Question

Write an equation (a) in slope intercept form and (b) in standard form for the line passing through (-2,6) and perpendicular to x + 3y = 7

Answer 1

Answer:

a) y = 3x+12

b) y-6 = 3(x+2)

Step-by-step explanation:

The equation of a line in slope-intercept form is expressed as y = mx+c

m is the slope or gradient

c is the intercept

We need to calculate the value of slope and intercept.

We will get the slope from the equation of line x+3y = 7

Rewriting the equation

3y = 7-x

y = 7/3 -x/3

M = -1/3

Since the equation if the unknown line is perpendicular to this line then Mm = -1 where m is the slope of the unknown line

m = -1/M

m = -1/(-1/3)

m = 3

To get c, we will substite the point given (-2,6) and the slope into the equation y = mx+c

6 = 3(-2)+c

6 = -6+c

c = 12

Substituting m= 3 and c = 12 into the standard form of the equation we have;

y = 3x+12 (This gives the required equation in its slope intercept form)

b) The standard form of a line is expressed as y-y1 = m(x-x1) where (x1,y1) are the points and m is the slope. On substituting the point {-2,6) and slope of 3 into this equation we will have:

y - 6 = 3(x-(-2))

y-6 = 3(x+2)

This gives the equation of the line in its standard form