Find the equation of line which pass through the point (3,4) and sum of whose intercepts on the axes is 14.
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4
Given co-ordinator is (3,4)
So, x=3 and y=4
,i.e, 3 and 4 are the roots of the equation.
p(x)= x^2 - (3+4)x + (3)(4)
= x^2 - 7x + 12=0
The above is the required equation
So, x=3 and y=4
,i.e, 3 and 4 are the roots of the equation.
p(x)= x^2 - (3+4)x + (3)(4)
= x^2 - 7x + 12=0
The above is the required equation
Answered by
3
Answer:
4x+3y=24
Step-by-step explanation:
Equation of a straight line in intercept form is
xa+yb=1⇒3a+4b=1 ...(i)
Given, a+b=14 ...(ii)
On solving (i) and (ii) we get
3a+414−a=1⇒a2−13a+42=0
⇒(a−7)(a−6)=0
⇒a=6andb=8
or a=7andb=7.
∴ Required eqns are 4x+3y=24orx+y=1.
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