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

\displaystyle \text{i) } (1, 3) \hspace{1.0cm} \text{ii) } (0, 5) \hspace{1.0cm} \text{iii) } (-5, 0) \hspace{1.0cm} \text{iv) } (5, 5) \hspace{1.0cm} \text{v) } (2, -1.5) \hspace{1.0cm} \text{vi) } (-2, -1.5)  

Answer:

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

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

\displaystyle 1-6+5=0 which is true.

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

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

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

\displaystyle 0-10+5=0 which is NOT true.

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

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

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

\displaystyle -5-0+5=0 which is true.

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

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

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

\displaystyle 5-10+5=0 which is true.

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

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

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

\displaystyle 2+3+5=0 which is NOT true.

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

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

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

\displaystyle -2+3+5=0 which is NOT true.

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

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Question 2: State, true or false:

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

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

\displaystyle \text{iii) the point } (8,7) \text{ lies on the line } y-7=0  

\displaystyle \text{iv) the point } (-3,0) \text{ lies on the line } x+3=0  

\displaystyle \text{v) if the point } (2,a) \text{ lies on the line } 2x-y=3 , \displaystyle \text{ then } a = 5  

Answer:

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

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

\displaystyle 1+1=0 which is NOT true.

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

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

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

\displaystyle 2-2=0 which is true.

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

\displaystyle \text{iii) Substituting } x=8 \text{ and } y = 7 \text{ in } y-7=0 \text{ we get }

\displaystyle (7)-7=0 which is true.

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

\displaystyle \text{iv) Substituting } x=-3 \text{ and } y = 0 \text{ in } x+3=0 \text{ we get }

\displaystyle (-3)+3=0 which is true.

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

\displaystyle \text{v) Substituting } x=2 \text{ and } y = a \text{ in } 2x-y=3 \text{ we get }

\displaystyle 2(2)-(a)=3 \Rightarrow a = 1  

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

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

Answer:

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

\displaystyle 2(k)- \frac{6}{3} =7  

\displaystyle \Rightarrow k = 4.5  

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

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

Answer:

\displaystyle \text{Substituting } x=3 \text{ and } y = -k \text{ in } 9x+4y=3 \text{ we get }

\displaystyle 9(3)+4(-k)=3  

\displaystyle \Rightarrow k = 6  

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

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

Answer:

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

\displaystyle \frac{3m}{5} - \frac{2(2m-1)}{3} +1=0  

\displaystyle 9m-10(2m-1)+15=0  

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

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

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

Answer:

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

Let the coordinates of the point \displaystyle P\text{ be } (x, y)  

Therefore

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

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

\displaystyle \text{Therefore } P = (2, 0)  

\displaystyle \text{Substituting } x=2 \text{ and } y = 0 \text{ in } 3x-5y=6 \text{ we get }

\displaystyle 3(2)-5 \times (0) = 6  

\displaystyle 6= 6 which is true.

\displaystyle \text{Therefore point } (2, 0) satisfies the equation\displaystyle 3x-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 \displaystyle y=3x-2 bisects the join of \displaystyle (a, 3) \text{ and } (2, -5) , find the value of \displaystyle a

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

Answer:

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

Let the coordinates of the point \displaystyle P\text{ be } (x, y)  

Therefore

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

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

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

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

\displaystyle (-1)=3 \Big( \frac{a+2}{2} \Big)-2  

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

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

Let the coordinates of the point \displaystyle P\text{ be } (x, y)  

Therefore

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

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

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

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

\displaystyle 4-6( \frac{k-1}{2} )+11=0  

\displaystyle 8-6(k-1)+22=0 \Rightarrow k =6  

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Question 8: 

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

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

Answer:

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

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

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

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

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

Answer:

\displaystyle \text{Ratio: } m_1:m_2 = 2:3  

Let the coordinates of the point \displaystyle P\text{ be } (x, y)  

Therefore

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

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

\displaystyle \text{Therefore } P = (0, 3)  

\displaystyle \text{Substituting } x=0 \text{ and } y = 3 \text{ in } x-5y+15=0 \text{ we get }

\displaystyle (0)-5(3)+15=0  

\displaystyle -15+15=0 which is true.

\displaystyle \text{Therefore point } (0, 3) \text{ 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 \displaystyle (5, -4) \text{ and } (2,2) is divided by the point \displaystyle Q \text{ in the } \text{Ratio } 1:2 . Does the line \displaystyle x-2y=0 contain \displaystyle Q ?

Answer:

\displaystyle \text{Ratio: } m_1:m_2 = 2:1  

Let the coordinates of the point \displaystyle Q\text{ be } (x, y)  

Therefore

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

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

\displaystyle \text{Therefore } Q = (4, -2)  

\displaystyle \text{Substituting } x=4 \text{ and } y = -2 \text{ in } x-2y=0 \text{ we get }

\displaystyle (4)-2(-2)=0  

\displaystyle 0=0 which is true.

\displaystyle \text{Therefore point } Q(4, -2) \text{ satisfies the equation } x-2y=0 and therefore lies on the equation.

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

Answer:

Solving the two equations: \displaystyle 4x+3y=1 \text{ and } 3x-y+9=0  

\displaystyle y =3x+9  

Substituting

\displaystyle 4x+3(3x+9)=1  

\displaystyle 13x=-26 \Rightarrow x=-2  

\displaystyle \text{Therefore } y = 3(-2)+9=3  

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

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

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

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

Answer:

Solving the two equations first: \displaystyle 2x+5y=1 \text{ and } x-3y=6  

\displaystyle x=3y+6  

\displaystyle \text{Substituting } 2(3y+6)+5y=1  

\displaystyle 6y+12+5y=1  

\displaystyle 11y = 11 \Rightarrow y = -1  

\displaystyle \text{Therefore } x = 3(-1)+6=3  

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

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

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

\displaystyle (3)+5(-1)+2=0  

\displaystyle 10=0 Hence the lines are concurrent.