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JEE Main 2024
Straight Lines
Straight Lines and Pair of Straight Lines
Hard

Question

Let A(a,b),B(3,4)A(a, b), B(3,4) and C(6,8)C(-6,-8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a+3,7b+5)P(2 a+3,7 b+5) from the line 2x+3y4=02 x+3 y-4=0 measured parallel to the line x2y1=0x-2 y-1=0 is

Options

Solution

Key Concepts and Formulas

  • Euler Line: The orthocentre (HH), centroid (GG), and circumcentre (OO) of a triangle are collinear, and this line is called the Euler line.
  • Centroid Division: The centroid (GG) divides the line segment joining the orthocentre (HH) and the circumcentre (OO) in the ratio 2:12:1, i.e., HG:GO=2:1HG:GO = 2:1.
  • Distance Formula (Parallel to a Line): The distance between a point (x1,y1)(x_1, y_1) and a line ax+by+c=0ax + by + c = 0 measured parallel to the line y=mx+dy = mx + d is given by ax1+by1+ca+bm1+m2\left| \frac{ax_1 + by_1 + c}{a+bm} \right| \sqrt{1+m^2}.

Step-by-Step Solution

Step 1: Apply the Centroid Division Formula

Since the centroid G(a,b)G(a, b) divides the line segment joining the orthocentre H(6,8)H(-6, -8) and the circumcentre O(3,4)O(3, 4) in the ratio 2:12:1, we can use the section formula: a=2(3)+1(6)2+1=663=0a = \frac{2(3) + 1(-6)}{2+1} = \frac{6-6}{3} = 0 b=2(4)+1(8)2+1=883=0b = \frac{2(4) + 1(-8)}{2+1} = \frac{8-8}{3} = 0 Therefore, A=G(0,0)A = G(0, 0).

Step 2: Determine the Coordinates of Point P

Given that P(2a+3,7b+5)P(2a+3, 7b+5) and a=0a = 0 and b=0b = 0, we have: P(2(0)+3,7(0)+5)=P(3,5)P(2(0)+3, 7(0)+5) = P(3, 5)

Step 3: Calculate the Slope of the Parallel Line

The distance is measured parallel to the line x2y1=0x - 2y - 1 = 0. We rewrite this line in slope-intercept form to find its slope: 2y=x12y = x - 1 y=12x12y = \frac{1}{2}x - \frac{1}{2} The slope of this line is m=12m = \frac{1}{2}.

Step 4: Apply the Distance Formula (Parallel to a Line)

We need to find the distance of the point P(3,5)P(3, 5) from the line 2x+3y4=02x + 3y - 4 = 0 measured parallel to the line x2y1=0x - 2y - 1 = 0. Using the distance formula with x1=3x_1 = 3, y1=5y_1 = 5, a=2a = 2, b=3b = 3, c=4c = -4, and m=12m = \frac{1}{2}: Distance=2(3)+3(5)42+3(12)1+(12)2\text{Distance} = \left| \frac{2(3) + 3(5) - 4}{2 + 3(\frac{1}{2})} \right| \sqrt{1 + \left(\frac{1}{2}\right)^2} Distance=6+1542+321+14\text{Distance} = \left| \frac{6 + 15 - 4}{2 + \frac{3}{2}} \right| \sqrt{1 + \frac{1}{4}} Distance=177254\text{Distance} = \left| \frac{17}{\frac{7}{2}} \right| \sqrt{\frac{5}{4}} Distance=34752\text{Distance} = \left| \frac{34}{7} \right| \frac{\sqrt{5}}{2} Distance=34752\text{Distance} = \frac{34}{7} \cdot \frac{\sqrt{5}}{2} Distance=1757\text{Distance} = \frac{17 \sqrt{5}}{7}

Common Mistakes & Tips

  • Remember the correct ratio for the Euler line: HG:GO=2:1HG:GO = 2:1. Confusing this ratio will lead to incorrect coordinates for the centroid.
  • Be careful when substituting values into the distance formula, especially the slope of the parallel line.
  • Double-check your arithmetic to avoid errors in the calculations.

Summary

We used the properties of the Euler line and the centroid to determine the coordinates of the centroid. Then we found the coordinates of point PP. Finally, we applied the formula for the distance between a point and a line, measured parallel to another line, to find the desired distance. The distance of the point P(3,5)P(3, 5) from the line 2x+3y4=02x + 3y - 4 = 0 measured parallel to the line x2y1=0x - 2y - 1 = 0 is 1757\frac{17\sqrt{5}}{7}.

Final Answer

The final answer is 1757\boxed{\frac{17 \sqrt{5}}{7}}, which corresponds to option (C).

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