# what is the equation of the line through (4,-1) and whose segment intercepted between the axes in the fourth quadrant is equal to 2sqrt17?my professor in analytic geometry gave this question to me...

what is the equation of the line through (4,-1) and whose segment intercepted between the axes in the fourth quadrant is equal to 2sqrt17?

my professor in analytic geometry gave this question to me as a takehome quiz.from the book of analytic geometry by demetrio quirino and jose mijares.i hope you can help me.thank you!

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In the fourth quadrant x is positive and y is negative. The length of the segment between the fourth quadrant is `2sqrt17` .

Let the x intercepts be A(a,0) and y intercept be B(0,b). Now the length between AB is `2sqrt17` .

Length AB;

`sqrt[(a-0)^2+(0-b)^2] = 2sqrt17`

`a^2+b^2 = 4*17` ----(1)

If we consider this line as y = mx+c

Slope of the line (m) = (0-b)/(a-0) = -b/a

y intercept of the line (c) = b

Now we can write the equation of the line as;

`y = -b/ax+b`

This line passes through (4,-1)

`-1 = -b/a*4+b`

`-1 = b(1-4/a)`

`-1 = b((a-4)/a)`

`b = a/(4-a)`

substitute b = a/(4-a) to equation (1)

`a^2+b^2 = 4*17`

`a^2+a^2/(4-a)^2 = 68`

`a^2(4-a)^2+a^2 = 68(4-a)^2`

`a^2(16-8a+a^2)+a^2 = 68(16-8a+a^2)`

`16a^2-8a^3+a^4+a^2 = 68*16-68*8a+68a^2`

`a^4-8a^3+a^2(16+1-68)+68*8a-68*16 = 0`

`a^4-8a^3-51a^2+544a-1088 = 0`

This equation has 4 roots. Accordingly we have 4 values 'a' and 4 values for 'b'. They are the x and y intercepts of the **four quadrants.** We must get the value that refers to fourth quadrant. So a>0 and b<0.

`-51a^2+544a-1088`

`= (a-(-544+sqrt(544^2-4*(-51)*(-1088)))/(2*(-51)))(a-(-544-sqrt(544^2-4*(-51)*(-1088)))/(2*(-51)))`

`= (a-8)(a-8/3)`

`a^4-8a^3-51a^2+544a-1088 = 0`

`a^3(a-8)+(a-8)(a-8/3) = 0`

`(a-8)(a^3-a-8/3) = 0`

It is clear that a = 8 is a root of the equation. Now for a = 8 if we get a negative value for 'b' we are in the fourth quadrant.

b = `a/(4-a)`

= `8/(4-8)`

= -2

So b = -2 preferably the answer that we want.

So the equation of the line is;

`y = -(-2/8)x+(-2)`

`y = (1/4)x-2`

For further clearance i will draw the graph of the line. The dot shows that the line passes through (4,-1) which is the given data.

**Sources:**