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 $x-2y+5=0$ :

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.

$\\$

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 $a = 5$ .

$\\$

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

$\\$

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

$\\$

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 $m =\frac{25}{11}$

$\\$

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 equation $3x-5y=6$ .

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

$\\$

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

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

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

$\\$

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

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

$\\$

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.

$\\$

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 equation $x-2y=0$ and therefore lies on the equation.

$\\$

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

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

$\\$

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 $2(3y+6)+5y=1$

$6y+12+5y=1$

$11y = 11 \Rightarrow y = -1$

Therefore $x = 3(-1)+6=3$

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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Class 10: Equation of a Line – Exercise 14(a)

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