You need to find the equation of the line AB such that:
`y - y_A = m_(AB)(x - x_A)`
`m_(AB)` denotes the slope of the line AB.
`m_(AB) = (y_B - y_A)/(x_B - x_A) =gtm_(AB) = (10+2)/(5+1) =gt m_(AB) = 12/6 =gt m_(AB) = 2`
y + 2 = 2(x + 1)
You need to open the brackets and to move the terms to the left side:
y + 2 - 2x - 2 = 0 => y - 2x = 0
You need to remember the formula of distance from point M`(x_M,y_M) ` to the line ax+by+c = 0.
`d = |ax_M + by_M + c|/sqrt(a^2 + b^2)`
Hence, the distance from C(0,5) to the line AB is:
`d = |-2*0 + 1*5 + 0|/sqrt((-2)^2 + 1^2)`
`` d = `5/sqrt5 ` => d =`5sqrt5/5` => d = `sqrt 5`
Hence, the distance from C(0,5) to the line AB is d=`sqrt 5` .