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

Coordinates Quiz 2

mathematics

 In the given figure the coordinates of triangle $PQR$ are $\left( -1,-17 \right), \left( -23,19 \right)$ and $R\left( -41,-32 \right)$. Write the center of triangle PQR. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9cfcff9b-6b2a-4ba1-81b8-6dd8e74a974c4630669658292031078.png"/>

 Use the centroid of a triangle formula and substitute given coordinates in it.

  $\text{The given vertices are }P\left( -1,-17 \right), Q\left( -23,19 \right)\text{ and }R\left( -41,-32 \right) \\ \text{Here }\left( x_{3},{{y}_{3}} \right)=\left( -41,-32 \right) \\ \text{Coordinates of centroid }=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \\ \Rightarrow \left( \dfrac{-1-23-41}{3},\dfrac{-17+19-32}{3} \right) \\ \Rightarrow \left( -\dfrac{65}{3},-\dfrac{30}{3} \right) \\ \Rightarrow \left( -\dfrac{65}{3},-10 \right)$

 Let $\left( m,\,n \right)$ be the midpoint of line joining points $\left( 2,3 \right)$ and $\left( -60,-80 \right)$and$\left( p,q \right)$ be the midpoint of the line joining points $\left( 26,42 \right)$and $\left( 16,48 \right)$. Write the value of $m+q$.

 Using midpoint formula, find points $\left( m,n \right)$ and $\left( p,q \right)$ Then write value of $m+q.$

  $\left( m,n \right)=\text{ midpoint of line joining points }\left( 2,3 \right)\text{ and }\left( -60,-80 \right) \\ \Rightarrow \left( \dfrac{2-60}{2},\dfrac{3-80}{2} \right) \\ \Rightarrow \left( m,n \right)=\left( -29,-\dfrac{77}{2} \right)  -----(i) \\ \left( p,q \right)=\text{ midpoint of line joining points }\left( 26,42 \right)\text{ and }\left( 16,48 \right) \\ \Rightarrow \left( \dfrac{26+16}{2},\dfrac{42+48}{2} \right) \\ \Rightarrow \left( p,q \right)=\left( 24,45 \right)  ----(ii) \\ \text{Using equation (i), }m=-29 \\ \text{Using equation (ii), }q=45 \\ \therefore m+q=-29+45=16$

 If $J$ be the sum of x-coordinates of points $\left( \dfrac{-9}{2},6 \right)$ and $\left( 6,\dfrac{5}{9} \right)$and$K$be the product of $y$-coordinates of the points $\left( \dfrac{6}{7},9 \right)$and $\left( \dfrac{15}{11},2 \right)$. Then, in which quadrant, does the point $\left( J,K \right)$  lie?

 Write values of J and K by using the given coordinates. Then check in which quadrant $\left( J,K \right)$ lies.

  $J=\text{ Sum of } “x” \text{ coordinates of points }\left( -\dfrac{9}{2},6 \right)\text{ and }\left( 6,\dfrac{5}{9} \right) \\ \Rightarrow J=-\dfrac{9}{2}+6=\dfrac{3}{2} \\ k=\text{ Product of }y-\text{coordinates of points }\left( \dfrac{6}{7},9 \right)\text{ and }\left( \dfrac{15}{11},2 \right) \\ \Rightarrow k=9\times 2=18 \\ \therefore \text{ Point }\left( J,K \right)=\left( \dfrac{3}{2},18 \right) \\ \text{Clearly, }\left( J,K \right)\text{ lies in the first quadrant.}$

 For the points $u\left( 2,0 \right),v\left( 12,3 \right)$and $w\left( -1,-9 \right)$. Write the value of $uv\,\,+vw.$

 Using distance formula, find $uv$ and $vw.$ Then write value of $uv+vw.$

  $\text{Given points are }u\left( 2,0 \right),v\left( 12,3 \right)\text{ and }w\left( -1,-9 \right) \\ uv=\text{ Distance between } “u” \text{ and } “v” =\sqrt{{{\left( 12-2 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}=\sqrt{100+9}=\sqrt{109} \\ vw=\text{ distance between } “v”\text{ and } “w” =\sqrt{{{\left( -1-12 \right)}^{2}}+{{\left( -9-3 \right)}^{2}}}=\sqrt{169+144}=\sqrt{313} \\ \therefore uv+vw=\sqrt{109}+\sqrt{313}$

 If $\left( \sqrt{\dfrac{1-\cos \dfrac{9p}{2}}{1+\cos \dfrac{9p}{2}}} \right)=\left( 1,0 \right)$, then write the value of $P.$

 First compare $x$ coordinates on both sides. Then use trigonometric formulae to solve further.

  $\text{Given, }\left( \sqrt{\dfrac{1-\cos \dfrac{9p}{2}}{1+\cos \dfrac{9p}{2}}} \right)=\left( 1,0 \right) \\ \text{Compare }“x”\text{ coordinates } \\ \Rightarrow \sqrt{\dfrac{1-\cos \dfrac{9p}{2}}{1+\cos \dfrac{9p}{2}}}=1 \\ \Rightarrow \sqrt{\dfrac{2{{\sin }^{2}}\dfrac{9p}{4}}{2{{\cos }^{2}}\dfrac{9p}{4}}}=1 \\ \Rightarrow \tan \dfrac{9p}{4}=1 \\ \text{Solve further } \\ \Rightarrow \tan \dfrac{9p}{4}=\tan \dfrac{\pi }{4} \\ \Rightarrow \dfrac{9p}{4}=\dfrac{\pi }{4} \\ \Rightarrow p=\dfrac{\pi }{9}$