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exercise 12 b q 8 (eq. of a straight line)
Asked by nilesh.dhote74 | 17 May, 2020, 10:00: PM
The equation of a line is 2x + 3y = 6 ... (1)
and it intesects the y - axis at A.
So, the coordinates of A(0, y).

From (1), 2(0) + 3y = 6 → 3y = 6 → y = 2.

(i) The coordinates of A are (0, 6).

(ii) The required line is perpendicular to the given line.

From (1), 3y = -2x + 6 → y = (-2/3)x + 2

Slop of required line = 3/2

By Point - slope form, Equation of required line is
y - 6 = 3/2(x - 0)
→ y - 6 = (3/2)x
→ 2y - 12 = 3x
→ 3x - 2y = -12

Answered by Yasmeen Khan | 17 May, 2020, 11:57: PM

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