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प्रश्न
Find the equation of a line drawn perpendicular to the line `x/4 + y/6 = 1`through the point, where it meets the y-axis.
उत्तर
Equation of line AB,
`"x"/4 + "y"/6 = 1` or 3x + 2y = 12,
2y = −3x + 12
y = `(-3)/2 "x" + 12/2`
Slope of line AB = `(-3)/2`
AB ⊥ BC,
∴ Slope of BC = `2/3`
∵ The line intersects the y-axis, hence the point is B(0, 6).
∴ Equation of line BC
y – 6 = `2/3 ("x" - 0)`
or 3y – 18 = 2x
or 2x – 3y + 18 = 0
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