  Question

Find the equation of the straight line passing through the  origin and bisecting the portion of the line ax + by + c = 0 intercepted between the coordinate axes.

Solution

The equation of the line passing through the origin is y = mx Let the line ax+by+c=0 meet the coordinate axes at A and B. So, the coordinates of A and B are A (−ca,0) and B (0,−cb). Now, the midpoint of AB is (−c2a,−c2b). Clearly, (−c2a,−c2b) lies on the line y = mx ∴ −c2b=m×−c2a ⇒ m=ab Hence, the equation of the required line is y=abx ⇒ ax−by=0 P.Q. The area of the triangle formed by the coordinate axes and a line is 6 square units and the length of the hypotenues is 5 units. Find the equation of the line. The equation of the line with intercepts a and b is xa+yb=1 The line intersects the axes at A (a, 0) and B (0, b) Let O be the origin It is given that the area of the triangle formed by the coordinate axes and the line is 6 square units. ∴ Area of ΔOAB=12|0A×OB| ⇒ 6=12.|ab| ⇒ |ab|=12 ⇒ |a|b|=3×4                     ...(i) Here, the length of the hypotenuse is 5 units, i.e., AB = 5 ∴ OA2+OB2=AB2 ⇒ a2+b2=25                     ...(ii) Let us find the possible values of a and b from (i) and (ii). Using equation (i) 144b2+b2=25 ⇒ b4−25b2+144=0 ⇒ b2=16, 9 ⇒ b±4, ±3 Now, from equation (i) For b=±4, |a|=3⇒b=4, a=3, −3 and b=−4, a=3, −3 For b=±3, |a|=4⇒b=3, a=4, −4 and b=−3, a=4, 4 Hence, the possible equations of the lines are x3+y4=1, x−3+y4=1, x3+y−4 =1, x−3+y−4=1, x4+y3=1, x4+ y3=1, x4+y3=1 and x−4+y3=1 or 4x+3y=12 or 4x−3y=−12 or −4x+3y=−12 or −4x−3y=12 or 3x +4y=12 or x−4y=−12 or −3x+4y =−12 or −3x−4y=12 MathematicsRD SharmaStandard XI

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