
What is the number of points on the line 3x + 4y = 5, which are at a distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta ,\theta \in R\] from the point (1, 3)?
(a). 1
(b). 2
(c). 3
(d). Infinite
Answer
600.9k+ views
Hint: Find the shortest distance of the point (1, 3) from the line 3x + 4y = 5 using the formula \[\dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]. Then find if the distance \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta ,\theta \in R\] is smaller than, equal to, or greater than this distance and conclude the answer.
Complete step-by-step answer:
We need to find the number of points on the line 3x + 4y = 5, which are at a distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta ,\theta \in R\] from the point (1, 3).
For this, we first find the shortest distance between the point (1, 3) and 3x + 4y = 5.
The shortest distance of a point \[({x_0},{y_0})\] from a line \[ax + by + c = 0\] is given as follows:
\[d = \dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]
The standard form of the line 3x +4y = 5 is given as follows:
\[3x + 4y - 5 = 0\]
Now, the distance of the point (1, 3) from this line is given as follows:
\[d = \dfrac{{|3(1) + 4(3) - 5|}}{{\sqrt {{3^2} + {4^2}} }}\]
Simplifying, we have:
\[d = \dfrac{{|3 + 12 - 5|}}{{\sqrt {9 + 16} }}\]
\[d = \dfrac{{|10|}}{{\sqrt {25} }}\]
\[d = \dfrac{{10}}{5}\]
\[d = 2..............(1)\]
Now, we find the value of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \].
We know that \[{\sec ^2}\theta = 1 + {\tan ^2}\theta \] and also, we know that \[{\text{cose}}{{\text{c}}^2}\theta = 1 + {\cot ^2}\theta \], hence, we have:
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta + 2(1 + {\cot ^2}\theta )\]
Simplifying, we have:
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta + 2 + 2{\cot ^2}\theta \]
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta = 3 + {\tan ^2}\theta + 2{\cot ^2}\theta \]
The squares of tangent and cotangent functions are positive and hence, we have:
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \geqslant 3\]
Hence, we observe that the distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \] is greater than the shortest distance between the point (1, 3) and the line 3x + 4y = 5, hence, there are two points on either side of the shortest distance, that equals the distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \].
Hence, the correct answer is option (b).
Note: You can also express the point on the line 3x + 4y = 5 as \[\left( {x,\dfrac{{5 - 3x}}{4}} \right)\] and equate the distance between two points as \[\sqrt {{{({x_1},{y_1})}^2} + {{({y_2} - {y_1})}^2}} \] to find the number of values of x.
Complete step-by-step answer:
We need to find the number of points on the line 3x + 4y = 5, which are at a distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta ,\theta \in R\] from the point (1, 3).
For this, we first find the shortest distance between the point (1, 3) and 3x + 4y = 5.
The shortest distance of a point \[({x_0},{y_0})\] from a line \[ax + by + c = 0\] is given as follows:
\[d = \dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]
The standard form of the line 3x +4y = 5 is given as follows:
\[3x + 4y - 5 = 0\]
Now, the distance of the point (1, 3) from this line is given as follows:
\[d = \dfrac{{|3(1) + 4(3) - 5|}}{{\sqrt {{3^2} + {4^2}} }}\]
Simplifying, we have:
\[d = \dfrac{{|3 + 12 - 5|}}{{\sqrt {9 + 16} }}\]
\[d = \dfrac{{|10|}}{{\sqrt {25} }}\]
\[d = \dfrac{{10}}{5}\]
\[d = 2..............(1)\]
Now, we find the value of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \].
We know that \[{\sec ^2}\theta = 1 + {\tan ^2}\theta \] and also, we know that \[{\text{cose}}{{\text{c}}^2}\theta = 1 + {\cot ^2}\theta \], hence, we have:
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta + 2(1 + {\cot ^2}\theta )\]
Simplifying, we have:
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta + 2 + 2{\cot ^2}\theta \]
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta = 3 + {\tan ^2}\theta + 2{\cot ^2}\theta \]
The squares of tangent and cotangent functions are positive and hence, we have:
\[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \geqslant 3\]
Hence, we observe that the distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \] is greater than the shortest distance between the point (1, 3) and the line 3x + 4y = 5, hence, there are two points on either side of the shortest distance, that equals the distance of \[{\sec ^2}\theta + 2{\text{cose}}{{\text{c}}^2}\theta \].
Hence, the correct answer is option (b).
Note: You can also express the point on the line 3x + 4y = 5 as \[\left( {x,\dfrac{{5 - 3x}}{4}} \right)\] and equate the distance between two points as \[\sqrt {{{({x_1},{y_1})}^2} + {{({y_2} - {y_1})}^2}} \] to find the number of values of x.
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