Respuesta :
Answer:
(a)
Given: A line passes through the points (3, -2) and (6, 2)
Point slope form: An equation of line passing through two points [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] is given by:
[tex]y -y_1=m(x-x_1`)[/tex] .....[1] where m is the slope of the line.
Calculate first the slope of the line:
Slope(m) = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
Substitute the given points;
[tex]m = \frac{2-(-2)}{6-3}=\frac{2+2}{3} =\frac{4}{3}[/tex]
Substitute the value of m in [1] ;
[tex]y - (-2) = \frac{4}{3}(x-3)[/tex]
[tex]y+2=\frac{4}{3}(x-3)[/tex] ......[1]
therefore, the equation of line in point slope form is, [tex]y+2=\frac{4}{3}(x-3)[/tex]
(b)
to find the standard form of the equation [1]
Multiply both sides by 3 in [1] we get;
[tex]3(y+2) = 4(x-3)[/tex]
using distributive property; [tex]a\cdot (b+c) = a\cdot b + a\cdot c[/tex]
3y + 6 = 4x -12
Subtract 3y to both sides we get;
3y + 6 -3y = 4x - 12 - 3y
Simplify:
6 = 4x - 3y -12
Subtract 6 from both sides we get;
0 = 4x - 3y -12-6
Simplify:
4x - 3y - 18 =0
Therefore, the standard form of the equation is; 4x - 3y - 18 =0
