Class 10: Equation of a Line

$\displaystyle \textbf{Question 1: }\text{Which of the following points lie on the line }x-2y+5=0:$
$\displaystyle \text{i) }(1,3)\qquad \text{ii) }(0,5)\qquad \text{iii) }(-5,0)\qquad \text{iv) }(5,5)\qquad \text{v) }(2,-1.5)\qquad \text{vi) }(-2,-1.5)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Substituting }x=1\text{ and }y=3\text{ in }x-2y+5=0,$
$\displaystyle 1-2(3)+5=1-6+5=0$
$\displaystyle \therefore (1,3)\text{ satisfies the equation and lies on the line.}$
$\displaystyle \text{ii) Substituting }x=0\text{ and }y=5\text{ in }x-2y+5=0,$
$\displaystyle 0-2(5)+5=0-10+5=-5\neq0$
$\displaystyle \therefore (0,5)\text{ does not satisfy the equation and does not lie on the line.}$
$\displaystyle \text{iii) Substituting }x=-5\text{ and }y=0\text{ in }x-2y+5=0,$
$\displaystyle -5-2(0)+5=-5+5=0$
$\displaystyle \therefore (-5,0)\text{ satisfies the equation and lies on the line.}$
$\displaystyle \text{iv) Substituting }x=5\text{ and }y=5\text{ in }x-2y+5=0,$
$\displaystyle 5-2(5)+5=5-10+5=0$
$\displaystyle \therefore (5,5)\text{ satisfies the equation and lies on the line.}$
$\displaystyle \text{v) Substituting }x=2\text{ and }y=-1.5\text{ in }x-2y+5=0,$
$\displaystyle 2-2(-1.5)+5=2+3+5=10\neq0$
$\displaystyle \therefore (2,-1.5)\text{ does not satisfy the equation and does not lie on the line.}$
$\displaystyle \text{vi) Substituting }x=-2\text{ and }y=-1.5\text{ in }x-2y+5=0,$
$\displaystyle -2-2(-1.5)+5=-2+3+5=6\neq0$
$\displaystyle \therefore (-2,-1.5)\text{ does not satisfy the equation and does not lie on the line.}$
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$\displaystyle \textbf{Question 2: }\text{State, True or False:}$
$\displaystyle \text{i) The line }\frac{x}{2}+\frac{y}{3}=0\text{ passes through the point }(2,3)$
$\displaystyle \text{ii) The line }\frac{x}{2}+\frac{y}{3}=0\text{ passes through the point }(4,-6)$
$\displaystyle \text{iii) The point }(8,7)\text{ lies on the line }y-7=0$
$\displaystyle \text{iv) The point }(-3,0)\text{ lies on the line }x+3=0$
$\displaystyle \text{v) If the point }(2,a)\text{ lies on the line }2x-y=3,\text{ then }a=5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Substituting }x=2\text{ and }y=3\text{ in }\frac{x}{2}+\frac{y}{3}=0,$
$\displaystyle \frac{2}{2}+\frac{3}{3}=1+1=2\neq0$
$\displaystyle \therefore (2,3)\text{ does not satisfy the equation. Hence, the statement is False.}$
$\displaystyle \text{ii) Substituting }x=4\text{ and }y=-6\text{ in }\frac{x}{2}+\frac{y}{3}=0,$
$\displaystyle \frac{4}{2}+\frac{-6}{3}=2-2=0$
$\displaystyle \therefore (4,-6)\text{ satisfies the equation. Hence, the statement is True.}$
$\displaystyle \text{iii) Substituting }y=7\text{ in }y-7=0,$
$\displaystyle 7-7=0$
$\displaystyle \therefore (8,7)\text{ lies on the line }y-7=0.\text{ Hence, the statement is True.}$
$\displaystyle \text{iv) Substituting }x=-3\text{ in }x+3=0,$
$\displaystyle -3+3=0$
$\displaystyle \therefore (-3,0)\text{ lies on the line }x+3=0.\text{ Hence, the statement is True.}$
$\displaystyle \text{v) Substituting }x=2\text{ and }y=a\text{ in }2x-y=3,$
$\displaystyle 2(2)-a=3$
$\displaystyle 4-a=3$
$\displaystyle a=1$
$\displaystyle \therefore a\neq5.\text{ Hence, the statement is False.}$
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$\displaystyle \textbf{Question 3: }\text{The line }2x-\frac{y}{3}=7\text{ passes through the point }(k,6). \text{ Calculate }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=k\text{ and }y=6\text{ in }2x-\frac{y}{3}=7,$
$\displaystyle 2k-\frac{6}{3}=7$
$\displaystyle 2k-2=7$
$\displaystyle 2k=9$
$\displaystyle k=\frac{9}{2}=4.5$
$\displaystyle \therefore \text{The required value of }k\text{ is }4.5.$
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$\displaystyle \textbf{Question 4: }\text{For what value of }k\text{ will the point }(3,-k)\text{ lie on the line }9x+4y=3?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=3\text{ and }y=-k\text{ in }9x+4y=3,$
$\displaystyle 9(3)+4(-k)=3$
$\displaystyle 27-4k=3$
$\displaystyle -4k=-24$
$\displaystyle k=6$
$\displaystyle \therefore \text{The required value of }k\text{ is }6.$
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$\displaystyle \textbf{Question 5: }\text{The line }\frac{3x}{5}-\frac{2y}{3}+1=0\text{ contains the point }(m,2m-1).\text{ Calculate }m.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=m\text{ and }y=2m-1\text{ in }\frac{3x}{5}-\frac{2y}{3}+1=0,$
$\displaystyle \frac{3m}{5}-\frac{2(2m-1)}{3}+1=0$
$\displaystyle 9m-10(2m-1)+15=0$
$\displaystyle 9m-20m+10+15=0$
$\displaystyle -11m+25=0$
$\displaystyle m=\frac{25}{11}$
$\displaystyle \therefore \text{The required value of }m\text{ is }\frac{25}{11}.$
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$\displaystyle \textbf{Question 6: }\text{Does the line }3x-5y=6\text{ bisect the join of }(5,-2)\text{ and }(-1,2)?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The midpoint divides the line segment in the ratio }1:1.$
$\displaystyle \text{Midpoint }P=\left(\frac{5+(-1)}{2},\frac{-2+2}{2}\right)=(2,0)$
$\displaystyle \text{Substituting }x=2\text{ and }y=0\text{ in }3x-5y=6,$
$\displaystyle 3(2)-5(0)=6$
$\displaystyle 6=6$
$\displaystyle \therefore (2,0)\text{ lies on the given line.}$
$\displaystyle \therefore \text{The line }3x-5y=6\text{ bisects the join of the given points.}$
$\\$

$\displaystyle \textbf{Question 7:}$
$\displaystyle \text{i) The line }y=3x-2\text{ bisects the join of }(a,3)\text{ and }(2,-5).\text{ Find }a.$
$\displaystyle \text{ii) The line }x-6y+11=0\text{ bisects the join of }(8,-1)\text{ and }(0,k).\text{ Find }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) The midpoint divides the line segment in the ratio }1:1.$
$\displaystyle P=\left(\frac{a+2}{2},\frac{3+(-5)}{2}\right)=\left(\frac{a+2}{2},-1\right)$
$\displaystyle \text{Substituting }x=\frac{a+2}{2}\text{ and }y=-1\text{ in }y=3x-2,$
$\displaystyle -1=3\left(\frac{a+2}{2}\right)-2$
$\displaystyle 1=\frac{3(a+2)}{2}$
$\displaystyle 2=3a+6$
$\displaystyle 3a=-4$
$\displaystyle a=-\frac{4}{3}$
$\displaystyle \therefore \text{The required value of }a\text{ is }-\frac{4}{3}.$
$\displaystyle \text{ii) The midpoint divides the line segment in the ratio }1:1.$
$\displaystyle P=\left(\frac{8+0}{2},\frac{-1+k}{2}\right)=\left(4,\frac{k-1}{2}\right)$
$\displaystyle \text{Substituting }x=4\text{ and }y=\frac{k-1}{2}\text{ in }x-6y+11=0,$
$\displaystyle 4-6\left(\frac{k-1}{2}\right)+11=0$
$\displaystyle 8-6(k-1)+22=0$
$\displaystyle 30-6k+6=0$
$\displaystyle 36-6k=0$
$\displaystyle k=6$
$\displaystyle \therefore \text{The required value of }k\text{ is }6.$
$\\$

$\displaystyle \textbf{Question 8:}$
$\displaystyle \text{i) The point }(-3,2)\text{ lies on the line }ax+3y+6=0.\text{ Calculate }a.$
$\displaystyle \text{ii) The line }y=mx+8\text{ contains the point }(-4,4).\text{ Calculate }m.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Substituting }x=-3\text{ and }y=2\text{ in }ax+3y+6=0,$
$\displaystyle a(-3)+3(2)+6=0$
$\displaystyle -3a+12=0$
$\displaystyle a=4$
$\displaystyle \therefore \text{The required value of }a\text{ is }4.$
$\displaystyle \text{ii) Substituting }x=-4\text{ and }y=4\text{ in }y=mx+8,$
$\displaystyle 4=m(-4)+8$
$\displaystyle -4m=-4$
$\displaystyle m=1$
$\displaystyle \therefore \text{The required value of }m\text{ is }1.$
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$\displaystyle \textbf{Question 9: }\text{The point }P\text{ divides the join of }(2,1)\text{ and }(-3,6)\text{ in the ratio }2:3.\text{ Does }P\text{ lie on the line }x-5y+15=0?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }P\text{ divides the line segment in the ratio }2:3,$
$\displaystyle x=\frac{2(-3)+3(2)}{2+3}=0$
$\displaystyle y=\frac{2(6)+3(1)}{2+3}=3$
$\displaystyle \therefore P=(0,3)$
$\displaystyle \text{Substituting }x=0\text{ and }y=3\text{ in }x-5y+15=0,$
$\displaystyle 0-5(3)+15=0$
$\displaystyle -15+15=0$
$\displaystyle \therefore (0,3)\text{ satisfies the equation }x-5y+15=0\text{ and lies on the given line.}$
$\\$

$\displaystyle \textbf{Question 10: }\text{The line segment joining }(5,-4)\text{ and }(2,2)\text{ is divided by }Q\text{ in the ratio }1:2.\text{ Does the line }x-2y=0\text{ contain }Q?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }Q\text{ divides the line segment in the ratio }1:2,$
$\displaystyle x=\frac{1(2)+2(5)}{1+2}=4$
$\displaystyle y=\frac{1(2)+2(-4)}{1+2}=-2$
$\displaystyle \therefore Q=(4,-2)$
$\displaystyle \text{Substituting }x=4\text{ and }y=-2\text{ in }x-2y=0,$
$\displaystyle 4-2(-2)=4+4=8\neq0$
$\displaystyle \therefore Q(4,-2)\text{ does not satisfy }x-2y=0.$
$\displaystyle \therefore \text{The line }x-2y=0\text{ does not contain }Q.$
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$\displaystyle \textbf{Question 11: }\text{Find the point of intersection of the lines }4x+3y=1\text{ and }3x-y+9=0.\text{ If this point lies on the line }(2k-1)x-2y=4,\text{ find }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equations are }4x+3y=1\text{ and }3x-y+9=0.$
$\displaystyle \text{From }3x-y+9=0,$
$\displaystyle y=3x+9$
$\displaystyle \text{Substituting }y=3x+9\text{ in }4x+3y=1,$
$\displaystyle 4x+3(3x+9)=1$
$\displaystyle 4x+9x+27=1$
$\displaystyle 13x=-26$
$\displaystyle x=-2$
$\displaystyle y=3(-2)+9=3$
$\displaystyle \therefore \text{The point of intersection is }(-2,3).$
$\displaystyle \text{Substituting }x=-2\text{ and }y=3\text{ in }(2k-1)x-2y=4,$
$\displaystyle (2k-1)(-2)-2(3)=4$
$\displaystyle -4k+2-6=4$
$\displaystyle -4k-4=4$
$\displaystyle -4k=8$
$\displaystyle k=-2$
$\displaystyle \therefore \text{The required value of }k\text{ is }-2.$
$\\$

$\displaystyle \textbf{Question 12: }\text{Show that the lines }2x+5y=1,\;x-3y=6\text{ and }x+5y+2=0\text{ are concurrent.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First solve the equations }2x+5y=1\text{ and }x-3y=6.$
$\displaystyle \text{From }x-3y=6,\;x=3y+6$
$\displaystyle \text{Substituting in }2x+5y=1,$
$\displaystyle 2(3y+6)+5y=1$
$\displaystyle 6y+12+5y=1$
$\displaystyle 11y=-11$
$\displaystyle y=-1$
$\displaystyle x=3(-1)+6=3$
$\displaystyle \therefore \text{The point of intersection of the first two lines is }(3,-1).$
$\displaystyle \text{Substituting }x=3\text{ and }y=-1\text{ in }x+5y+2=0,$
$\displaystyle 3+5(-1)+2=0$
$\displaystyle 3-5+2=0$
$\displaystyle 0=0$
$\displaystyle \therefore (3,-1)\text{ lies on the third line.}$
$\displaystyle \therefore \text{The three lines are concurrent.}$
$\\$