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

Slope Quiz 2

mathematics

 The slope of the tangent to the circle ${x^2} + {y^2} = 5$ at the point $\left( {1,2} \right)$ is

 The radius line joining the origin and the given point will be perpendicular to the tangent at that point.

The slope of the line joining the center and the point $\left( {1,2} \right)$ is, $m = \dfrac{{2 - 0}}{{1 - 0}} = 2$ $ \\ $ Therefore, slope of the tangent at the point $\left( {1,2} \right)$ will be, $m' = \dfrac{{ - 1}}{m} = \dfrac{{ - 1}}{2}$

 The angle between the lines $2x + 3y - 1 = 0$ and $5x + y = 0$ is

 The angle between the lines would be the difference of the angles made by the lines with the x-axis.

Slope of first line, ${m_1} =  - \dfrac{a}{b} =  - \dfrac{2}{3}$ $ \\ $ Angle made with the x-axis, ${\theta _1} = {\tan ^{ - 1}}{m_1} = {\tan ^{ - 1}}\left( { - \dfrac{2}{3}} \right)$ $ \\ $ Slope of second line, ${m_2} =  - \dfrac{a}{b} =  - 5$ $ \\ $ Angle made with the x-axis, ${\theta _2} = {\tan ^{ - 1}}{m_2} = {\tan ^{ - 1}}\left( { - 5} \right)$ $ \\ $ Angle between the lines, $\theta  = \left| {{\theta _2} - {\theta _1}} \right| = \left| {{{\tan }^{ - 1}}\left( { - 5} \right) - {{\tan }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right| \\ {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) \\ $ Thus, $ \\ \left| {{{\tan }^{ - 1}}\left( { - 5} \right) - {{\tan }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right| = \left| {{{\tan }^{ - 1}}\left( {\dfrac{{\left( { - 5} \right) - \left( { - \dfrac{2}{3}} \right)}}{{1 + \left( { - 5} \right)\left( { - \dfrac{2}{3}} \right)}}} \right)} \right| \\  = \left| {{{\tan }^{ - 1}}\left( {\dfrac{{ - 5 + \dfrac{2}{3}}}{{1 + \dfrac{{10}}{3}}}} \right)} \right| = \left| {{{\tan }^{ - 1}}\left( {\dfrac{{ - \dfrac{{13}}{3}}}{{\dfrac{{13}}{3}}}} \right)} \right| = \left| {{{\tan }^{ - 1}}\left( { - 1} \right)} \right| = \left| { - {{\tan }^{ - 1}}\left( 1 \right)} \right| = {\tan ^{ - 1}}\left( 1 \right) = {45^o}  $

 What is the acute angle made by the line $5x - y - 3 = 0$ with the y-axis?

 The slope of a line can be used to find the angle made by the line with the x-axis. Using this we can find the angle made by the y-axis.

The slope of the line is $m =  - \dfrac{a}{b} = 5$ $ \\ $ The angle made by the line with the positive x-axis is $\theta  = {\tan ^{ - 1}}m = {\tan ^{ - 1}}5$ $ \\ $ The slope is negative which means the angle is obtuse. $ \\ $ The acute angle made by the line with the positive y-axis will be $\theta ' = \dfrac{\pi }{2} - \theta  = \dfrac{\pi }{2} - {\tan ^{ - 1}}5 = {\cot ^{ - 1}}5$ $ \\ $ We used the formula ${\tan ^{ - 1}}\theta  + {\cot ^{ - 1}}\theta  = \dfrac{\pi }{2}$.

 The points $A\left( {2,4} \right)$, $B\left( {5,3} \right)$ and $C\left( {6,k} \right)$ form a right triangle right-angled at $B$. The value of $k$ is

 The slope of the line $AB$ will be perpendicular to the line $BC$. The product of their slopes will be equal to $ - 1$.

Since the triangle is right-angled at $B$, $Slop{e_{AB}} \times Slop{e_{BC}} =  - 1$ $ \\ \Rightarrow \dfrac{{3 - 4}}{{5 - 2}} \times \dfrac{{k - 3}}{{6 - 5}} =  - 1 \\   \Rightarrow \dfrac{{ - 1}}{3} \times \dfrac{{k - 3}}{1} =  - 1 \\   \Rightarrow k - 3 = 3 \Rightarrow k = 6 \\  $ 

 The mid-points of the sides $AB$, $BC$ and $AC$ of a triangle are $\left( {\dfrac{3}{2},7} \right)$, $\left( {2,5} \right)$ and $\left( {\dfrac{5}{2},8} \right)$ respectively. The equation of the side $AC$ is

 Use the mid-point theorem of triangles to find slope of $AC$.

Let the equation of $AC$ be $y = mx + c$. $ \\ $ From the Mid-point Theorem of triangles, line $AC$ is parallel to the line joining the mid-points of line $AB$ and $BC$. Therefore, the slope of $AC$ is equal to the slope of the line joining the mid-points of line $AB$ and $BC$. Thus, $ \\ Slop{e_{AC}} = m = \dfrac{{5 - 7}}{{2 - \dfrac{3}{2}}} = \dfrac{{ - 2}}{{\dfrac{{4 - 3}}{2}}} =  - 4 \\ $ Thus the equation becomes $y =  - 4x + c$. This passes through the point $\left( {\dfrac{5}{2},8} \right)$. $ \\  \Rightarrow 8 =  - 4 \times \dfrac{5}{2} + c \\   \Rightarrow 8 =  - 10 + c \\   \Rightarrow c = 18 \\  $  Thus the equation of $AC$ is $y =  - 4x + 18 \Rightarrow 4x + y = 18$