Question

The equation of the line joining the point \[\left( {3,5} \right)\] to the point of intersection of the lines \[4x + y - 1 = 0\] and \[7x - 3y - 35 = 0\] is equidistant from the points \[\left( {0,0} \right)\] and \[\left( {8,34} \right)\]
1. True
2. False
3. Nothing can be said
4. None of these

Answer 1

Hint:
In this question, we have to check if the given statement is true or not. by using the concept of equation of line. Here, firstly, we will find the point of intersection and the slope of line.to find the equation of line Then, find distance from the points \[\left( {0,0} \right)\] and \[\left( {8,34} \right)\] to the equation of line

Formula used:
The equation of line formula are given as
1. slope \[ = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
2. \[y - {y_1} = m\left( {x - {x_1}} \right)\]
3. \[d = \dfrac{{\left| {a{x_1} + b{y_1} + {c_1}} \right|}}{{\sqrt {{a^2} + {b^2}} }}\]

Complete step-by-step solution:
The given lines are \[4x + y - 1 = 0\] (1)
And \[7x - 3y - 35 = 0\] (2)
Firstly, we will solve equation (1) and (2) by substituting the value of \[y = 1 - 4x\] from equation (1) into equation (2), we get
\[\begin{array}{l}7x - 3\left( {1 - 4x} \right) - 35 = 0\\7x - 3 + 12x - 35 = 0\\19x - 38 = 0\\x = 2\end{array}\]
Now, we will substitute \[x = 2\] in equation (1) to find \[y\], we get
\[\begin{array}{l}4\left( 2 \right) + y - 1 = 0\\8 + y - 1 = 0\\y = - 7\end{array}\]
So, the point of intersection \[\left( {x,y} \right)\] is \[\left( {2, - 7} \right)\].
Further, we will find the slope of the line joining the points \[\left( {3,5} \right)\] and \[\left( {2, - 7} \right)\].
To find the slope we will use the formula slope\[ = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] where \[\left( {{x_1},{y_1}} \right)\]is \[\left( {3,5} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is \[\left( {2, - 7} \right)\], we get
\[\begin{array}{l}slope = \dfrac{{ - 7 - 5}}{{2 - 3}}\\slope = \dfrac{{ - 12}}{{ - 1}}\\slope = 12\end{array}\]
Furthermore, we will find the equation of line joining the points \[\left( {3,5} \right)\] and \[\left( {2, - 7} \right)\] by using the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\] where \[\left( {{x_1},{y_1}} \right)\]is \[\left( {3,5} \right)\] and \[m\] is slope that is \[12\],we get
\[\begin{array}{l}y - 5 = 12\left( {x - 3} \right)\\y - 5 = 12x - 36\\12x - y + 5 - 36 = 0\\12x - y - 31 = 0{\rm{ }}......(3)\end{array}\]
Now, we will find the distance between the equation of line (3) from the point \[\left( {0,0} \right)\] by using the formula \[d = \dfrac{{\left| {a{x_1} + b{y_1} + {c_1}} \right|}}{{\sqrt {{a^2} + {b^2}} }}\] where \[\left( {a,b,c} \right)\] is \[\left( {12, - 1, - 31} \right)\] and \[\left( {x,y} \right) = \left( {0,0} \right)\], we get
\[\begin{array}{l}d = \dfrac{{\left| {12 \times 0 - 1 \times 0 - 31} \right|}}{{\sqrt {{{\left( {12} \right)}^2} + {{\left( { - 1} \right)}^2}} }}\\d = \dfrac{{\left| {0 + 0 - 31} \right|}}{{\sqrt {144 + 1} }}\\d = \dfrac{{31}}{{\sqrt {145} }}{\rm{ }}......(4)\end{array}\]
Further, we will find the distance between the equation of line (3) from the point \[\left( {8,34} \right)\] by using the formula \[d = \dfrac{{\left| {a{x_1} + b{y_1} + {c_1}} \right|}}{{\sqrt {{a^2} + {b^2}} }}\] where \[\left( {a,b,c} \right)\] is \[\left( {12, - 1, - 31} \right)\] and \[\left( {x,y} \right) = \left( {8,34} \right)\], we get
\[\begin{array}{l}d = \dfrac{{\left| {12 \times 8 - 1 \times 34 - 31} \right|}}{{\sqrt {{{\left( {12} \right)}^2} + {{\left( { - 1} \right)}^2}} }}\\d = \dfrac{{\left| {96 - 34 - 31} \right|}}{{\sqrt {144 + 1} }}\\d = \dfrac{{31}}{{\sqrt {145} }}{\rm{ }}......(5)\end{array}\]
Thus, the distance in equation (4) and equation (5) are same so, the given statement is true
Hence, the correct option is (1)

Note
In these types of questions, we should remember the formulas of slope of line and equation of line. Also as in such questions we have to remember how to solve the system of linear equations in two variables.