}   =   \frac{1}{-3}   =   \frac{-1}{3}

Therefore equation of \displaystyle BC :

\displaystyle y - (-3) =   \frac{-1}{3}   ( x-2) \hspace{0.5cm} \Rightarrow 3y + 9 = - x + 2 \hspace{0.5cm} \Rightarrow x + 3y + 7 = 0

\displaystyle \text{ Slope of } CA = m_3 =   \frac{4-(-2)}{1-(-1)}   =   \frac{6}{2}   = 3

Therefore equation of \displaystyle CA :

\displaystyle y - 4 = 3 ( x-1) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0

ii)      Given points \displaystyle A(0,1), B(2,0) and \displaystyle C(-1, -2)

\displaystyle \text{ Slope of } AB = m_1 =   \frac{0-1}{2-0}   =   \frac{-1}{2}  

Therefore equation of \displaystyle AB :

\displaystyle y - 1 = \frac{-1}{2} ( x-0) \hspace{0.5cm} \Rightarrow x+2y-2 = 0

\displaystyle \text{ Slope of } BC = m_2 =   \frac{-2-0}{-1-2}   =   \frac{-2}{-3}   =   \frac{2}{3}

Therefore equation of \displaystyle BC :

\displaystyle y - 0 =   \frac{2}{3}   ( x-2) \hspace{0.5cm} \Rightarrow 2x-3y-4=0 

\displaystyle \text{ Slope of } CA = m_3 =   \frac{1-(-2)}{0-(-1)}   =   \frac{3}{1}   = 3

Therefore equation of \displaystyle CA :

\displaystyle y - 1 = 3 ( x-0) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0

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Question 3: Find the equations of the medians of a triangle, the the coordinates of the vertices are \displaystyle (-1, 6), (- 3, - 9) and \displaystyle (5, -8) .

Answer:

2021-01-10_11-30-15Given points \displaystyle A(-1,6), B(-3,-9) and \displaystyle C(5,-8) . Please refer to the figure shown.

Let \displaystyle D, E \ \&  \ F  be the mid point of \displaystyle AB, BC and \displaystyle CA respectively. Therefore,

\displaystyle D = \Big(   \frac{-3-1}{2}   ,   \frac{-9+6}{2}   \Big ) = \Big( -2,   \frac{-3}{2}   \Big )

\displaystyle E = \Big(   \frac{5-3}{2}   ,   \frac{-8-9}{2}   \Big ) = \Big( 1,   \frac{-17}{2}   \Big )

\displaystyle F = \Big(   \frac{-1+5}{2}   ,   \frac{6-8}{2}   \Big ) = ( 2, -1)

\displaystyle \text{ Slope of } AE = m_1 =   \frac{\frac{-17}{2}-6}{1-(-1)}   =   \frac{-29}{4}  

Therefore equation of \displaystyle AE :

\displaystyle y - 6 =   \frac{-29}{4}   ( x-(-1)) \hspace{0.5cm} \Rightarrow 4y-24=-29x-29 \hspace{0.5cm} \Rightarrow 29x+4y+5 = 0

\displaystyle \text{ Slope of } BF = m_2 =   \frac{-1-(-9)}{2-(-3)}   =   \frac{8}{5}  

Therefore equation of \displaystyle BF :

\displaystyle y - (-9) =   \frac{8}{5}   ( x-(-3)) \hspace{0.5cm} \Rightarrow 5y+45 = 8x+ 24 \hspace{0.5cm} \Rightarrow 8x-5y-21=0 

\displaystyle \text{ Slope of } CD = m_3 =   \frac{\frac{-3}{2} - (-8)}{-2-5}   =   \frac{-13}{14}    

Therefore equation of \displaystyle CD :

\displaystyle y - (-8) =   \frac{-13}{14}   ( x-5) \hspace{0.5cm} \Rightarrow 14y + 112 = - 13x +65 \hspace{0.5cm} \Rightarrow 13x + 14y + 47 = 0

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Question 4: Find the equations to the diagonals of the rectangle the equations of whose sides are \displaystyle x = a, x = a', y = b and \displaystyle y = b' .

2021-01-10_11-30-31Answer:

Please refer to the figure shown. Therefore the four points are \displaystyle A(a, b), B ( a', b), C( a', b'), D( a , b')

\displaystyle \text{ Slope of } AC = m_3 =   \frac{b'-b}{a'-a}    

Therefore equation of \displaystyle AC :

\displaystyle y - b=   \frac{b'-b}{a'-a}   ( x-a)

\displaystyle \Rightarrow (a'-a)y-a'b+ab=(b'-b)x-ab'+ab

\displaystyle \Rightarrow (b'-b)x-(a'-a)y - ab'+a'b = 0

\displaystyle \Rightarrow (a'-a)y - ( b'-b)x = ba'-ab'

\displaystyle \text{ Slope of } BD = m_3 =   \frac{b'-b}{a-a'}    

Therefore equation of \displaystyle BD :

\displaystyle y - b=   \frac{b'-b}{a-a'}   ( x-a')

\displaystyle \Rightarrow (a-a')y - ab + ba' = ( b'-b) x - a'b' + a'b

\displaystyle \Rightarrow (b-b')x - ( a - a') y + ab - a'b'=0

\displaystyle \Rightarrow  ( a - a') y + (b-b')x = a'b' - ab

\displaystyle \\

Question 5: Find the equation of the side \displaystyle BC of the \displaystyle \triangle ABC whose vertices are \displaystyle A(- 1, - 2), B (0, 1) and \displaystyle C (2,0) respectively. Also, find the equation  of the median  through \displaystyle A ( -1, -2) .

Answer:

Given points \displaystyle A(-1,-2), B(0,1) and \displaystyle C(2,0)

\displaystyle \text{ Slope of } BC = m_1 =   \frac{0-1}{2-0}   =   \frac{-1}{2}  

Therefore equation of \displaystyle BC :

\displaystyle y - 1 =   \frac{-1}{2}   ( x-0) \hspace{0.5cm} \Rightarrow 2y-2=-x \hspace{0.5cm} \Rightarrow x+ 2y - 2 = 0

\displaystyle \text{ Mid point  D  of } BC = \Big(   \frac{0+2}{2}   ,   \frac{1+0}{2}   \Big)= \Big( 1,   \frac{1}{2}   \Big)

\displaystyle \text{ Slope of } AD = m_2 =   \frac{\frac{1}{2} -(-2)}{1-(-1)}   =   \frac{5}{4}  

Therefore equation of \displaystyle AD :

\displaystyle y - (-2) =   \frac{5}{4}   ( x-(-1))

\displaystyle \Rightarrow y+2 =   \frac{5}{4}   ( x+1) \Rightarrow 4y+ 8 = 5x + 5 \Rightarrow 5x - 4y - 3

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Question 6: Using the concept of the equation of a line, prove that the three points \displaystyle (-2, -2) (8, 2)   and \displaystyle ( 3, 0)   are collinear.

Answer:

Given points \displaystyle A(-2,-2), B(8,2) and \displaystyle C(3,0)

\displaystyle \text{ Slope of } AC = m_1 =   \frac{0-(-2)}{3-(-2)}   =   \frac{2}{5}  

Therefore equation of \displaystyle AC

\displaystyle y - (-2) =   \frac{2}{5}   ( x-(-2))

\displaystyle \Rightarrow y+2 = \frac{2}{5} (x+2) \hspace{0.5cm} \Rightarrow 5y+10 = 2x + 4 \hspace{0.5cm} \Rightarrow 2x - 5y - 6 = 0

Now check if \displaystyle B( 8, 2) satisfy the equation

\displaystyle \therefore 2 ( 8) - 5(2) - 6 = 0

\displaystyle \Rightarrow 16-10-6 = 0

Therefore \displaystyle A, B, C are collinear.

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Question 7: Prove that the line \displaystyle y -x+2=0 , divides the join of points \displaystyle (3,-1) and \displaystyle (8,9) in the ratio \displaystyle 2:3 .

Answer:

Let \displaystyle y -x+2=0 , divides the join of points \displaystyle (3,-1) and \displaystyle (8,9) at a point P in the ratio of \displaystyle k:1

\displaystyle \therefore P = \Big(   \frac{3+8k}{k+1}   ,   \frac{-1+9k}{k+1}   \Big)

Since \displaystyle P lies on \displaystyle y - x+2 = 0 , it should satisfy the equation.

\displaystyle \therefore   \frac{-1+9k}{k+1}   -   \frac{3+8k}{k+1}   + 2 = 0

\displaystyle \Rightarrow -1 + 9k - 3 - 8 k + 2 k + 2 = 0

\displaystyle \Rightarrow 3k - 2 = 0

\displaystyle \Rightarrow k =   \frac{2}{3}  

Hence the line \displaystyle y -x+2=0 , divides the join of points \displaystyle (3,-1) and \displaystyle (8,9) in the ratio \displaystyle 2:3 .

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Question 8: Find the equation to the straight line which bisects the distance  between the points \displaystyle (a, b), (a', b') and also bisects the distance between the points \displaystyle ( -a, b) and \displaystyle (a', -b') .

Answer:

Given points \displaystyle A(a, b), B(a', b') and \displaystyle C( -a, b) and \displaystyle D(a', -b')

\displaystyle \text{ Mid point  P  of } AB = \Big(   \frac{a+a'}{2}   ,   \frac{b+b'}{2}   \Big)

 \displaystyle \text{ Mid point  Q  of } CD = \Big(   \frac{-a+a'}{2}   ,   \frac{b-b'}{2}   \Big)

\displaystyle \text{ Slope of } PQ = m_1 =   \frac{\frac{b-b'}{2} - \frac{b+b'}{2} }{\frac{-a+a'}{2} - \frac{a+a'}{2}}   =   \frac{b-b'-b-b'}{-a+a'-a-a'}   =   \frac{-2b'}{-2a}   =   \frac{b'}{a}  

Therefore equation of \displaystyle PQ :

\displaystyle y -   \frac{b+b'}{2}   =   \frac{b'}{a}   \Big( x-   \frac{a+a'}{2}   \Big)

\displaystyle \Rightarrow 2y - b - b' =   \frac{b'}{a}   ( 2x - a - a')

\displaystyle \Rightarrow 2ay - ab - ab' = 2b'x-ab'-a'b'

\displaystyle \Rightarrow 2b'x-2ay+ab-a'b'=0

\displaystyle \Rightarrow 2ay - 2b'x = ab - a'b'

\displaystyle \\

Question 9: In what ratio is the line joining the points \displaystyle (2, 3) and \displaystyle (4, -5) divided by the line passing through the points \displaystyle (6,8) and \displaystyle (-3, -2) .

Answer:

Given points: \displaystyle (6,8) and \displaystyle (-3, -2)

\displaystyle \text{ Slope of } AB =   \frac{-2-8}{-3-6}   =   \frac{-10}{-9}   =   \frac{10}{9}  

Therefore equation of \displaystyle AB :

\displaystyle y - 8 =   \frac{10}{9}   ( x - 6) \hspace{0.5cm} \Rightarrow 9y - 72 = 10 x - 60 \hspace{0.5cm} \Rightarrow 10x - 9 y + 12 = 0

Let \displaystyle 10x - 9 y + 12 = 0 divides the line joining \displaystyle (2, 3) and \displaystyle (4, -5) in the ratio of \displaystyle k:1

\displaystyle \therefore point of intersection \displaystyle = \Big(    \frac{4k+2}{k+1}   ,   \frac{-5k+3}{k+1}   \Big)

This point is on line \displaystyle 10x - 9 y + 12 = 0

\displaystyle \therefore 10 \Big(   \frac{4k+2}{k+1}   \Big ) - 9 \Big(   \frac{-5k+3}{k+1}   \Big) + 12 = 0

\displaystyle \Rightarrow 40k+20+45k-27+12k+12=0

\displaystyle \Rightarrow 97k + 5 = 0

\displaystyle \Rightarrow k = -   \frac{5}{97}  

Hence the required ratio is \displaystyle 5:97 externally.

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Question 10: The vertices of a  quadrilateral are \displaystyle A(-2,6),8(1,2),C(10, 4) and \displaystyle D(7, 8) . Find the equations of its diagonals.

Answer:

Given points: \displaystyle A(-2,6), B(1,2) , C(10,4), D(7,8)

\displaystyle \text{ Slope of } AC =   \frac{4-6}{10-(-2)}   =   \frac{-2}{12}   =   \frac{-1}{6}  

Therefore equation of \displaystyle AC :

\displaystyle y - 6 =   \frac{-1}{6}   ( x - (-2)) \hspace{0.5cm} \Rightarrow 6y-36=-x-2 \hspace{0.5cm} \Rightarrow x+6y-34=0

\displaystyle \text{ Slope of } BD =   \frac{8-2}{7-1}   =   \frac{6}{6}   = 1

Therefore equation of \displaystyle BD :

\displaystyle y - 2 =   \displaystyle 1 ( x - 1) \hspace{0.5cm} \Rightarrow x-y+1=0 

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Question 11: The length \displaystyle L (in centimeters) of a copper rod is a linear function of its Celsius temperature \displaystyle C .  In an experiment,  if \displaystyle L = 124.942 when \displaystyle C = 20 , and \displaystyle L = 125.134 when \displaystyle C = 110 , express \displaystyle L in terms of \displaystyle C

Answer:

Given two points \displaystyle ( 124.942, 20) and \displaystyle (125.134, 110)

\displaystyle \text{ Slope of  line} =   \frac{110-20}{125.134-124.942}   =   \frac{90}{0.192}  

Therefore the equation of line:

\displaystyle y - 20 =   \frac{90}{0.192}   ( x - 124.942)

\displaystyle \Rightarrow 0.192 y - 3.84 = 90x - 11244.78

\displaystyle \Rightarrow 90x = 0.192y + 11240.78

\displaystyle \Rightarrow x =   \frac{0.192}{90}   y +   \frac{11240.78}{90}  

\displaystyle \Rightarrow x =   \frac{4}{1875}   y + 124.898

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Question 12: The owner of  a milk store finds that he can sell \displaystyle 980 liters milk each week at (\displaystyle 14 per liter and \displaystyle 1220 liters of milk each week at Rs. \displaystyle 16 per liter. Assuming a linear relationship between selling price and demand, how many liters can he sell weekly at Rs. \displaystyle 17 per liter.

Answer:

Given two points \displaystyle ( 14, 980) and \displaystyle (16, 1220)

\displaystyle \text{ Slope of  line} =   \frac{1220-980}{16-14}   =   \frac{240}{2}   = 120

Therefore the equation of line:

\displaystyle y - 980 = 120(x-14)   \hspace{0.5cm} \Rightarrow y = 120x - 700

When \displaystyle x = 17, y = 120 \times 17 - 700 = 1340

Hence the owner of the milk store would be able to sell \displaystyle 1340 liters of milk at Rs. \displaystyle 17 / liter.

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Question 13: Find the equation of the bisector of \displaystyle \angle A of the triangle whose vertices are \displaystyle A (4,3), B(0,0) and \displaystyle C( 2, 3) .

Answer:

Let \displaystyle AD be the bisector of \displaystyle \angle BAC .

We know, \displaystyle \frac{AB}{AC}   =   \frac{BD}{DC}    

Now \displaystyle AB = \sqrt{(4-0)^2 + ( 3-0)^2} = \sqrt{25} = 5

\displaystyle AC = \sqrt{(4-2)^2 + ( 3-3)^2} = \sqrt{4} = 2

\displaystyle \therefore   \frac{BD}{DC}   =   \frac{5}{2}  

Therefore coordinates of \displaystyle D = \Big(   \frac{2 \times 5 + 0 \times 2}{5+2}   ,   \frac{3 \times 5 + 0 \times 2}{5+2}   \Big) = \Big(   \frac{10}{7}   ,   \frac{15}{7}   \Big)

\displaystyle \text{ Slope of } AD =   \frac{\frac{15}{7}-3}{\frac{10}{7}-4}   =   \frac{15-21}{10-28}   =   \frac{-6}{-18}   =   \frac{1}{3}  

Therefore equation of \displaystyle AD :

\displaystyle y - 3 =   \frac{1}{3}   ( x - 4) \hspace{0.5cm} \Rightarrow 3y-9=x-4 \hspace{0.5cm} \Rightarrow x-3y+5=0

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Question 14: Find the equations to the straight lines which go through the origin and trisect the portion of the straight line \displaystyle 3x+ y = 12 which is intercepted between the axes of coordinates.

Answer:

Given line \displaystyle 3x+y =12

x-intercept \displaystyle = A ( 4, 0)

y-intercept \displaystyle = B ( 0, 12

\displaystyle C divides \displaystyle BC in the ratio of \displaystyle 2:1

Coordinates of \displaystyle C = \Big(   \frac{2 \times 4 + 1 \times 0}{2+1}   ,   \frac{2 \times 0 + 1 \times 12}{2+1}   \Big) = \Big(   \frac{8}{3}   , 4 \Big)

\displaystyle D divides \displaystyle BC in the ratio of \displaystyle 1:2

Coordinates of \displaystyle C = \Big(   \frac{2 \times 4 + 1 \times 0}{1+2}   ,   \frac{2 \times 0 + 1 \times 12}{1+2}   \Big) = \Big(   \frac{8}{3}   , 4 \Big)

\displaystyle \text{ Slope of } OC =   \frac{4-0}{\frac{8}{3}-0}   =   \frac{3}{2}  

Therefore equation of \displaystyle OC :

\displaystyle y - 0 =   \frac{3}{2}   ( x - 0) \hspace{0.5cm} \Rightarrow 2y=3x 

\displaystyle \text{ Slope of } OD =   \frac{8-0}{\frac{4}{3}-0}   = 6  

Similarly, equation of \displaystyle OD :

\displaystyle y - 0 =  6 ( x - 0) \hspace{0.5cm} \Rightarrow y=6x 

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Question 15: Find the equations of the diagonals of the square formed by the lines \displaystyle x=0, y=0, x=1 and \displaystyle y = 1 .

Answer:2021-01-10_11-30-51

\displaystyle \text{ Slope of } AC =   \frac{1-0}{1-0}   = 1  

Therefore equation of \displaystyle AC :

\displaystyle y - 0 = 1 ( x - 0) \hspace{0.5cm} \Rightarrow y=x 

\displaystyle \text{ Slope of } BD =   \frac{1-0}{0-1}   = -1  

Similarly, equation of \displaystyle OD :

\displaystyle y - 0 =  -1 ( x - 1) \hspace{0.5cm} \Rightarrow x+y = 1