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10 Questions 40 Marks 10 Mins
Concept:
Two line equations are parallel when their tangent is equal to eachother.
Let,
y = m1x + c ------ (1)
y = m2x + c ------ (2)
If these lines are parallel. Then,
m1 = m2.
Given:
4x + 6y - 1 = 0 and 2x + ky - 7 = 0
Calculation:
After arranging the given equation (1) & (2)
y = -\(\frac{4}{6}\)x + \(\frac{1}{6}\) ------ (3)
y = -\(\frac{2}{k}\)x + \(\frac{7}{k}\) ------ (4)
After comparing with standard equation,
m1 = -\(\frac{4}{6}\) And m2 = -\(\frac{2}{k}\)
If these lines are parallel. Then,
m1 = m2
⇒ -\(\frac{4}{6}\) = - \(\frac{2}{k}\)
⇒ 4k = 12
⇒ k = 3
Additional Information
Two line equations are perpendicular when the product of their tangent is (-1).
Let,
y = m1x + c ------ (1)
y = m2x + c ------ (2)
If these lines are parallel. Then,
m1 × m2 = -1
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The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 7 = 0 represents parallel lines is
k = 3
Explanation;
Hint:
Slope of 4x + 6y – 1 = 0
6y = – 4x + 1
⇒ y = `(-4)/6x + 1/6`
Slope = `(-4)/6 = (-2)/3`
Slope of 2x + ky – 7 = 0
ky = – 2x + 7
y = `(-2)/"k"x + 7/"k"`
Slope of a line = `(-2)/"k"`
Since the lines are parallel
`(-2)/3 = (-2)/"k"`
– 2k = – 6
k = `6/2`
= 3
Concept: Simultaneous Linear Equations
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