Q106 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
The distance of the point $(3,4)$ from the line $3 x+2 y+7=0$ measured along the line parallel to $y-2 x+7=0$ is equal to
- A$\dfrac{23 \sqrt{5}}{7}$
- B$4 \sqrt{5}$
- C$3 \sqrt{5}$
- Correct answerD$\dfrac{24 \sqrt{5}}{7}$
Solution
The given point is $P(3, 4)$ and the line is $L: 3x + 2y + 7 = 0$. The line along which the distance is measured is parallel to $y - 2x + 7 = 0$, which has a slope $m = 2$.
The equation of the line passing through $(3, 4)$ with slope $m = 2$ is given by $y - 4 = 2(x - 3)$, which simplifies to $y = 2x - 2$ or $2x - y - 2 = 0$.
To find the intersection point $Q$ of the line $3x + 2y + 7 = 0$ and $2x - y - 2 = 0$, we solve the system:
$3x + 2y = -7$
$4x - 2y = 4$
Adding the two equations gives $7x = -3$, so $x = -3/7$. Substituting $x$ into $2x - y - 2 = 0$ gives $y = 2(-3/7) - 2 = -6/7 - 14/7 = -20/7$. Thus, $Q = (-3/7, -20/7)$.
The distance $PQ$ is $\sqrt{(3 - (-3/7))^2 + (4 - (-20/7))^2} = \sqrt{(24/7)^2 + (48/7)^2}$.
$PQ = \sqrt{\dfrac{576}{49} + \dfrac{2304}{49}} = \sqrt{\dfrac{2880}{49}} = \dfrac{\sqrt{576 \times 5}}{7} = \dfrac{24 \sqrt{5}}{7}$.
Answer: $\dfrac{24 \sqrt{5}}{7}$
The equation of the line passing through $(3, 4)$ with slope $m = 2$ is given by $y - 4 = 2(x - 3)$, which simplifies to $y = 2x - 2$ or $2x - y - 2 = 0$.
To find the intersection point $Q$ of the line $3x + 2y + 7 = 0$ and $2x - y - 2 = 0$, we solve the system:
$3x + 2y = -7$
$4x - 2y = 4$
Adding the two equations gives $7x = -3$, so $x = -3/7$. Substituting $x$ into $2x - y - 2 = 0$ gives $y = 2(-3/7) - 2 = -6/7 - 14/7 = -20/7$. Thus, $Q = (-3/7, -20/7)$.
The distance $PQ$ is $\sqrt{(3 - (-3/7))^2 + (4 - (-20/7))^2} = \sqrt{(24/7)^2 + (48/7)^2}$.
$PQ = \sqrt{\dfrac{576}{49} + \dfrac{2304}{49}} = \sqrt{\dfrac{2880}{49}} = \dfrac{\sqrt{576 \times 5}}{7} = \dfrac{24 \sqrt{5}}{7}$.
Answer: $\dfrac{24 \sqrt{5}}{7}$
Hence, the correct answer is option: 'D'.