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

Midpoint formula Quiz 2

mathematics

 If $\text{C}(5,6)$ is the midpoint of the line joining the points $\text{A}(a+1,b)$ and $\text{B}(-b,a-1)$, then the values of $a$ and $b$ are satisfying in which of the following expressions?

 First find the coordinates of midpoint of line AB using formula $\left( \dfrac{~{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ and then equate both the midpoint coordinates to find the value of $a$ and $b$. 

  $\text{Mid-point }=\left( \dfrac{~{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\ \Rightarrow (5,6)=\left( \dfrac{a+1-b}{2},\dfrac{b+a-1}{2} \right) \\ \Rightarrow \dfrac{a+1-b}{2}=5,\dfrac{b+a-1}{2}=6 \\ \Rightarrow a-b=9,a+b=13 \\ \Rightarrow a=11,b=2 \\\Rightarrow 2ab=2\times 11\times 2=44$  

 $(3,-2)$ is the midpoint of $(1,x)$ and $(5,7)$, find the value of $x$.

 Using the midpoint formula, $\left( x,y \right)=\left[ \dfrac{\left( x_1+x_2 \right)}{2},\dfrac{\left( y_1+y_2 \right)}{2} \right]$, and taking the unknown variable as $x$, we can solve further for the final answer.

  $\text{Using the midpoint formula, }\left( x,y \right)=\left[ \dfrac{\left( x_1+x_2 \right)}{2},\dfrac{\left( y_1+y_2 \right)}{2} \right] \\ \text{Now, }  \left( 3,\text{ -2} \right)=\left[ \dfrac{\left( x_1+x_2 \right)}{2},\dfrac{\left( y_1+y_2 \right)}{2} \right] \\ \text{(3,-2)=}\left[ \dfrac{5+1}{2},\dfrac{7+x}{2} \right] \\ \text{(3,-2)=}\left[ 3,\dfrac{7+x}{2} \right] \\ -2=\dfrac{7+x}{2} \\ x=-11 $ 

 Find the midpoint of the line which is formed by joining $(5,2)$ and the intersecting point of the lines $x-y=1$ and $3y-2x=5$.

 First we will find the intersection point from the two given lines, then by using the midpoint formula, $\left( x,y \right)=\left[ \dfrac{\left( x_1+x_2 \right)}{2},\dfrac{\left( y_1+y_2 \right)}{2} \right]$, we will get the midpoint.

  $\text{Point of intersection of }x-y=1\text{ and }3y-2x=5\text{ is } \\ 3x-3y+3=0 \\ 2x-3y+5=0 \\ x=2,y=3 \\ \text{Midpoint, }\left( x,y \right)=\left[ \dfrac{\left( \text{5 }+\text{ 2} \right)}{2},\dfrac{\left( \text{2 }+\text{ 3} \right)}{2} \right]\text{ =}\left( \dfrac{7}{2},\dfrac{5}{2} \right)$

 Find the equation of line PQ whose y-intercept is $-5$ and $PQ\parallel ST$ where S and T are the mid-points of BC and CD respectively. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bf17c5b0-89a5-46ac-a5f6-60462b904d5c410788562576413180.png"/>

 First find the coordinates of S and T using the midpoint formula and then slope of ST to get the required equation.

  $\text{Let }y=mx+c\text{ be the required equation of line PQ, } \text{where }c=-5 \\ \text{Coordinates of S }=\left( {{x}_{4}},{{y}_{4}} \right)=\left( \dfrac{~{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)=\left( \dfrac{4-2}{2},\dfrac{9-1}{2} \right)=\left( 1,4 \right) \\ \text{Coordinates of T }=\left( {{x}_{5}},{{y}_{5}} \right)=\left( \dfrac{~{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)=\left( \dfrac{4+6}{2},\dfrac{9-5}{2} \right)=\left( 5,2 \right)\\m=\dfrac{{{y}_{5}}-{{y}_{4}}}{{{x}_{5}}-{{x}_{4}}}=\dfrac{2-4}{5-1}=\dfrac{-1}{2}\text{        }\left[ \because \text{ PQ}\parallel \text{ST}\Rightarrow \text{Slope of PQ = Slope of ST} \right]\\\Rightarrow \text{ Equation of PQ is  }y=\dfrac{-1}{2}x-5\Rightarrow x+2y+10=0$ 

 The point $(k,7)$ passes through the line $2x+2y=-1$ and the point $(a,b)$ falls exactly between $(3,4)$ and $(k,7)$, find the value of $k$.

 Using the midpoint formula, $\left( x,y \right)=\left[ \dfrac{\left( x_1+x_2 \right)}{2},\dfrac{\left( y_1+y_2 \right)}{2} \right]$, we will get a relation between $k$ and $a$, then as the point passes through the given line, then we will put the value of $k$ there to get the required value.

  $ \dfrac{3+k}{2}=a \\ \Rightarrow 3+k=2a\ldots (1) \\ \dfrac{4+7}{2}=b \\ b=\dfrac{11}{2}\ldots (2) \\ \text{Given k passes through the line }2x+2y=-1 \\ \text{From the above two equations, } 3+k+11+1=0,k=-15$ 