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This is one of our many quizzes on Distance formula. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Distance formula Quiz 2

mathematics

 Find the length of the longest line that can be drawn in an A14 size paper $($i.e., $3\times 4$units$)$ which passes through the origin and one of its breadth lies on the line $x=2$?

 First find the coordinates of the four vertices of the rectangle then by using the distance formula$d=\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}}$, find the length of the diagonal.

  $\text{Draw the given four lines and we get a rectangle ABCD as follows:} \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/af72218e-cee6-433d-a0a1-86bd8bee881d3187020322990843182.png"/>$\\ \text{Length of the diagonals AC and BD, by distance formula,} \\  d=\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}} \\ \sqrt{{{(2+2)}^{2}}+{{(0-3)}^{2}}} \\ =\sqrt{{{4}^{2}}+{{3}^{2}}}=5$

 An ant starts from point $(2,3)$ travels to a sugar cube which is placed at $(4,5)$ and carries the sugar cube at $(6,7)$. Find the total distance travelled by the ant.

 First find the separate distance traveled by the ant by the distance formula, $d=\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}}$, then add both of them to find the total distance traveled. 

  The ant started from $(2,3)$, then traveled to $(4,5)$ then again traveled to $(6,7)$ now, the distance $d$ of a line segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is given by, $\\ d=\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}} \\ \Rightarrow \sqrt{{{(4-2)}^{2}}+{{(5-3)}^{2}}}+\sqrt{{{(6-4)}^{2}}+{{(7-5)}^{2}}} \\ \Rightarrow \sqrt{{{2}^{2}}+{{2}^{2}}}+\sqrt{{{2}^{2}}+{{2}^{2}}} \\ \Rightarrow \sqrt{8}+\sqrt{8}=2\sqrt{8}=4\sqrt{2}$  

 Find the radius of the circle ${{x}^{2}}~+{{y}^{2}}~-4x-1=0$, which passes through the point $\left( 1,2 \right)$.

 First convert the given equation in the standard form of a circle’s equation, then by comparing both, we will get the centre, then by distance formula $d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$ we will get the radius.

 $\text{Equation of the circle is }\\ {{x}^{2}}+{{y}^{2}}-4x-1=0  \\  {{x}^{2}}+{{y}^{2}}-4x-1+5=5  \\  {{x}^{2}}-4x+4+{{y}^{2}}=5  \\ {{\left( x-2 \right)}^{2}}+{{y}^{2}}=5  \\ \text{Comparing the above equation with the general equation of a circle,} \\ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}},\text{ we get,} \\ \text{Centre of the circle }C\left( 2,0 \right)\text{ and } \\ \text{radius = }\sqrt{{{(1-2)}^{2}}+{{(2-0)}^{2}}}=\sqrt{5}\text{ units}$ 

 A triangular carpet is to be made out, such that the three vertices of the carpet lie at $(0,1)$, $(2,3)$ and $(4,-1)$, what would be the length of the lace required to border the carpet?

 First find the length of the three sides by using $d=\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}}$, then add up all the three to get the total length.

 $\text{Using the distance formula, }d=\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}} \\ \text{Let the three vertices be “X” }(0,1), Y(2,3)\text{ and }Z(4,-1). \\ XY=\sqrt{{{(0-2)}^{2}}+{{(1-3)}^{2}}} \\ =\sqrt{8}=2\sqrt{2} \\ YZ=\sqrt{{{(2-4)}^{2}}+{{(3+1)}^{2}}} \\ =\sqrt{20}=2\sqrt{5} \\ ZX=\sqrt{{{(0-4)}^{2}}+{{(1+1)}^{2}}} \\ =\sqrt{20}=2\sqrt{5} \\ \text{Thus, the length of the lace required } \\ \Rightarrow XY+YZ+ZX=2\sqrt{2}+2\sqrt{5}+2\sqrt{5}=2(\sqrt{2}+2\sqrt{5)}\text{ units} $

 If the coordinated of the three vertices of a $\Delta ABC$are $\left( 6,-1 \right),\left( 1,3 \right)\text{ and }\left( a,8 \right)$ respectively, then find the value(s) of $x$ such that $AB=BC$.

 Use distance formula, $d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$find the individual lengths then equate both to get the value of x.

 $\begin{array}{*{35}{l}}   \text{AB }=\sqrt{{{\left( 6-1 \right)}^{2}}+{{\left( -1-3 \right)}^{2}}}  \\    \Rightarrow \sqrt{{{5}^{2}}+{{\left( -4 \right)}^{2}}}  \\   \Rightarrow \sqrt{41}  \\   \text{BC }=\sqrt{{{\left( 1-a \right)}^{2}}+{{\left( 3-8 \right)}^{2}}}  \\   \Rightarrow \sqrt{{{\left( 1-a \right)}^{2}}+25}  \\   \text{Therefore},\text{ AB=BC}  \\   \sqrt{41}=\sqrt{{{\left( 1-a \right)}^{2}}+25}  \\   {{\left( 1-a \right)}^{2}}=41-25  \\   a=1\pm 4  \\   a=-3,5  \end{array}$ 