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This is one of our many quizzes on Line intersections. Click on this to access more such competitive quizzes on Straight Lines in Coordinate Geometry.

Line intersections Quiz 2

mathematics

 Statement: $\dfrac{x}{a}+\dfrac{y}{b}=1$ is the normal form of a straight line.

 Despite the fact that $\dfrac{x}{a}+\dfrac{y}{b}=1$ Choose the appropriate alternative, as it is an intercept form of a straight line.

Checking the intercepts of given line which are the intersections of line and axis, $ \\ \dfrac{x}{a}+\dfrac{y}{b}=1 \\ $ It is clear that the given equation is in the intercept form of a straight line, because it contains $a$ and $b$ as $x$ and $y$ intercepts respectively and 1 as a constant term in RHS.

 Find the equation of line cutting of the y-axis at 5 units in the negative direction and make an angle $\tan \theta =\dfrac{9}{5}$.

 The slope and $y$ intercept of a line equation can be used to calculate the slope intercept. Substitute the values in the slope intercept form to get the required line equation.

Given $\text{tan}\theta \text{=}\dfrac{9}{5}$ which is equal to the slope of the line $m$. $ \\ $ From the question it is clear that the y-intercept is $-5$. $ \\ $ Now substitute the values in slope intercept form of a line equation. $ \\ \Rightarrow y=\dfrac{9}{5}x-5 \\  \Rightarrow 9x-5y-5=0$

 If the sum of $x$ intercept and $y$ intercept of a straight line is $12$ units and if it passes through the point $(4,-10)$. Then find the line equation?

 Develop two equations one from the sum of intercepts and other by substituting a given point in the line and then solve them with the other. Then use a quadratic equation to get the value of $a$ and $b$.

$\dfrac{x}{a}+\dfrac{y}{b}=1 \\  \text{From the question a+b=12}\text{.} \\  \Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=1 \\ $ Though it is passing from the point $(4,-10)$ $ \\ \dfrac{4}{a}+\dfrac{-10}{b}=1 \\  \Rightarrow 4b-10a=ab \\  \Rightarrow 4\left( 12-a \right)-10a=a\left( 12-a \right) \\ $ After the calculation part $ \\ \Rightarrow {{a}^{2}}-26a-48=0 \\  \Rightarrow \left( a-2 \right)\left( a-24 \right)=0 \\  \Rightarrow \text{ a=2,24} \\ $ Substituting in the first equation for $b$. $ \\ $ $b = 10, -12$ for the 2, 24 a values. Therefore line equations are, $ \\ \dfrac{x}{2}+\dfrac{y}{10}=1\text{ and }\dfrac{x}{24}-\dfrac{y}{12}=1$

 If a straight line $6x-9y=0$ intercepts a circle at the point $(0, 0)$ which is the centre of a circle and also has a radius of $5$ units. Then find out the other points in the circle intersected by line?

 Substitute the line equation in the circle equation to get the intersection points between the line and circle. 

From the question equation of circle is $ \\ {{\text{x}}^{2}}+{{y}^{2}}=25 \\ $ As the line $6x-9y=0$ is passing through ${{\text{x}}^{2}}+{{y}^{2}}=25$ $ \\ $ Substitute $x=\dfrac{3}{2}y$ in circle equation to get the intersected points. $ \\ {{\left( \dfrac{3}{2}y \right)}^{2}}+{{y}^{2}}=25 \\  \Rightarrow \dfrac{9}{4}{{y}^{2}}+{{y}^{2}}=25 \\  \Rightarrow {{y}^{2}}=25\left( \dfrac{4}{13} \right)=\dfrac{100}{13} \\  \Rightarrow y=\pm \dfrac{10}{\sqrt{13}} \\ $ By substituting in circle equation $ \\ \Rightarrow {{\text{x}}^{2}}=25-\dfrac{100}{13}=\dfrac{325-100}{13}=\dfrac{225}{13} \\  \Rightarrow x=\pm \dfrac{16}{\sqrt{13}} \\ $ Therefore intersected points are $\left( 0,0 \right),\left( \dfrac{16}{\sqrt{13}}, \dfrac{10}{\sqrt{13}} \right)$ and $\left( -\dfrac{16}{\sqrt{13}}, -\dfrac{10}{\sqrt{13}} \right)$

 Find the equation of the straight line if the magnitudes of both intercept forms are equal but opposite in sign and pass through the point $(3, 9)$

 As intercepts are equal in magnitude and opposite in sign i.e. $b=-a$. Now substitute this equation and the point in the basic intercept form of a straight line and then solve to get the line equation.

Given $b = -a$ $ \dfrac{x}{a}-\dfrac{y}{a}=1 \\  \Rightarrow x-y=a \\ $ As it is passing through the point $\left( 3,9 \right)$ $ \\ $ $a=-6$, Therefore line equation will be, $ \\ \dfrac{y}{6}-\dfrac{x}{6}=1 \\  \Rightarrow x-y+6=0$