If (a,b) be the image of point (16,-21) wr.t a line 6x - 7y = 158, then the quadratic
equation whose roots are a & ß, is
x2 – 3x + 28 = 0
x2 + 3x - 28 = 0
x2 - 7x + 12 = 0
x2 + 7 x – 12 = 0
Answers
Answer:
Step-by-step explanation:
Factoring x2-3x-28
The first term is, x2 its coefficient is 1 .
The middle term is, -3x its coefficient is -3 .
The last term, "the constant", is -28
Step-1 : Multiply the coefficient of the first term by the constant 1 • -28 = -28
Step-2 : Find two factors of -28 whose sum equals the coefficient of the middle term, which is -3 .
-28 + 1 = -27
-14 + 2 = -12
-7 + 4 = -3 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -7 and 4
x2 - 7x + 4x - 28
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-7)
Add up the last 2 terms, pulling out common factors :
4 • (x-7)
Step-5 : Add up the four terms of step 4 :
(x+4) • (x-7)
Which is the desired factorization
Answer:
Step-by-step explanation:
Solution :
`x^2+y^2+16x-24y+183=0`
`x^2+16x+64-64+y^2-24y+144-144+183=0`
`(x+8)^2+(y-12)^2+183-208=0`
`(x+8)^2+(y-12)^2-25=5^2`
Center(-8,12)
`4x+7y+13=0-(1)`
`7x+4y+k=0`
`-56+48+k=0`
`k=8`
`7x+4y+8=0-(2)`
From equation 1 and 2
`y=325/65=5`
`y=5`
`4x+7y+13=0`
9`4x+35+13=0`
`x=-12`
(-8,12),(-12,5),(h,k)
`(-8+h)/2=-12`
`h=-16`
`(12+k)/2=5`
`k=-2`
(-16,-2)
`(x+16)^2+(y+2)^2=25`
`x^2+y^2+32x+4y+256+4-25=0`
`x^2+y^2+32x+4y+235=0`.