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

Points on the coordinate plane Quiz 2

mathematics

 A Man starts from point$A\left( { - 2, - 2} \right)$ and walks to the point $B\left( { - 2,4} \right)$then again he walks to the point $C\left( {6,4} \right)$, find the displacement of the man from point $A\,to\,B$.

 First plot the given points on the graph and the use distance formula.

Plotting the graph for the given points $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/81e3390e-495b-4eb2-8f0a-5ce9059283912184648538074499749.png"/>$ \\ $  From the above points we can see the points form a Right angled triangle, where the distance between the point A and B is $ \\  \left( {A,B} \right) = \left( {4 - \left( { - 2} \right)} \right) \\    = \left( {4 + 2} \right) = 6units \\  $ And the distance between the point B and C will be $ \\ \left( {B,C} \right) = \left( {6 - \left( { - 2} \right)} \right) \\    = \left( {8 + 2} \right)  = 8units \\ $ Since the points form a Right angled triangle so the distance between A and C by using Pythagoras theorem, hence we can write $ \\ \left( {A,C} \right) = \sqrt {{6^2} + {8^2}}  \\    = \sqrt {36 + 64}  \\    = \sqrt {100}  = 10units \\  $

 Any of the points on the line $5x + 2y + 7 = 0$ will not lie in the

 Rearrange the equation and check for values of $y$ with respect to $x$.

$  5x + 2y + 7 = 0 \\   \Rightarrow y = \dfrac{{ - 5x - 7}}{2} \\  $ For all positive values of $x$, the equation will result in only negative values of $y$. Thus no point will lie in the 1<sup>st</sup> quadrant.

 The point of intersection of $3x + 2y - 3 = 0$ and $4x + y + 5 = 0$ is $\left( {h,k} \right)$. Then $\left| {\dfrac{k}{h}} \right|$ is

 Solve the system of equations by substitution method to find the intersecting point.

The intersecting point can be found by solving the system of equations using the substitution method. $ \\ 4x + y + 5 = 0 \Rightarrow y =  - 4x - 5 \\ $ Putting this in first equation, $ \\  3x + 2\left( { - 4x - 5} \right) - 3 = 0 \\   \Rightarrow 3x - 8x - 10 - 3 = 0 \\   \Rightarrow  - 5x - 13 = 0 \\   \Rightarrow x =  - \dfrac{{13}}{5} \\  $ And, $y =  - 4x - 5 = \dfrac{{4 \times 13}}{5} - 5 = \dfrac{{52 - 25}}{5} = \dfrac{{27}}{5}$ $ \\ $ Thus, $\left( {h,k} \right) = \left( { - \dfrac{{13}}{5},\dfrac{{27}}{5}} \right)$ $ \\ $ Therefore, $\left| {\dfrac{k}{h}} \right| = \dfrac{{27}}{{13}}$

 A point is given such that the ordinate is $5$ more than $3$ times abscissa. If we move the point $1$ unit to the right then ordinate becomes $4$ times the abscissa. The given point is

 Form equations with the given information and solve to get the point.

Let the given point be $\left( {h,k} \right)$. Then two equations are formed, $ k = 3h + 5$ and $k = 4\left( {h + 1} \right)$ $ \\ $ To solve, $ \\ 3h + 5 = 4\left( {h + 1} \right) \\   \Rightarrow 3h + 5 = 4h + 4 \\   \Rightarrow h = 1 \\  $ And, $k = 3h + 5 = 8$ $ \\ $ Thus the given point is $\left( {1,8} \right)$.

 If the points $\left( {1,4} \right)$, $\left( {5k,k} \right)$ and $\left( {7k,2k} \right)$ are collinear, then $k$ is

 The slope of two points taken together is equal for collinear points.

Slope is given as $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$<sub> </sub>Thus, $ \\   \dfrac{{2k - k}}{{7k - 5k}} = \dfrac{{k - 4}}{{5k - 1}} \\   \Rightarrow \dfrac{1}{2} = \dfrac{{k - 4}}{{5k - 1}} \\   \Rightarrow 5k - 1 = 2k - 8 \\   \Rightarrow 3k =  - 7 \\   \Rightarrow k =  - \dfrac{7}{3} \\  $