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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

 axby=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

 b425b2+144=0

 b2=16, 9

 b±4, ±3

Now, from equation (i)

For b=±4, |a|=3b=4, a=3,

3 and b=4, a=3, 3

For b=±3, |a|=4b=3, a=4,

4 and b=3, a=4, 4

Hence, the possible equations of the lines are

x3+y4=1, x3+y4=1, x3+y4

=1, x3+y4=1, x4+y3=1, x4+

y3=1, x4+y3=1 and x4+y3=1

or 4x+3y=12 or 4x3y=12 or

4x+3y=12 or 4x3y=12 or 3x

+4y=12 or x4y=12 or 3x+4y

=12 or 3x4y=12


Mathematics
RD Sharma
Standard XI

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