Bro I think you should search algerba calcautor and then you'll get the answers
Hello
<span>an equation for the line in point-slope form and general form is :
y = ax+b a : </span>slope ; the <span>Passing through (x' ; y')
</span>y' = ax'+b
y-y' =a(x-x') and : x' =2 y' = - 1
calculate a :
let : y = ax+b .....(D)
....<span>3y-x=7</span>....(D') or y = (1/3)x+7/3
.(D) perpendicular to(D') : slope (D) × slope (D') = -1
slope (D') = 1/3
slop(D)×(1/3) = -1
slope (D) = -3
equation for the line : y-y' =a(x-x')
y+1 =(-3) (x-2)
Answer:
Please see attachments for step by step guide and answers.
The origin is at the point (0,0). Therefore, the y-intercept will be 0. For two points to be perpendicular, the slopes must be opposite reciprocals of each other. .5 can be seen as 1/2, so the reciprocal is 2, and it is positive, making that 2 negative. Your equation would be y = -2x.
