Q8 of 112 Page 166

Find the equation of the straight line parallel to the line 3x – y + 7 = 0 and passing through the point (1, –2).

Here it’s given that the straight line is parallel to the line


3x – y + 7 = 0 and passing through the point (1, –2).


As the lines are parallel to each other their slopes are equal.


Slope of the given line 3x – y + 7 = 0 is



Equation of the line passing through the point(1,–2) is


(y–y1) = m(x–x1),where (x1,y1) is (1,–2)


⇒ (y–(–2)) = 3(x–1)


⇒ y + 2 = 3x–3


⇒ y + 2–3x + 3 = 0


⇒ –3x + y + 5 = 0


⇒ 3x–y–5 = 0 (multiplied by –1 on both sides of the equation)


Hence the equation of the line is 3x–y–5 = 0 .


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 10

Find the equation of the perpendicular bisector of the straight line segment joining the points (3, 4) and (–1, 2).