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प्रश्न
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
उत्तर
The given line is
x = 3y + 2 ...(1)
3y = x – 2
`y = 1/3 x - 2/3`
Slope of this line is `1/3`
The required line intersects the given line at right angle.
∴ Slope of the required line = `(-1)/(1/3) = -3`
The required line passes through (0, 0) = (x1, y1)
The equation of the required line is
y – y1 = m(x – x1)
y – 0 = –3(x – 0)
3x + y = 0 ...(2)
Point X is the intersection of the lines (1) and (2).
Using (1) in (2), we get,
9y + 6 + y = 0
`y = (-6)/10 = (-3)/5`
∴ x = 3y + 2
= `(-9)/5 + 2`
= `1/5`
Thus, the co-ordinates of the point X are `(1/5, (-3)/5)`
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