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MCQMediumJEE 2024Straight Line Equations

JEE Mathematics 2024 Question with Solution

Let A be the point of intersection of the lines 3x+2y=143x + 2y = 14 and 5x−y=65x - y = 6, and B be the point of intersection of the lines 4x+3y=84x + 3y = 8 and 6x+y=56x + y = 5. The distance of the point P(5,−2)P(5, -2) from the line AB is:

  • A

    132\frac{13}{2}

  • B

    88

  • C

    52\frac{5}{2}

  • D

    66

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Point P(5,−2)P(5, -2) and points A and B are intersections of the given pairs of lines.

Find: The distance of PP from line AB.

First find point A by solving

3x+2y=145x−y=6\begin{aligned} 3x + 2y &= 14 \\ 5x - y &= 6 \end{aligned}

Multiply the second equation by 22:

10x−2y=1210x - 2y = 12

Now add with the first equation:

3x+2y=1410x−2y=12\begin{aligned} 3x + 2y &= 14 \\ 10x - 2y &= 12 \end{aligned}

So,

13x=26  ⟹  x=213x = 26 \implies x = 2

Substituting in 3x+2y=143x + 2y = 14,

3(2)+2y=14  ⟹  6+2y=14  ⟹  2y=8  ⟹  y=43(2) + 2y = 14 \implies 6 + 2y = 14 \implies 2y = 8 \implies y = 4

Hence,

A=(2,4)A = (2, 4)

Using line through two points and point-line distance

Now find point B by solving

4x+3y=86x+y=5\begin{aligned} 4x + 3y &= 8 \\ 6x + y &= 5 \end{aligned}

Multiply the second equation by 33:

18x+3y=1518x + 3y = 15

Subtract the first equation:

14x=7  ⟹  x=1214x = 7 \implies x = \frac{1}{2}

Substituting in 4x+3y=84x + 3y = 8,

4(12)+3y=8  ⟹  2+3y=8  ⟹  3y=6  ⟹  y=24\left(\frac{1}{2}\right) + 3y = 8 \implies 2 + 3y = 8 \implies 3y = 6 \implies y = 2

Hence,

B=(12,2)B = \left(\frac{1}{2}, 2\right)

The slope of AB is

m=2−412−2=−2−32=43m = \frac{2 - 4}{\frac{1}{2} - 2} = \frac{-2}{-\frac{3}{2}} = \frac{4}{3}

Using point-slope form through A(2,4)A(2,4),

y−4=43(x−2)y - 4 = \frac{4}{3}(x - 2)

This gives

3y−12=4x−8  ⟹  4x−3y+4=03y - 12 = 4x - 8 \implies 4x - 3y + 4 = 0

Now use the distance formula from point P(5,−2)P(5,-2) to line 4x−3y+4=04x - 3y + 4 = 0:

d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

So,

d=∣4(5)+(−3)(−2)+4∣42+(−3)2=∣20+6+4∣16+9=305=6d = \frac{|4(5) + (-3)(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{\sqrt{16 + 9}} = \frac{30}{5} = 6

Therefore, the distance of the point P(5,−2)P(5,-2) from line AB is 66. The correct option is D.

Common mistakes

  • Finding the wrong intersection point for A or B by making an elimination sign error. This changes the equation of line AB completely. Solve each pair carefully and verify the point satisfies both original equations.

  • Using the slope formula incorrectly for points A(2,4)A(2,4) and B(12,2)B\left(\frac{1}{2},2\right). A wrong slope leads to a wrong line equation. Keep the subtraction order consistent in numerator and denominator.

  • Applying the point-to-line distance formula with incorrect signs in Ax1+By1+CAx_1 + By_1 + C. For the line 4x−3y+4=04x - 3y + 4 = 0, the coefficient of yy is −3-3, not 33. Substitute coefficients exactly as they appear.

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