Question

A line passes through (3, -2) and (6, 2). a. Write an equation for the line in point-slope form. b. Rewrite the equation in standard form using integers

Answer 1

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 $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$y -y_1=m(x-x_1`)$  .....[1] where m is the slope of the line.

Calculate first the slope of the line:

Slope(m) = $\frac{y_2-y_1}{x_2-x_1}$

Substitute the given points;

$m = \frac{2-(-2)}{6-3}=\frac{2+2}{3} =\frac{4}{3}$

Substitute the value of m in [1] ;

$y - (-2) = \frac{4}{3}(x-3)$

$y+2=\frac{4}{3}(x-3)$     ......[1]

therefore, the equation of line in point slope form is, $y+2=\frac{4}{3}(x-3)$    

(b)

to find the standard form of the equation [1]

Multiply both sides by 3  in [1] we get;

$3(y+2) = 4(x-3)$

using distributive property;   $a\cdot (b+c) = a\cdot b + a\cdot c$

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