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Find the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4units.
Welcome to AskiitiansGiven that,The equation of a line = (x/3) + (y/4) = 1It can be written as:4x + 3y -12 = 0 …(1)Compare the equation (1) with general line equation Ax + By + C = 0,we get the values A = 4, B = 3, and C = -12.Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.we know that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given byD = |Ax1+By1+ C|/ √A2+ B2Now, substitute the values in the above formula, we get:4 = |4a+0 + -12|/ √42+ 32⇒4 = |4a-12|/5⇒|4a-12| = 20⇒± (4a-12)= 20⇒ (4a-12)= 20 or -(4a-12) =20Therefore, it can be written as:(4a-12)= 204a = 20+124a = 32a = 8(or)-(4a-12) =20-4a +12 =20-4a = 20-12-4a= 8a= -2⇒ a= 8 or -2Hence, the required points on x axis are (-2, 0) and (8, 0).Thanks
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