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This is one of our many quizzes on Analytic geometry. Click on this to access more such competitive quizzes on 3D Coordinate Geometry.

Analytic geometry Quiz 2

mathematics

 What is $\sqrt{2}$  times the distance between $\left( 2,3 \right)$  and  $\left( 4,5 \right)$.

 Substituting points in distance formula and multiply with $\sqrt{2}$

  $\text{The distance between two points is given by }d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \\ \text{The two points are }(2, 3)\text{ and }(4, 5) \\ \text{By substituting we get,} \\ d=\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}}=\sqrt{8} \\ \therefore \sqrt{2}d=\sqrt{2}\times 2\sqrt{2}=4$

 What is the midpoint of  $\left( 2,6 \right)$  and $\left( 3,9 \right)$.

 Substituting two points in a midpoint formula.

  $\text{Midpoint between two points is given by }\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\ \text{Substituting the given values, we get} \\ \left( \dfrac{2+3}{2},\dfrac{6+9}{2} \right)=\left( \dfrac{5}{2},\dfrac{15}{2} \right)$

 What is the ratio of slopes of line passing through $\left( 1,2 \right)$, $\left( 3,1 \right)$ and line through $\left( 2,2 \right)$, $\left( 6,1 \right)$.

 Substitute two points in slope formula and divide them.

  $\text{Ratio of slopes is given by }\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right) \\ \text{Slope for first two points }(1, 2), (3, 1)\text{ is }\left( \dfrac{1-2}{3-1} \right)=-\dfrac{1}{2} \\ \text{Slope for second two points }(2, 2), (6, 1)\text{ is }\left( \dfrac{1-2}{6-2} \right)=-\dfrac{1}{4} \\ \text{Ratio of two slopes is given by }\dfrac{-\dfrac{1}{2}}{-\dfrac{1}{4}}=2$

 Find the square of the equation of line passing through the points $\left( 2,4 \right)$and $\left( 4,3 \right)$.

 Substitute values in equation of line formula $\left( y-{{y}_{1}} \right)=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$ and square on both sides.

  $\text{The equation of a line passing through two points is given by,} \\ \left( y-{{y}_{1}} \right)=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) \\ \Rightarrow \left( y-4 \right)=\dfrac{3-4}{4-2}\left( x-2 \right) \\ \therefore x-10=-2y \\ \text{Squaring on both sides,} \\ {{\left( x-10 \right)}^{2}}={{\left( -2y \right)}^{2}} \\ \Rightarrow {{x}^{2}}+100-20x=4{{y}^{2}} \\ \therefore {{y}^{2}}=\dfrac{{{x}^{2}}}{4}+25-5x$

 Find the midpoint between a point $\left( 16,10 \right)$ and the point of intersection of lines $\left( 1,1 \right)$, $\left( 7,7 \right)$   and  $\left( 4,8 \right)$, $\left( 10,2 \right)$  respectively.

 Find the equations of two lines using, $\left( y-{{y}_{1}} \right)=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$. Then find the intersection point of the two lines using, $(x,y)=\left[ \dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right]$. Next, find the midpoint of the two points using the midpoint formula.

  $\text{The equation of a line passing through two points is }\left( y-{{y}_{1}} \right)=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) \\ \text{Line through }\left( 1,1 \right), \left( 7,7 \right) \\ \left( y-1 \right)=\dfrac{7-1}{7-1}\left( x-1 \right) \\ \therefore y-x=0.........(i) \\ \text{Line through }\left( 4,8 \right), \left( 10,2 \right) \\ \left( y-8 \right)=\dfrac{2-8}{10-4}\left( x-4 \right) \\ y-8=-(x-4) \\ \therefore x+y-12=0......(ii) \\ \text{Intersection of equation }(i)\text{ and }(ii)\text{ is,} \\ (x,y)=\left[ \dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right] \\ \Rightarrow (x,y)=\left[ \dfrac{(1)(-12)-(1)(-1)}{(0)(1)-(1)(1)},\dfrac{(1)(-1)-(0)(-12)}{(0)(1)-(1)(1)} \right] \\ \Rightarrow (x,y)=(6,6) \\ \text{So, the midpoint between }\left( 16,10 \right)\text{ and }(6,6)\text{ is } \\ \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)=\left( \dfrac{16+6}{2},\dfrac{10+6}{2} \right)=\left( 11,8 \right)$