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Straight Lines and Pair of Straight Lines question

2019 · 12 Jan · Shift 1 · Q42
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  5. /2019 · 12 Jan · Shift 1 · Q42

Straight Lines and Pair of Straight Lines question

2019 · 12 Jan · Shift 1 · Q42

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β\betaβ), then β\betaβ equals :
  1. A
    353{{35} \over 3}335​
  2. B
    −-− 5
  3. C
    −353-{{35} \over 3}−335​
  4. D
    5
View written solutionFree

Correct answer: D

  1. Find the slope of the given line

The line is 2x−3y+17=02x-3y+17=02x−3y+17=0

Rearrange into slope-intercept form: −3y=−2x−17-3y=-2x-17−3y=−2x−17 y=23x+173y=\frac{2}{3}x+\frac{17}{3}y=32​x+317​

So its slope is m1=23m_1=\frac{2}{3}m1​=32​

  1. Use the condition for perpendicular lines

If two lines are perpendicular, then their slopes satisfy m1m2=−1m_1m_2=-1m1​m2​=−1

Hence the slope of the required line is m2=−1m1=−12/3=−32m_2=-\frac{1}{m_1}=-\frac{1}{2/3}=-\frac{3}{2}m2​=−m1​1​=−2/31​=−23​

  1. Find the slope of the line through (7,17)(7,17)(7,17) and (15,β)(15,\beta)(15,β)

Using slope formula: m=β−1715−7m=\frac{\beta-17}{15-7}m=15−7β−17​ m=β−178m=\frac{\beta-17}{8}m=8β−17​

Since this line is perpendicular to the given line, β−178=−32\frac{\beta-17}{8}=-\frac{3}{2}8β−17​=−23​

  1. Solve for β\betaβ

Multiply both sides by 888: β−17=8(−32)=−12\beta-17=8\left(-\frac{3}{2}\right)=-12β−17=8(−23​)=−12

So, β=17−12=5\beta=17-12=5β=17−12=5

  1. Check the options

The value is β=5\beta=5β=5

So the correct option is D.

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