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Q.Consider the line L1:3x−4y+12=0L_1: 3x - 4y + 12 = 0 and a point A(2,−3)A(2, -3).

(i) [2] Find the equation of the line passing through A and parallel to the given line L1L_1.
(ii) [1] Find the distance from the origin to the given line L1L_1.
Kerala DhseKerala DHSE Plus One Board 2024Subjective· 3mImportance★★★★★
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(i) A line parallel to L1L_1 has the same slope; use point-slope form through AA. (ii) Use the perpendicular distance formula from a point to a line.

(i) Rewrite L1:3x−4y+12=0L_1: 3x-4y+12=0 as y=3x+124y=\dfrac{3x+12}{4}, so its slope is m=34m=\dfrac34.

A line parallel to L1L_1 has the same slope. Using point-slope form through A(2,−3)A(2,-3):

y−(−3)=34(x−2).y-(-3) = \frac{3}{4}(x-2).

4(y+3)=3(x−2).4(y+3) = 3(x-2).

4y+12=3x−6.4y+12 = 3x-6.

3x−4y−18=0.3x-4y-18=0.

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