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प्रश्न
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
उत्तर
∵ x-intercept = 3
∴ The line passes through A(3, 0).
Line PQ: x – 7y + 5 = 0
or 7y = x + 5
or y = `1/7 "x" + 5/7`
Therefore slope of PQ = `1/7`
∵ PQ ⊥ AB
∴ Slope of line AB passing through A = –7
∴ Equation of line AB from point (3, 0),
y – 0 = –7(x – 3)
= –7x + 21
or 7x + y – 21 = 0
Second method: Any line perpendicular to ax + by + c = 0 bx – ay + k = 0
∴ Perpendicular to x – 7y + 5 = 0 7x + y + k = 0
This line passes through (3, 0).
∴ 7 x 3 + 0 + k = 0,
i.e. k = –21
∴ The equation of the required line is 7x + y – 21 = 0.
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