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This is one of our many quizzes on Types of lines. Click on this to access more such competitive quizzes on Geometry.

Types of lines Quiz 2

mathematics

 Suppose two planes travel in a straight line with trajectories $3x-8y+11=0$ and $8x+3y+11=0$ respectively. It was noticed that their trajectory intersects each other at point $\left( a,b \right)$. Then, find the point $\left( a,b \right)$.

 Solve the linear equation by replacing $(x,y)$ with $\left( a,b \right)$ and using the substitution method, obtain the value of $\left( a,b \right).$

 Trajectories $2x-8y=10$ and $8x+2y=10$ intersect each other at point $\left( a,b \right)$. $ \\ $ Then, the point $\left( a,b \right)$ will satisfy the lines $2x-8y=10$ and $8x+2y=10$. $ \\ $ So, from $2(a) - 8\left( b \right)\text{ = 10}$ $ \\  \Rightarrow a=4b+5.....\left( i \right) \\ $ Putting $\left( i \right)$ in equation $8a+2b=10$, $ \\  \Rightarrow 8\left( 4b+5 \right)\text{+}3b=11 \\  \Rightarrow 32b+40+3b=11 \\  \Rightarrow 35b=-29 \\  \Rightarrow b\text{ }=-\dfrac{29}{35} \\ $ Then, $ \\  \Rightarrow a=4\left( -\dfrac{29}{35} \right)+5 \\  \Rightarrow a=\left( \dfrac{59}{35} \right) \\ $ Point of intersection of trajectories is $\left( \dfrac{59}{35},-\dfrac{29}{35} \right)$ 

 An eagle glides from top of the tree towards its prey which was on ground. If the distance between the prey and the tree is $5\sqrt{3}\text{ m}$ and angle made by ground and the slant trajectory of an eagle is $60{}^\circ $ , then what is the height of the tree?

 Recall the value of $\text{tan}\theta $ in the right angle triangle and also the value of $\text{tan 60}{}^\circ $, to determine the height of the tree.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b81f798b-4cdb-456e-aed5-dbb5a46dd5723842242654824583920.png"/>$ \\ $ Distance between prey and tree is $5\sqrt{3}\text{ m}$ $ \\ $ Angle made by ground and the slant trajectory of eagle is $60{}^\circ$ $ \\ $ Let, the height of tree is $h$m $ \\ $ In right angle triangle, with height $h$, base $b$ and hypotenuse $H$ $ \\ \Rightarrow \text{tan}\theta \text{=}\dfrac{h}{b}; \\ \Rightarrow \text{tan 60}{}^\circ =\dfrac{h}{5\sqrt{3}} \\  \Rightarrow \sqrt{3}=\dfrac{h}{5\sqrt{3}} \\  \Rightarrow h=5\sqrt{3}\times \sqrt{3}\text{ m} \\  \Rightarrow h=5\times 3\text{ m} \\  \Rightarrow h=15\text{ m} \\ $ 

 A train has two stops only i.e. station A and station B. In between A and B, the station C lies. If the distance between the station A and station B is $150\text{ km}$ and the ratio of the distance between stations A and C and the distance between stations A and B is $2:5$ i.e. $\dfrac{\text{AC}}{\text{AB}}=\dfrac{2}{5}$. If stations A, B and C are collinear then, distance between station C and station B is

 Using the ratio of distance given in question and the fact that $\text{AO +}\text{ OB = AB}$, determine the distance between the station C and station B.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fa425c79-5568-4a42-b9a5-da62896dd8828650900918509230701.png"/> $ \\ $ The ratio of the distance between station A and C and the distance between station A and B is $\dfrac{2}{5}$ $ \\  \Rightarrow \dfrac{\text{AC}}{\text{AB}}=\dfrac{2}{5} \\ $  As, from the figure, $AC +{ CB = AB}$ $ \\  \Rightarrow \text{AC = AB - CB} \\  \Rightarrow \dfrac{\text{AB - CB}}{\text{AB}}=\dfrac{2}{5} \\ $ Distance between station A and station B is given to be $150 km$ $ \\  \Rightarrow \dfrac{\text{150 - CB}}{150}=\dfrac{2}{5} \\  \Rightarrow 5\left( \text{150 - CB} \right)=2\times 150 \\  \Rightarrow \text{750 - 5CB= 30}0 \\  \Rightarrow \text{5CB = 450 km} \\  \Rightarrow \text{CB = 90 km}\text{.} \\ $ 

 The income of two employees A and B are represented by the straight lines ${{y}_{1}}=20x+60\text{ }$ and ${{y}_{2}}=5x+240$, where ${{y}_{1}},{{y}_{2}}$ represents their income and x represents the number of units sold by them in a month. Then how many units they must sell in a month so that their income is the same.

 For the same income, make their income equal i.e. ${{y}_{1}}={{y}_{2}}$ and solve the equation to find out the number of units to be sold.

  $  \text{The income of employee A is }{{y}_{1}}=20x+60\text{ } \\  \text{and the income of employee B is }{{y}_{2}}=5x+240. \\  \text{For same income, } \\  \Rightarrow {{y}_{1}}={{y}_{2}} \\  \Rightarrow 20x+60=5\text{x+240} \\  \Rightarrow 20x-5x=\text{240-60} \\  \Rightarrow 15x=180 \\  \Rightarrow x=\dfrac{180}{15} \\  \Rightarrow x=12 \\  \text{Thus, they have to sell 12 units to make their income same} \\ $ 

 On the plotting graph, the map of the park was traced. The boundaries of the triangle are formed by lines $x+y=1$ , $y-x=1$ and $y=0$. Then area of triangle is

 Find the boundary points of the park by solving the linear equations. Recall the area of the triangle with given vertices and calculate the area of the park.

  The boundaries of the triangle are $x+y=1\text{, }y-x=1$ and $y=0$. $ \\ $ As, all three lines intersect each other, then boundaries of triangle will be the point satisfying the linear equations $x+y=1$ and $y=0$,  $ \\  \Rightarrow (0)+x=1 \\  \Rightarrow x=1 \\ $ Point satisfying the linear equations $y-x=1$ and $y=0$, $ \\  \Rightarrow (0)-x=1 \\  \Rightarrow x=-1 \\ $ Point satisfying the linear equations $y-x=1$ and $x+y=1$ $ \\ \Rightarrow \left( y-x \right)+\left( x+y \right)=1+1 \\  \Rightarrow 2y=2 \\  \Rightarrow y=1 \\  \Rightarrow x+(1)=1 \\  \Rightarrow x=0 \\ $ So, boundaries of park are $\left( -1,0 \right),\text{ }\left( 1,0 \right)$ and $\left( 0,1 \right)$. $ \\ $ If coordinates of triangle are $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then area of Triangle is equal to $\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right| \\ $ Then area of park whose vertices are $\left( -1,0 \right),\text{ }\left( 1,0 \right)$ and $\left( 0,1 \right)$ is, $ \\  \Rightarrow \dfrac{1}{2}\left| \left( -1 \right)\left( 0-1 \right)+\left( 1 \right)\left( 1-0 \right)+\left( 0 \right)\left( 0-0 \right) \right|\text{ c}{{\text{m}}^{2}} \\  \Rightarrow \dfrac{1}{2}\left| 1+1 \right|\text{ c}{{\text{m}}^{2}} \\  \Rightarrow \dfrac{1}{2}\times 2\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 1\text{ c}{{\text{m}}^{2}} \\ $ 