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

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

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

iii) (0, -a) and ( b, 0)           iv) (a, b) and (a + b , a -b)

v) ( a t_1, a/t_1) and ( a t_2, a/t_2)           vi) (a \cos \alpha, a\sin \alpha) and (a \cos \beta, a\sin \beta)

Answer:

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

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

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

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

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

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

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

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

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

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

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

\Rightarrow by + ab = ax

\Rightarrow ax-by-ab=0

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

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

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

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

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

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

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

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

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

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

\Rightarrow t_1t_2y- at_2 = -x + at_1

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

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

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

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

\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)

\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)

\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)

\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)

\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)

\\

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

Answer:

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

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

Therefore equation of AB :

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

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

Therefore equation of BC :

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

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

Therefore equation of CA :

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

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

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

Therefore equation of AB :

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

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

Therefore equation of BC :

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

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

Therefore equation of CA :

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

\\

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

Answer:

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

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

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

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

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

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

Therefore equation of AE :

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

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

Therefore equation of BF :

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

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

Therefore equation of CD :

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

\\

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

Answer:

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

2021-01-10_11-30-31Slope of AC = m_3 = \frac{b'-b}{a'-a}  

Therefore equation of AC :

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

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

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

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

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

Therefore equation of BD :

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

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

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

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

\\

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

Answer:

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

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

Therefore equation of BC :

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

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

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

Therefore equation of AD :

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

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

\\

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

Answer:

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

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

Therefore equation of AC

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

\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 B( 8, 2) satisfy the equation

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

\Rightarrow 16-10-6 = 0

Therefore A, B, C are collinear.

\\

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

Answer:

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

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

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

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

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

\Rightarrow 3k - 2 = 0

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

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

\\

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

Answer:

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

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

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

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 PQ :

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

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

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

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

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

\\

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

Answer:

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

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

Therefore equation of AB :

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

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

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

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

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

\Rightarrow 97k + 5 = 0

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

Hence the required ratio is 5:97 externally.

\\

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

Answer:

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

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

Therefore equation of AC :

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

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

Therefore equation of BD :

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

\\

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

Answer:

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

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

Therefore the equation of line:

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

\Rightarrow 0.192 y - 3.84 = 90x - 11244.78

\Rightarrow 90x = 0.192y + 11240.78

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

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

\\

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

Answer:

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

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

Therefore the equation of line:

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

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

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

\\

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

Answer:

2021-01-10_11-30-40Let AD be the bisector of \angle BAC .

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

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

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

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

Therefore coordinates of 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)

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 AD :

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

\\

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

Answer:

2021-01-10_11-30-59Given line 3x+y =12

x-intercept = A ( 4, 0)

y-intercept = B ( 0, 12

C divides BC in the ratio of 2:1

Coordinates of 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)

D divides BC in the ratio of 1:2

Coordinates of 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)

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

Therefore equation of OC :

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

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

Similarly, equation of OD :

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

\\

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

Answer:

2021-01-10_11-30-51Slope of AC = \frac{1-0}{1-0} = 1  

Therefore equation of AC :

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

Slope of BD = \frac{1-0}{0-1} = -1  

Similarly, equation of OD :

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