Question

Line A passes through the point (-8, 3) and is perpendicular to the line represented by the
equation y = -4r + 9. Line A can be expressed as an equation of the form
Y = mx +b
Part A
What is the value of m for line A?
Part B.
What is the value of b for line A?​

Answer 1

Answer:

Part A) Value of slope for line A is m = 1/4

Part B) Value of b for line A is = 5

Answer 2

Answer:

here slope of the line A ,m= 0.25

b=5

Step-by-step explanation:

given that the coordinate of line A as (-8,3).and line perpendicular to it as y=-4r+9.to find the equation of line A we know that equation is$\frac{(y1 -y0)}{(x1 - 0)} \frac{(y - y0)}{(x - x0)} \\ that \: \: is \\ slope \: of \: line \: a \: = \frac{(y - y0)}{(x - x0)} \: equation \: (1) \\ slope \: is \: negative \: of \: reciprocal \: \\of \: slope \: of \: line \:perpendicular \: to \: a$$we \: know \: that \: slope \: of\\ \: given \: line \: is \: - 4 \\ on \: comparing \: it \: \\with \: linear \: equation \: y = mx + b\\ m = - 4 \\ the \: \\ slope \: of \: a \: \: is \frac{ 1}{4} \\ equation(1)\\becomes \: \\ \frac{ 1}{4} = \frac{(y - y0)}{(x - x0)} \\ given \: a \: point \: (x0.y0) \: in \: a \: ( - 8.3) \\ \frac{ 1}{4} = \frac{(y -3)}{(x + 8)} \\ that \: is \: \\ 4(y - 3) = x + 8 \\ 4y = x + 8 + 12 \\ 4y = x + 20 \\ y = \frac{x}{4} + 5. \\ then \: slope \: of \: a \: (m) = \frac{1}{4} . \\ \:and \: value \: of \: b = 5$

definitions:

we know that equation of line is y=mx+bwere,

y= y coordinate variable of line which is considered

x= x coordinate m=slope of the line

b=-m(x0+y0).

\frac{ (y1-y0)}{(x-z0)} = \frac{(y - y0)}{(x - x0)} \\ given \: a \: point \: (x0.y0)were,(x0ny0) and (x1,y1) are the point in the line.

to find the equation of a line atleat 2 points are required or one point and slope of a line .

#SPJ2