Find the equation of the straight line passing through the point (0, 1) which is perpendicular to the line y = -2x+2

This is about equation of a straight line in slope intercept form.

Equation of the straight line is y =

y = mx + c

where m is slope and c is y-intercept

A line passing through the point (0, 1)

This line is perpendicular to the line y = -2x + 2

m = -2

Thus, slope of new line = -1/-2 = 1/2

y - y1 = m(x - x1)

This gives;

y - 1 = (1/2)(x - 0)

y - 1 =

y = + 1

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y = x + 1

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

y = - 2x + 2 is in this form with slope m = - 2

given a line with slope m then the slope of a line perpendicular to it is

= - = - =

the point (0, 1 ) is the y- intercept ⇒ c = 1

y = x + 1 ← equation of perpendicular line

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

i cant see the graph

step-by-step explanation: