Write an equation of a line that passes through the point (2, 3) and is parallel to the line y = 3 over 2x + 5.
1 answer:
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Answer:
D. 13
Step-by-step explanation:
well since it's a quadratic equation, it can be written in the form of ax²+bx+c=0. So we have a=1, b=b and c= -30. Also the roots are given by the equation-
x= -b± /2a. So all I did was put the number from the choices which would satisfy the sqrt part and that turned out to be 13 since
=
= 17. So b=13.
Answer:
1) y=⅚x -2⅓
2) y=8/3x -5
Step-by-step explanation:
<u>Point-slope form:</u>
y=mx+c, where m is the gradient and c is the y-intercept.
Parallel lines have the same gradient.
Gradient of given line=
Thus, m=⅚
Susbt. m=⅚ into the equation,
y= ⅚x +c
Since the line passes through the point (4, 1), (4, 1) must satisfy the equation. Thus, substitute (4, 1) into the equation to find c.
When x=4, y=1,
1= ⅚(4) +c
Thus the equation of the line is .
The gradients of perpendicular lines= -1.
Gradient of given line= -⅜
-⅜(gradient of line)= -1
gradient of line
= -1 ÷ (-⅜)
= -1 ×(-8/3)
=
When x=3, y=3,
Thus the equation of the line is .