Class 10: Equation of a Line – Exercise 14(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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.

$\displaystyle \\$

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 $\displaystyle a = 5$ .

$\displaystyle \\$

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle 9(3)+4(-k)=3$

$\displaystyle \Rightarrow k = 6$

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

$\displaystyle \\$

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:

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

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

$\displaystyle 9m-10(2m-1)+15=0$

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \\$

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:

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

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

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

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

$\displaystyle \\$

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.

$\displaystyle \\$

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.

$\displaystyle \\$

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 $\displaystyle k$