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प्रश्न
Find the equation of the straight line which has Y-intercept equal to 4/3 and is perpendicular to 3x – 4y + 11 = 0.
उत्तर
Equation of the given line is
3x - 4y + 11 = 0
Slope of this line y = mx + c
4y = 3x + 11
y = `(3)/(4) x + (11)/(4)`
m1 = `(3)/(4)`
Let m2 be the slope of the line which is perpendicular toi the given line then
m1m2 = -1
`(3)/(4)m_2` = -1
⇒ m2 = `-(4)/(3)`.
Also, Y-intercept c = `(4)/(3)`.
Equation of the required line
y = m2x + c
y = `(-4)/(3) x + (4)/(3)`
⇒ 3y = -4x + 4
⇒ 4x + 3y - 4 = 0.
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