Question

If the slope of a line passing through the point \[A\left( {3,{\rm{ }}2} \right)\] be \[\dfrac{3}{4}\] then the points on the line which are \[5\] units away from A are
A. \[\left( {5,{\rm{ }}5} \right),\left( { - {\rm{ }}1, - 1} \right)\]
B. \[\left( {7,{\rm{ }}5} \right),{\rm{ }}\left( { - 1, - 1} \right)\]
C. \[\left( {5,{\rm{ }}7} \right),\left( { - {\rm{ }}1, - 1} \right)\]
D. \[\left( {7,{\rm{ }}5} \right),\left( {1,{\rm{ }}1} \right)\]

Answer 1

Hint: In this question, we need to find the points on the line(s) that goes through the point \[A\left( {3,{\rm{ }}2} \right)\]and has a slope \[\dfrac{3}{4}\] and the points on the line are \[5\] units away from A. To accomplish this, we will follow the relationship between the x and y axes of the line equation while also satisfying the provided requirements.

Formula Used: Given a point on a line and the slope of a line, we have to use the point-slope form:
\[y - y1 = m\left( {x - x1} \right)\]

Complete step by step solution: We have been given that the, equation of line with a slope of \[\dfrac{3}{4}\]
And passing through a point \[{\rm{A}}(3,2)\]
And by slope equation, we can write the equation as,
\[y - 2 = \dfrac{3}{4}(x - 3)\]
On multiplying the denominator \[{\rm{4}}\] with terms on the left hand side of the equation, we get
\[4{\rm{ }}y - 8 = 3{\rm{ }}x - 9\]
And now on multiplying the numerator \[{\rm{3}}\] with terms on the right hand side of the equation, we get
\[3{\rm{ }}x = 1 + 4{\rm{ }}y + 11\]
Now, let us assume the point is
\[{\rm{B}}\left( {{{\rm{x}}_1},{{\rm{y}}_1}} \right)\]
We have been given that the distance between \[{\rm{A}}\] and \[{\rm{B}}\] are \[5\]
Now, by distance formula we have
\[{\left( {{{\rm{x}}_1} - 3} \right)^2} + {\left( {{{\rm{y}}_1} - 2} \right)^2} = 25\]
Square the above terms using square formula, we get
\[{\rm{x}}_1^2 + 9 - 6{{\rm{x}}_1} + {\rm{y}}_1^2 + 4 - 4{{\rm{y}}_1} = 25\]
\[{\rm{x}}_1^2 + {\rm{y}}_1^2 - 6{{\rm{x}}_1} - 4{{\rm{y}}_1} - 12 = 0\]--- (1)
We know that point \[{\rm{B}}\] lies on the line \[3{\rm{x}} = 1 + 4{\rm{y}}\]
Consequently,
\[{{\rm{x}}_1} = \dfrac{{4{{\rm{y}}_1} + 1}}{3}\]
Now, on substituting the value of \[{{\rm{x}}_1} = \dfrac{{4{{\rm{y}}_1} + 1}}{3}\] in the equation (1), we get
\[{\left( {\dfrac{{4{{\rm{y}}_1} + 1}}{3}} \right)^2} + {\rm{y}}_1^2 - 6\left( {\dfrac{{4{{\rm{y}}_1} + 1}}{3}} \right) - 4{{\rm{y}}_1} - 12 = 0\]
On solving the above, we obtain
\[\dfrac{{16{\rm{y}}_1^2 + 1 + 8{{\rm{y}}_1}}}{9} + {\rm{y}}_1^2 - 8{{\rm{y}}_1} - 2 - 4{{\rm{y}}_1} - 12 = 0\]
Now, we have to multiply the denominator with the other terms of the equation, we get\[16{\rm{y}}_1^2 + 1 + 8{{\rm{y}}_1} + 9{\rm{y}}_1^2 - 72{{\rm{y}}_1} - 18 - 36{{\rm{y}}_1} - 108 = 0\]
On simplifying the above equation, we get
\[25{\rm{y}}_1^2 - 100{{\rm{y}}_1} - 125 = 0\]
On further simplification, it becomes
\[{\rm{y}}_1^2 - 4{{\rm{y}}_1} - 5 = 0\]
Now, we have to expand the equation in order to factorize,
\[{\rm{y}}_1^2 - 5{{\rm{y}}_1} + {{\rm{y}}_1} - 5 = 0\]
Now, we have to factor the above equation
\[\left( {{y_1} - 5} \right)\left( {{y_1} + 1} \right) = 0\]
Now, we have to solve for \[{y_1}\] we get
\[ \Rightarrow {y_1} = - 1,5\]
Now, from the equation (1)
\[3{{\rm{x}}_1} = 1 + 4{{\rm{y}}_1}\]
Now, it becomes,
\[3{{\rm{x}}_1} = 1 - 4{\rm{ and }}3{{\rm{x}}_1} = 1 + 20\]
We have now obtained the value of \[{{\rm{x}}_1}\]
\[{{\rm{x}}_1} = - 1{\rm{ and }}{{\rm{x}}_1} = 7\]
Let us substitute the values we previously obtained, we get
Point \[{\rm{B}}(7,5)\] and \[( - 1, - 1)\]
The points on the line which are \[5\] units away from ‘\[A\]’ are \[\left( {7,{\rm{ }}5} \right),{\rm{ }}\left( { - 1, - 1} \right)\]

Option ‘B’ is correct

Note: Students must use extreme caution while inserting the values of the x and y-intercepts in the equation of the line while simultaneously satisfying the required requirements. Furthermore, when a point is on a line, it satisfies the line's equation.