Question

Given the family of lines, $(3x + 4y + 6) + \lambda (x + y + 2) = 0$. The line of the family is situated at the greatest distance from the point $P\left( {2,3} \right)$ has equation:
A) $4x + 3y + 8$
B) $5x + 3y + 10 = 0$
C) $15x + 8y + 30 = 0$
D) None of these

Answer 1

Hint: First we have to split the given equation.
Then we find the point and the equation of the line which is perpendicular to this line because it will be located at the greatest distance.
Finally we get the answer.

Formula used: Equation of the line, $\dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{x - {x_1}}}{{{x_2} - {x_1}}}$
Equation of the line which slope is given,$y = mx + b$

Complete step-by-step answer:
It is given in the question that $(3x + 4y + 6) + \lambda (x + y + 2) = 0$
We split the given equation as ${L_1}:3x + 4y + 6 = 0...\left( 1 \right)$
${L_2}:x + y + 2 = 0....\left( 2 \right)$
From equation \[\left( 2 \right)\] we get-
$x + y = - 2$
Therefore $x = y - 2$
Substituting the value of $x$ in equation \[\left( 1 \right)\]we get
${L_1}:3x + 4y + 6$$ = 0$
$3(y - 2) + 4y + 6 = 0$
On multiplying the term, we get
$3y - 6 + 4y + 6 = 0$
Let us cancel the same term but different sign and adding the $y$ term we get,
${\text{6y = 0}}$
Hence, $y = 0$
Putting $y = 0$ in equation \[\left( 2 \right)\] we get
${L_2}:x + y + 2$$ = 0$
$x + 0 + 2 = 0$
So, $x = - 2$
Hence it is clear that the point of intersection of the two lines is ($ - 2,0$)
Here the point $P$ is given whose coordinate is $(2,3)$
Now, equation of the line from $\left( { - 2,0} \right)$and $\left( {2,3} \right)$is
Here $\left( {{x_1},{y_1}} \right) = \left( { - 2,0} \right)$
$\left( {{x_2},{y_2}} \right) = \left( {2,3} \right)$
Equation of the line, $\dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{x - {x_1}}}{{{x_2} - {x_1}}}$
Substitute the above point in the formula,
$\dfrac{{y - 0}}{{3 - 0}} = \dfrac{{x + 2}}{{2 + 2}}$
On simplifying,
$\dfrac{y}{3} = \dfrac{{x + 2}}{4}$
We can write $y$ terms in RHS and remaining as in LHS
$y = \dfrac{3}{4}(x + 2)$
By doing cross multiplication we get
$4y = 3x + 6$
So $y = \dfrac{{3x}}{4} + \dfrac{6}{4}$
As we know when the equation of a line is given in $y = mx + b$ form the slope of the line is $m$
So here the slope of the line is $\dfrac{3}{4}$
Here we have to find the line which is situated at the greatest distance and that line will be the line perpendicular to the line$4y = 3x + 6$.
Therefore slope of the required line =$ - \dfrac{4}{3}$ since it is perpendicular to the line $4y = 3x + 6$
Hence the required equation of the line $y = - \dfrac{4}{3}$$(x + 2)$
By doing cross multiplication we get-
$3y + 4x + 8 = 0$
The required equation of the line is $3y + 4x + 8 = 0$
We can rewrite it as, $4x + 3y + 8 = 0$

So, the correct option is A.

Note: It is to be noted that there are two formulas of finding the slope of the straight line.
The tangent of the inclination is the slope of a line and it is denoted by \[m\] and if the inclination of a line is $\theta $
Any kind of straight line of horizontal line segment has no x intercept.
Each point on a number line is assigned a number. In the same way, each point in a plane is assigned a pair of numbers.
The coordinates for the origin are $\left( {0,0} \right)$.