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Reference Material for Preparation

Question 1: Which of the following points lie on the line x-2y+5=0 :

i) (1, 3)  ii) (0, 5)  iii) (-5, 0)  iv) (5, 5)  v) (2, -1.5)  vi) (-2, -1.5)

Answer:

i) Substituting x=1 \ and \ y = 3 in  x-2y+5=0 we get

(1)-2 \times (3)+5 = 0

1-6+5=0 which is true.

Therefore point (1, 3) satisfies the equation x-2y+5=0 and therefore lies on the equation.

ii) Substituting x=0 \ and \ y = 5  in x-2y+5=0 we get

(0)-2 \times (5)+5 = 0

0-10+5=0  which is NOT true.

Therefore point (0, 5) does not satisfies the equation x-2y+5=0 and therefore does not lies on the equation.

iii) Substituting x=-5 \ and \ y = 0  in x-2y+5=0 we get

(-5)-2 \times (0)+5 = 0

-5-0+5=0  which is true.

Therefore point (-5, 0) satisfies the equation x-2y+5=0 and therefore  lies on the equation.

iv) Substituting x=5 \ and \ y = 5  in x-2y+5=0 we get

(5)-2 \times (5)+5 = 0

5-10+5=0  which is true.

Therefore point (5, 5) satisfies the equation x-2y+5=0 and therefore  lies on the equation.

v) Substituting x=2 \ and \ y = -1.5  in x-2y+5=0 we get

(2)-2 \times (-1.5)+5 = 0

2+3+5=0  which is NOT true.

Therefore point (2, -1.5) does not satisfies the equation x-2y+5=0 and therefore does not lies on the equation.

vi) Substituting x=-2 \ and \ y = -1.5  in x-2y+5=0 we get

(-2)-2 \times (-1.5)+5 = 0

-2+3+5=0  which is NOT true.

Therefore point (-2, -1.5) does not satisfies the equation x-2y+5=0 and therefore does not lies on the equation.

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Question 2: State, true or false:

i) the line \frac{x}{2}+\frac{y}{3}=0 passes through the point (2,3) 

ii) the line \frac{x}{2}+\frac{y}{3}=0  passes through the point (4,-6) 

iii) the point (8,7)  lies on the line y-7=0 

iv) the point (-3,0)   lies on the line x+3=0 

v) if the point (2,a)  lies on the line 2x-y=3 , then a = 5 

Answer:

i) Substituting x=2 \ and \ y = 3 in \frac{x}{2}+\frac{y}{3}=0 we get

\frac{2}{2}+\frac{3}{3}=0

1+1=0 which is NOT true.

Therefore point (2, 3) satisfies the equation \frac{x}{2}+\frac{y}{3}=0 and therefore the equation does not pass through the given point.

ii) Substituting x=4 \ and \ y = -6 in \frac{x}{2}+\frac{y}{3}=0 we get

\frac{4}{2}+\frac{-6}{3}=0

2-2=0 which is true.

Therefore point (4, -6) satisfies the equation \frac{x}{2}+\frac{y}{3}=0 and therefore the equation does not pass through the given point.

iii) Substituting x=8 \ and \ y = 7 in y-7=0   we get

(7)-7=0  which is true.

Therefore point (8, 7) satisfies the equation y-7=0   and therefore the equation passes through the given point.

iv) Substituting x=-3 \ and \ y = 0 in x+3=0   we get

(-3)+3=0  which is true.

Therefore point (-3, 0) satisfies the equation x+3=0   and therefore the equation passes through the given point.

v) Substituting x=2 \ and \ y = a in 2x-y=3   we get

2(2)-(a)=3 \Rightarrow a = 1  

Therefore point (2, a) satisfies the equation 2x-y=3  if a=1  and not when  a = 5 .

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Question 3: The line given by the equation 2x-\frac{y}{3}=7  passes through the point (k,6)  ; calculate the value of k  .

Answer:

Substituting x=k \ and \ y = 6 in  2x-\frac{y}{3}=7  we get

2(k)-\frac{6}{3}=7

\Rightarrow k = 4.5

Therefore if point (k, 6) satisfies the equation 2x-\frac{y}{3}=7  then k = 4.5

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Question 4: For what value of k   will the point (3, -k)  lie on the line 9x+4y=3  ?

Answer:

Substituting x=3 \ and \ y = -k in 9x+4y=3  we get

9(3)+4(-k)=3

\Rightarrow k = 6

Therefore if point (3, -k) satisfies the equation 9x+4y=3  then k = 6

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Question 5: The line \frac{3x}{5}-\frac{2y}{3}+1=0  contains the point (m, 2m-1)  ; calculate the value of m  .

Answer:

Substituting x=m \ and \ y = 2m-1 in  \frac{3x}{5}-\frac{2y}{3}+1=0  we get

\frac{3m}{5}-\frac{2(2m-1)}{3}+1=0 

9m-10(2m-1)+15=0 

-11m+25=0 \Rightarrow m = \frac{25}{11} 

Therefore if point (m, 2m-1) satisfies the equation \frac{3x}{5}-\frac{2y}{3}+1=0  then m =\frac{25}{11}

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Question 6: Does the line 3x-5y=6  bisect the join of (5, -2) \ and \ (-1, 2)  ?

Answer:

Ratio for being a midpoint: m_1:m_2 = 1:1

Let the coordinates of the point P \ be \ (x, y)

Therefore

x = \frac{1 \times 5+1 \times (-1)}{1+1} = 2

y = \frac{1 \times (-2)+1 \times (2)}{1+1}  = 0

Therefore P = (2, 0)

Substituting x=2\ and \ y = 0 in  3x-5y=6  we get

3(2)-5 \times (0) = 6

6= 6 which is true.

Therefore point (2, 0) satisfies the equation3x-5y=6 .

Hence the mid point is on the given line and bisects the given points.

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Question 7:

i) The line y=3x-2  bisects the join of (a, 3) \ and \ (2, -5)  , find the value of a  .

ii) The line x-6y+11=0  bisects the join of (8,-1) \ and \ (0,k)  . Find the value of k  .

Answer:

i) Ratio for being a midpoint: m_1:m_2 = 1:1

Let the coordinates of the point P \ be \ (x, y)

Therefore

x = \frac{1 \times a+1 \times (2)}{1+1}

y = \frac{1 \times (3)+1 \times (-5)}{1+1} = -1

Therefore P = (\frac{a+2}{2}, -1)

Substituting x=\frac{a+2}{2} \ and \ y = -1 in y=3x-2  we get

(-1)=3(\frac{a+2}{2})-2 

\Rightarrow a = -\frac{4}{3}

ii) Ratio for being a midpoint: m_1:m_2 = 1:1

Let the coordinates of the point P \ be \ (x, y)

Therefore

x = \frac{1 \times 8+1 \times (0)}{1+1} = 4

y = \frac{1 \times (-1)+1 \times (k)}{1+1}  = \frac{k-1}{2}

Therefore P = (4,\frac{k-1}{2})

Substituting x=4 \ and \ y =\frac{k-1}{2} in x-6y+11=0  we get

4-6(\frac{k-1}{2})+11=0 

8-6(k-1)+22=0 \Rightarrow k =6 

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Question 8: 

i) The point (-3, 2)  lies on the line ax+3y+6=0  , calculate the value of a  .

ii) The line y=mx+8  contains the point (-4, 4)  , calculate the value of m  .

Answer:

i) Substituting x=-3 \ and \ y =2 in  ax+3y+6=0  we get

a(-3)+3(2)+6=0 \Rightarrow a = 4

ii) Substituting x=-4 \ and \ y =4 in y=mx+8  we get

4=m(-4)+8 \Rightarrow m=1

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Question 9: The point P  divides the join of (2,1) \ and \ (-3,6)  in the ratio of 2:3  . Does P  lie on the line x-5y+15=0  ?

Answer:

Ratio: m_1:m_2 = 2:3

Let the coordinates of the point P \ be \ (x, y)

Therefore

x = \frac{2 \times (-3)+3 \times (2)}{2+3} = 0 

y = \frac{2 \times (6)+3 \times (1)}{2+3}  = 3

Therefore P = (0, 3)

Substituting x=0 \ and \ y = 3 in x-5y+15=0  we get

(0)-5(3)+15=0 

-15+15=0 which is true.

Therefore point (0, 3) satisfies the equation x-2y+5=0 and therefore lies on the equation.

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Question 10: The line segment joining the points (5, -4) \ and \ (2,2)  is divided by the point Q  in the ratio 1:2 . Does the line x-2y=0  contain Q  ?

Answer:

Ratio: m_1:m_2 = 2:1

Let the coordinates of the point Q \ be \ (x, y)

Therefore

x = \frac{1 \times (2)+2 \times (5)}{1+2} = 4 

y = \frac{1 \times (2)+2 \times (-4)}{1+2}  = -2

Therefore Q = (4, -2)

Substituting x=4 \ and \ y = -2 in  x-2y=0  we get

(4)-2(-2)=0 

0=0 which is true.

Therefore point Q(4, -2) satisfies the equationx-2y=0  and therefore  lies on the equation.

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Question 11: Find the point of intersection of the lines 4x+3y=1  and 3x-y+9=0  . If this point lies on the line (2k-1)x-2y=4  , find the value of k .

Answer:

Solving the two equations: 4x+3y=1  and 3x-y+9=0 

y =3x+9 

Substituting

4x+3(3x+9)=1 

13x=-26 \Rightarrow x=-2 

Therefore y = 3(-2)+9=3 

Hence the point of intersection = (-2, 3) 

Substituting x=-2 \ and \ y = 3 in (2k-1)x-2y=4  we get

(2k-1)(-2)-2(3)=4 \Rightarrow k = -2

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Question 12: Show that the lines 2x+5y=1 , x-3y=6   and x+5y+2=0  are concurrent.

Answer:

Solving the two equations first: 2x+5y=1 , and x-3y=6  

x=3y+6 

Substituting 2(3y+6)+5y=1

6y+12+5y=1 

11y = 11 \Rightarrow y = -1 

Thereforex = 3(-1)+6=3 

Hence the point of intersection of the two lines is (3, -1)  

If  (3,-1)  satisfies x+5y+2=0 , then the three lines are concurrent.

Substituting x=3 \ and \ y = -1 in  x+5y+2=0  we get

(3)+5(-1)+2=0 

10=0 Hence the lines are concurrent.

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