Find the equation of the line parallel to 2x+3y+11=0 and the sum of its intercepts on the axis is 15.
ans quickly, this ques was hung up for many days, but no-one answered it
Answers
Answered by
8
Given, a line l is parallel to a straight line m whose equation is 2x+3y+11=0.
Since;the line l is parallel to the the line m, its equation will be 2x+3y+C=0.
◆In slope intercept form, the equation of the line l would be x/a + y/b = 1 ---- (2).
Given a+b=15 or a=15-b.
Therefore , the equation (2) can be written as
x/(15-b) + y/b = 1
=)bx + (15-b)y = (15-b)b
=) bx + (15-b)y - (15-b)b = 0
Comparing the above with eq. (1)
=) b = 2k, and (15-b)=3k for some factor k.
Solving for b, we get b = 6 and therefore a = 15-6 = 9.
Thus, the equation of the line l is x/9 + y/6 =1 or 2x .
Hope it Helps u
Since;the line l is parallel to the the line m, its equation will be 2x+3y+C=0.
◆In slope intercept form, the equation of the line l would be x/a + y/b = 1 ---- (2).
Given a+b=15 or a=15-b.
Therefore , the equation (2) can be written as
x/(15-b) + y/b = 1
=)bx + (15-b)y = (15-b)b
=) bx + (15-b)y - (15-b)b = 0
Comparing the above with eq. (1)
=) b = 2k, and (15-b)=3k for some factor k.
Solving for b, we get b = 6 and therefore a = 15-6 = 9.
Thus, the equation of the line l is x/9 + y/6 =1 or 2x .
Hope it Helps u
Similar questions