Answer
Verified
39k+ views
Hint: Two lines are said to be in perpendicular if \[{m_1}{m_2} = - 1\] where \[{m_1},{m_2}\] are the slopes of the given two lines. This statement is also known as the condition of perpendicularity of two straight lines.
Complete step-by-step answer:
Given line is \[y = 3x - 1\] but , we know that the general equation of the line is \[y = mx + c\] where \[m\] is the slope of the equation.
So, slope of the line \[y = 3x - 1\] is \[{m_1} = 3\] and
Let \[{m_2}\] be the slope of the required line
By the condition of perpendicularity,\[{m_1}{m_2} = - 1\]
i.e. \[3 \times {m_2} = - 1\]
\[\therefore {m_2} = \dfrac{{ - 1}}{3}\]
The line perpendicular to the line \[y = {m_1}x + c\] is given by\[y = {m_2}x + c\].
i.e. \[y = \dfrac{{ - 1}}{3}x + c\]
But this line is passing through \[\left( {1,2} \right)\]. So, it must satisfy the equation if we put the point \[\left( {1,2} \right)\]in the line.
\[
\Rightarrow 2 = \dfrac{{ - 1}}{3}\left( 1 \right) + c \\
\Rightarrow 6 = - 1 + 3c \\
\Rightarrow 3c = 7 \\
\therefore c = \dfrac{7}{3} \\
\]
So, the required line is
\[
y = \dfrac{{ - 1}}{3}x + \dfrac{7}{3} \\
y = \dfrac{{ - x + 7}}{3} \\
3y = - x + 7 \\
\therefore x + 3y - 7 = 0 \\
\]
Thus, the line passing through the point \[\left( {1,2} \right)\]and perpendicular to the line \[y = 3x - 1\] is \[x + 3y - 7 = 0\].
Therefore, the answer is option B. \[x + 3y - 7 = 0\].
Note: The above problem can be done by a using a direct formula i.e. equation of line passing through the point \[\left( {{x_1},{y_1}} \right)\] and perpendicular to the line \[y = mx + c\] is given by the equation\[x - {x_1} + m\left( {y - {y_1}} \right) = 0\].
Complete step-by-step answer:
Given line is \[y = 3x - 1\] but , we know that the general equation of the line is \[y = mx + c\] where \[m\] is the slope of the equation.
So, slope of the line \[y = 3x - 1\] is \[{m_1} = 3\] and
Let \[{m_2}\] be the slope of the required line
By the condition of perpendicularity,\[{m_1}{m_2} = - 1\]
i.e. \[3 \times {m_2} = - 1\]
\[\therefore {m_2} = \dfrac{{ - 1}}{3}\]
The line perpendicular to the line \[y = {m_1}x + c\] is given by\[y = {m_2}x + c\].
i.e. \[y = \dfrac{{ - 1}}{3}x + c\]
But this line is passing through \[\left( {1,2} \right)\]. So, it must satisfy the equation if we put the point \[\left( {1,2} \right)\]in the line.
\[
\Rightarrow 2 = \dfrac{{ - 1}}{3}\left( 1 \right) + c \\
\Rightarrow 6 = - 1 + 3c \\
\Rightarrow 3c = 7 \\
\therefore c = \dfrac{7}{3} \\
\]
So, the required line is
\[
y = \dfrac{{ - 1}}{3}x + \dfrac{7}{3} \\
y = \dfrac{{ - x + 7}}{3} \\
3y = - x + 7 \\
\therefore x + 3y - 7 = 0 \\
\]
Thus, the line passing through the point \[\left( {1,2} \right)\]and perpendicular to the line \[y = 3x - 1\] is \[x + 3y - 7 = 0\].
Therefore, the answer is option B. \[x + 3y - 7 = 0\].
Note: The above problem can be done by a using a direct formula i.e. equation of line passing through the point \[\left( {{x_1},{y_1}} \right)\] and perpendicular to the line \[y = mx + c\] is given by the equation\[x - {x_1} + m\left( {y - {y_1}} \right) = 0\].
Recently Updated Pages
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
The number of ways in which 5 boys and 3 girls can-class-12-maths-JEE_Main
Find dfracddxleft left sin x rightlog x right A left class 12 maths JEE_Main
Distance of the point x1y1z1from the line fracx x2l class 12 maths JEE_Main
In a box containing 100 eggs 10 eggs are rotten What class 12 maths JEE_Main
dfracddxex + 3log x A ex cdot x2x + 3 B ex cdot xx class 12 maths JEE_Main
Other Pages
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
What is the volume of water that must be added to a class 11 chemistry JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
When an electron and a proton are placed in an electric class 12 physics JEE_Main