Question

Write an equation in point slope form that passes through the given points

Answer 1

1. $y+2 = 3(x-2)$

Point-slope form of the equation of a straight line is:

$y-y_0 = m(x-x_0)$ (1)

The two points in this case are:

$(x_0,y_0)=(2,-2)\\(x_1,y_1)=(5,7)$

Slope of the line is given by:

$m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{7-(-2)}{5-2}=\frac{9}{3}=3$

Substituting into eq.(1), we find:

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

2. $y-4 = \frac{3}{4}(x-6)$

Point-slope form of the equation of a straight line is:

$y-y_0 = m(x-x_0)$ (1)

The two points in this case are:

$(x_0,y_0)=(6,4)\\(x_1,y_1)=(2,1)$

Slope of the line is given by:

$m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{1-4}{2-6}=\frac{3}{4}$

Substituting into eq.(1), we find:

$y-4 = \frac{3}{4}(x-6)$

3. $y+3x=3$

Standard form of the equation of a straight line is:

$ax+bx=c$

with a, b, c integer numbers.

Let's start by finding the point slope form first.

Point-slope form of the equation of a straight line is:

$y-y_0 = m(x-x_0)$ (1)

The two points in this case are:

$(x_0,y_0)=(0,3)\\(x_1,y_1)=(2,-3)$

Slope of the line is given by:

$m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{-3-3}{2-0}=-\frac{6}{2}=-3$

Substituting into eq.(1), we find:

$y-3 = -3(x-0)$

Now we can re-arrange the equation to re-write it in standard form:

$y-3 = -3(x-0)\\y-3 = -3x\\y-3+3x=0\\y+3x=3$

4. $y-x=-2$

Standard form of the equation of a straight line is:

$ax+bx=c$

with a, b, c integer numbers.

Let's start by finding the point slope form first.

Point-slope form of the equation of a straight line is:

$y-y_0 = m(x-x_0)$ (1)

The two points in this case are:

$(x_0,y_0)=(1,-2)\\(x_1,y_1)=(4,2)$

Slope of the line is given by:

$m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{2-(-1)}{4-1}=\frac{3}{3}=1$

Substituting into eq.(1), we find:

$y+1 = x-1$

Now we can re-arrange the equation to re-write it in standard form:

$y+1=x-1\\y+1-x=-1\\y-x=-2$