Question

A line \[4x + y = 1\] passes through the point \[A\left( {2, - 7} \right)\] meets the lines \[BC\] whose equation is \[3x - 4y + 1 = 0\] at the point \[B\]. The equation of the line \[AC\] so that \[AB = AC\], is
A. \[52x + 89y + 519 = 0\]
B. \[52x + 89y - 519 = 0\]
C. \[89x + 52y + 519 = 0\]
D. \[89x + 52y - 519 = 0\]

Answer 1

Hint: In this question, we first find the slope of the line \[4x + y = 1\] and the second line \[3x - 4y + 1 = 0\] , and then we find the angle between these two lines by the formula \[\theta = {\tan ^{ - 1}}\left| {\dfrac{{{m_1} - m}}{{1 + {m_1}m}}} \right|\] and finally substitute the given points in it to get the desired result

Formula Used:
1. \[\theta = {\tan ^{ - 1}}\left| {\dfrac{{{m_1} - m}}{{1 + {m_1}m}}} \right|\]
Where \[m\] is the slope of the line
2. \[y - {y_1} = m\left( {x - {x_1}} \right)\]

Complete step-by-step solution:
Given that
 \[
  4x + y = 1...\left( 1 \right) \\
  3x - 4y + 1 = 0...\left( 2 \right)
 \]
Rewrite equation \[\left( 1 \right)\] as:
\[
  y = 1 - 4x \\
  y = - 4x + 1 \\
 \]
Now compare the above equation with the standard equation of the line \[y = mx + b\]:
\[m = - 4\]
Rewrite equation \[\left( 2 \right)\] as:
\[
   - 4y = - 3x - 1 \\
   - 4y = - \left( {3x + 1} \right) \\
  4y = 3x + 1 \\
  y = \dfrac{3}{4}x + \dfrac{1}{4} \\
 \]
Now compare the above equation with the standard equation of the line \[y = mx + b\]:
\[m = \dfrac{3}{4}\]

Image: Triangle ABC with A vertex (2,-7)
Now we have two slopes which is \[m = \dfrac{3}{4}\] and \[m = - 4\] of \[BC\] and \[AB\]
We know the formula \[\theta = {\tan ^{ - 1}}\left| {\dfrac{{{m_1} - m}}{{1 + {m_1}m}}} \right|\]
For line \[AB\] and \[BC\]:
\[\alpha = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \left( {\dfrac{3}{4} \times \left( { - 4} \right)} \right)}}} \right|\]
By taking \[\tan \] on both sides, we get
\[
  \tan \,\alpha = \tan \left( {{{\tan }^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \left( {\dfrac{3}{4} \times \left( { - 4} \right)} \right)}}} \right|} \right) \\
  \tan \,\alpha = \left| {\dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \left( {\dfrac{3}{4} \times \left( { - 4} \right)} \right)}}} \right|...\left( 3 \right)
 \]
Let us assume that \[{m_1} = \dfrac{3}{4}\]be the slope of \[BC\] and \[m\] be the slope of \[AC\]
\[
  \theta = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \left( {\dfrac{3}{4} \times m} \right)}}} \right| \\
   = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \dfrac{3}{4}m}}} \right|
 \]
By taking \[\tan \] on both sides, we get
\[
  \tan \,\theta = \tan \left( {{{\tan }^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \left( {\dfrac{3}{4}m} \right)}}} \right|} \right) \\
  \tan \,\theta = \left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \dfrac{3}{4}m}}} \right|...\left( 4 \right)
 \]
Now we are given that \[AB = AC\]
So, \[\tan \,\alpha = \tan \,\theta \]
Substitute the value from equation \[\left( 3 \right)\] and \[\left( 4 \right)\]:
\[
  \left| {\dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \dfrac{3}{4} \times \left( { - 4} \right)}}} \right| = \left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \dfrac{3}{4}m}}} \right| \\
  \left| {\dfrac{{\dfrac{3}{4} + 4}}{{1 + 3\left( { - 1} \right)}}} \right| = \left| {\dfrac{{\dfrac{{3 - 4m}}{4}}}{{\dfrac{{4 + 3m}}{4}}}} \right| \\
  \left| {\dfrac{{\dfrac{{3 + 16}}{4}}}{{1 - 3}}} \right| = \left| {\dfrac{{3 - 4m}}{{4 + 3m}}} \right| \\
  \left| {\dfrac{{19}}{{4 \times \left( { - 2} \right)}}} \right| = \left| {\dfrac{{3 - 4m}}{{4 + 3m}}} \right|
 \]
Further solving,
\[
  \left| {\dfrac{{19}}{{ - 8}}} \right| = \left| {\dfrac{{3 - 4m}}{{4 + 3m}}} \right| \\
  \dfrac{{19}}{8} = \dfrac{{3 - 4m}}{{4 + 3m}} \\
  19\left( {4 + 3m} \right) = 8\left( {3 - 4m} \right) \\
  76 + 57m = 24 - 32m
 \]
Furthermore,
\[
  57m + 32m = 24 - 76 \\
  89m = - 52 \\
  m = \dfrac{{ - 52}}{{89}}
 \]
Also given that the line passes through the point \[A\left( {2, - 7} \right)\]
We know the formula of equation of line is \[y - {y_1} = m\left( {x - {x_1}} \right)\]
Substituting all the values in above formula:
\[y + 7 = \dfrac{{ - 52}}{{89}}\left( {x - 2} \right)\]
Multiply 89 on both sides:
\[
  89\left( {y + 7} \right) = - 52\left( {x - 2} \right) \\
  89y + 623 = - 52x + 104 \\
  89y + 52x + 623 - 104 = 0 \\
  89y + 52x + 519 = 0
 \]

Hence, option (A) is correct

Note: Students must be careful while finding the slope of the given line. Also, students must be careful while finding the angle between these two lines and also substitute the values of slope in the formula correctly to get the desired result.