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

Terms in geometry Quiz 2

mathematics

 Which one of the following is a point on the line $y=3\sqrt{x}+9$  .

 Substitute x ordinate values of each option in the given equation of line to get ‘y’ value.

Given equation is $y=3\sqrt{x}+9$, A point satisfying the given equation represents that the point is on a given line. Substituting (4, 15), we get $y=3\sqrt{4}+9=6+9=15$ By substituting each option point in the equation we can say that option (b) is correct.

Find $b$ when $h=30$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bb039049-e9dc-4a94-a711-9269b1078ce56028467967499220807.png"/>

 Use co-exterior and interior angle properties.

According to co-exterior and interior angle property $\angle b=\angle f$ and $\angle f=\angle g$ then $\angle b=\angle g$ $\begin{align} &nbsp;&nbsp;& \angle h+\angle g=180{}^\circ \\ &nbsp;& \Rightarrow 30{}^\circ +\angle b=180{}^\circ \\ &nbsp;& \therefore \angle b=150{}^\circ \\ \end{align}$

 Find which of the following points are collinear and determine the distance between the points $B\ and\text{ C}$.

 First find the slopes of AB, BC and AC if they are equal then the points are collinear. Now, by using the distance formula, find the distance between the two points.

Find slope between the points AB, BC and AC so that if the slopes of three lines are equal then they are collinear. $ \\ \text{slope}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\ $ For points in option $(b)$ $ \\ $ $A(3,2),B\left( 6,5 \right),C(8,7)$ slopes $ \\ AB=\dfrac{5-2}{6-3}=\dfrac{3}{3}=1; BC=\dfrac{7-5}{8-6}=\dfrac{2}{2}=1; AC=\dfrac{7-2}{8-3}=\dfrac{5}{5}=1$ Slopes are equal so points are collinear By using distances between formula $\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$ $BC=\sqrt{{{(8-6)}^{2}}+{{(7-5)}^{2}}}=2\sqrt{2}$

 What is  ${{\dfrac{1}{4}}^{th}}$of the area of a circle with radius  $\dfrac{\sqrt{\sqrt{3}}\sqrt{\sqrt{4}}}{2}$.

 Use the formula of area of circle $A=\pi {{r}^{2}}$ by substituting the given radius and then divide the area by 4 to get the required answer.

Circle Area $=\pi {{r}^{2}}$ ${{\dfrac{1}{4}}^{th}}$area of circle is given as, $\dfrac{1}{4}\times \pi \times {{\left( \dfrac{\sqrt{\sqrt{3}}\sqrt{\sqrt{4}}}{2} \right)}^{2}}=\dfrac{\sqrt{3}\pi }{8}$

 Check the position of point $\left( 1,1 \right)$ with respect to line formed by points $A\left( 1,3 \right),B\left( 5,-1 \right)$

 Find the equation of line from the given points. Then use the relative position of point with respect to line condition for given point and origin to get the required answer.

Now, finding the equations of all lines AB using the two point equation $\begin{align} &nbsp;&nbsp;& \text{Equation of AB} \\ &nbsp;& \Rightarrow y+1=\dfrac{3+1}{1-5}\left( x-5 \right) \\ &nbsp;& \Rightarrow -y-1=x-5 \\ &nbsp;& \Rightarrow x+y-4=0 \\ \end{align}$ If there is line equation $ax+by+c=0$ and two points $P\left( {{x}_{1}},{{y}_{1}} \right),Q\left( {{x}_{2}},{{y}_{2}} \right)$ then (i) $P,Q$ lie on same side of line if $\dfrac{-\left( a{{x}_{1}}+b{{y}_{1}}+c \right)}{a{{x}_{2}}+b{{y}_{2}}+c}<0$ (i) $P,Q$ lie on opposite side of line if $\dfrac{-\left( a{{x}_{1}}+b{{y}_{1}}+c \right)}{a{{x}_{2}}+b{{y}_{2}}+c}>0$ Take the points $\left( 1,1 \right),\left( 0,0 \right)$ and apply the above formula then $\Rightarrow \dfrac{-\left( 1+1-4 \right)}{0+0-4}=\dfrac{2}{-4}<0$ So, point $\left( 1,1 \right)$ lies on same side of origin with respect to line formed by points $A\left( 1,3 \right),B\left( 5,-1 \right)$