1) Find the equation of a line with slope [tex]\( m = 3 \)[/tex] that passes through the point [tex]\( (7, 5) \)[/tex]. Write your answer in [tex]\( y = mx + b \)[/tex] form.



Answer :

To find the equation of a line with a given slope \( m = 3 \) that passes through the point \( (7, 5) \), we can use the point-slope form of a linear equation. The point-slope form is:

[tex]\[ y - y_1 = m(x - x_1) \][/tex]

where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope of the line. Here, \( (x_1, y_1) = (7, 5) \) and \( m = 3 \).

Step 1: Write the point-slope form using the given values:

[tex]\[ y - 5 = 3(x - 7) \][/tex]

Step 2: Distribute the slope \( m = 3 \) on the right-hand side:

[tex]\[ y - 5 = 3x - 21 \][/tex]

Step 3: Solve for \( y \) to put the equation into the slope-intercept form \( y = mx + b \):

[tex]\[ y = 3x - 21 + 5 \][/tex]
[tex]\[ y = 3x - 16 \][/tex]

Therefore, the equation of the line in the form \( y = mx + b \) is:

[tex]\[ y = 3x - 16 \][/tex]

The \( y \)-intercept \( b \) is \(-16\).

So, the final equation of the line is:

[tex]\[ y = 3x - 16 \][/tex]