
If the equation \[y = mx + c\] and \[xcos\alpha + ysin\alpha = p\;\] represents the same straight line, then
A. \[\;\;\;p = c\sqrt {1 + {m^2}} \;\;\]
B. \[\;\;\;\;\;c = p\sqrt {1 + {m^2}} \]
C. \[\;\;cp = \sqrt {1 + {m^2}} \]
D. \[{p^2} + {c^2} + {m^2} = 1\]
Answer
160.8k+ views
Formula Used:Equation of straight line is given by:
\[y = mx + c\]
Where
m is slope straight line
c is y intercept of straight line
\[{\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1\]
Complete step by step solution:Given: equation \[y = mx + c\] and \[xcos\alpha + ysin\alpha = p\;\] represents the same straight line
\[y = mx + c\]……………………….(i)
\[xcos\alpha + ysin\alpha = p\;\]….………(ii)
Equation two can be written as
\[xcos\alpha + ysin\alpha = p\;\]
\[ysin\alpha = p\; - xcos\alpha \]
\[y = \;\dfrac{p}{{sin\alpha }} - \dfrac{{xcos\alpha }}{{sin\alpha }}\]
\[y = \; - \dfrac{{xcos\alpha }}{{sin\alpha }} + \dfrac{p}{{sin\alpha }}\]
Now compare this with equation one
Slope of straight line =\[m = - \dfrac{{cos\alpha }}{{sin\alpha }}\]
Y intercept of straight line =\[c = \dfrac{p}{{sin\alpha }}\]
\[sin\alpha = \dfrac{p}{c}\]
\[cos\alpha = - msin\alpha \]
\[cos\alpha = - m\dfrac{p}{c}\]
Now use
\[{\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1\]
We get
\[{( - m\dfrac{p}{c})^2} + {(\dfrac{p}{c})^2} = 1\]
\[\dfrac{{{m^2}{p^2}}}{{{c^2}}} + \dfrac{{{p^2}}}{{{c^2}}} = 1\]
\[\dfrac{{{m^2}{p^2} + {p^2}}}{{{c^2}}} = 1\]
\[{p^2}({m^2} + 1) = {c^2}\]
\[\;\;\;\;\;c = p\sqrt {1 + {m^2}} \]
Option ‘B’ is correct
Note: In this question compare it is given that both equation represent the same straight line that is why we compare two given equation after rearranging it.
\[{\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1\] this relation is a standard relation of trigonometry
Trigonometry is a branch of mathematics in which we study the relation between angles and side of triangle.
Right angle triangle is used in study of trigonometry.
Right angle triangle is a triangle in which one angle is \[90\]degree and sum of other two angle is also \[90\] degree.
Every line makes some angle with axes. Slope of a line is also equal to the tan of angle which lines make with axes. Slope is also known as gradient.
Product of slope of two perpendicular line is and slope of two equal lines are equal.
\[y = mx + c\]
Where
m is slope straight line
c is y intercept of straight line
\[{\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1\]
Complete step by step solution:Given: equation \[y = mx + c\] and \[xcos\alpha + ysin\alpha = p\;\] represents the same straight line
\[y = mx + c\]……………………….(i)
\[xcos\alpha + ysin\alpha = p\;\]….………(ii)
Equation two can be written as
\[xcos\alpha + ysin\alpha = p\;\]
\[ysin\alpha = p\; - xcos\alpha \]
\[y = \;\dfrac{p}{{sin\alpha }} - \dfrac{{xcos\alpha }}{{sin\alpha }}\]
\[y = \; - \dfrac{{xcos\alpha }}{{sin\alpha }} + \dfrac{p}{{sin\alpha }}\]
Now compare this with equation one
Slope of straight line =\[m = - \dfrac{{cos\alpha }}{{sin\alpha }}\]
Y intercept of straight line =\[c = \dfrac{p}{{sin\alpha }}\]
\[sin\alpha = \dfrac{p}{c}\]
\[cos\alpha = - msin\alpha \]
\[cos\alpha = - m\dfrac{p}{c}\]
Now use
\[{\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1\]
We get
\[{( - m\dfrac{p}{c})^2} + {(\dfrac{p}{c})^2} = 1\]
\[\dfrac{{{m^2}{p^2}}}{{{c^2}}} + \dfrac{{{p^2}}}{{{c^2}}} = 1\]
\[\dfrac{{{m^2}{p^2} + {p^2}}}{{{c^2}}} = 1\]
\[{p^2}({m^2} + 1) = {c^2}\]
\[\;\;\;\;\;c = p\sqrt {1 + {m^2}} \]
Option ‘B’ is correct
Note: In this question compare it is given that both equation represent the same straight line that is why we compare two given equation after rearranging it.
\[{\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1\] this relation is a standard relation of trigonometry
Trigonometry is a branch of mathematics in which we study the relation between angles and side of triangle.
Right angle triangle is used in study of trigonometry.
Right angle triangle is a triangle in which one angle is \[90\]degree and sum of other two angle is also \[90\] degree.
Every line makes some angle with axes. Slope of a line is also equal to the tan of angle which lines make with axes. Slope is also known as gradient.
Product of slope of two perpendicular line is and slope of two equal lines are equal.
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