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JEE Main 2023
Straight Lines
Straight Lines and Pair of Straight Lines
Easy

Question

Let PSPS be the median of the triangle with vertices P(2,2)P(2, 2), Q(6,1)Q(6, -1) and R(7,3)R(7, 3). The equation of the line passing through (1,1)(1, -1) band parallel to PS is :

Options

Solution

Key Concepts and Formulas

  • Midpoint Formula: The midpoint MM of a line segment with endpoints (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).
  • Slope Formula: The slope mm of a line passing through points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
  • Parallel Lines: Parallel lines have equal slopes.
  • Point-Slope Form: The equation of a line passing through (x0,y0)(x_0, y_0) with slope mm is yy0=m(xx0)y - y_0 = m(x - x_0).

Step-by-Step Solution

1. Identify the vertices and the median

We are given the vertices P(2,2)P(2, 2), Q(6,1)Q(6, -1), and R(7,3)R(7, 3). PSPS is the median, meaning SS is the midpoint of QRQR.

2. Calculate the coordinates of point S

We use the midpoint formula to find the coordinates of SS, the midpoint of QRQR. S=(6+72,1+32)=(132,22)=(132,1)S = \left( \frac{6+7}{2}, \frac{-1+3}{2} \right) = \left( \frac{13}{2}, \frac{2}{2} \right) = \left( \frac{13}{2}, 1 \right) Thus, S=(132,1)S = \left( \frac{13}{2}, 1 \right).

3. Calculate the slope of median PS

We use the slope formula with points P(2,2)P(2, 2) and S(132,1)S\left( \frac{13}{2}, 1 \right) to find the slope of PSPS. mPS=121322=11342=192=29m_{PS} = \frac{1 - 2}{\frac{13}{2} - 2} = \frac{-1}{\frac{13 - 4}{2}} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9} The slope of PSPS is 29-\frac{2}{9}.

4. Determine the slope of the required line

The line we are looking for is parallel to PSPS, so it has the same slope. mreq=mPS=29m_{req} = m_{PS} = -\frac{2}{9}

5. Find the equation of the required line

We use the point-slope form of a line with the point (1,1)(1, -1) and slope 29-\frac{2}{9}. y(1)=29(x1)y - (-1) = -\frac{2}{9}(x - 1) y+1=29(x1)y + 1 = -\frac{2}{9}(x - 1) Multiply both sides by 9 to eliminate the fraction: 9(y+1)=2(x1)9(y + 1) = -2(x - 1) 9y+9=2x+29y + 9 = -2x + 2 Rearrange to get the equation in the form Ax+By+C=0Ax + By + C = 0: 2x+9y+92=02x + 9y + 9 - 2 = 0 2x+9y+7=02x + 9y + 7 = 0

Common Mistakes & Tips

  • Carefully handle negative signs, especially when using the slope and midpoint formulas. Double-check each calculation.
  • Remember that parallel lines have the same slope, while perpendicular lines have slopes that are negative reciprocals of each other.
  • When rearranging the equation into the standard form, ensure all terms are moved correctly, and the signs are handled appropriately.

Summary

We found the midpoint of QRQR to determine the coordinates of SS. Then, we calculated the slope of the median PSPS. Since the required line is parallel to PSPS, it has the same slope. Finally, using the point-slope form with the given point (1,1)(1, -1) and the calculated slope, we derived the equation of the line. The equation of the line is 2x+9y+7=02x + 9y + 7 = 0.

The final answer is \boxed{2x + 9y + 7 = 0}, which corresponds to option (D).

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