Given the system of straight lines a $(2x + y - 3) + b(3x + 2y - 5) = 0$, the line of the system situated farthest from the point $(4, -3)$ has the equation
- 4x + 11y - 15 = 0
- 7x + y - 8 = 0
- 4x + 3y - 7 = 0
- 3x - 4y + 1 = 0
The Correct Option is D
Solution and Explanation
The required line will also pass through this point.
Further, the line will be farthest from point $(4,-3)$
if it is in direction perpendicular to line joining $(1,1)$ and $(4,-3)$.
$\therefore$ The equation of the required line is
$y-1=\frac{-1}{\frac{-3-4}{4-1}}(x-1)$
$\Rightarrow 3 x-4 y+1=0$
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)