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Exercise 12.1 · Q4

Q.Find the equation of the line passing through the point (2,2)(2, 2) and cutting off intercepts on the axes, whose sum is 9.

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Write the line in intercept form xa+yb=1\frac{x}{a}+\frac{y}{b}=1 with a+b=9a+b=9, force it through (2,2)(2,2), and solve the resulting quadratic in aa.

Intercept form: xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1, where a,ba,b are the xx- and yy-intercepts.

  1. Set up the intercept-form equation with the given sum condition. Let the xx-intercept be aa and yy-intercept be bb, with

a+b=9   ⟹   b=9−aa+b = 9 \ \implies\ b = 9-a

xa+y9−a=1\frac{x}{a} + \frac{y}{9-a} = 1

  1. Force the line through (2,2)(2,2).

2a+29−a=1\frac{2}{a} + \frac{2}{9-a} = 1

  1. Clear denominators by multiplying through by a(9−a)a(9-a):

2(9−a)+2a=a(9−a)2(9-a) + 2a = a(9-a)

  1. Expand both sides.

18−2a+2a=9a−a2   ⟹   18=9a−a218 - 2a + 2a = 9a - a^2 \ \implies\ 18 = 9a - a^2

  1. Rearrange into standard quadratic form.

a2−9a+18=0a^2 - 9a + 18 = 0

  1. Factor. Looking for two numbers with product 1818 and sum −9-9: −3-3 and −6-6. (a−3)(a−6)=0   ⟹   a=3 or a=6(a-3)(a-6) = 0 \ \implies\ a=3 \ \text{or}\ a=6 …

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