Answer:
The point-slope form an equation for the line through the points (6, 10) and (12, 14) is [tex]y=\frac{2}{3} x+6[/tex]
Solution:
The slope - intercept form equation of line is given as
y=mx+c --- eqn (1)
Where m is the slope of the line. The coefficient of “x” is the value of slope of the line.
Where slope of the line which is passes through [tex]\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) and \left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)[/tex] is given as
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex] --- eqn 2
From question given that two points are (6, 10), (12,14).
Hence we get [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
By substituting the values in equation (2),
[tex]m=\frac{14-10}{12-6}[/tex]
On simplifying above term,
[tex]m=\frac{4}{6}=\frac{2}{3}[/tex]
On substituting the value of m in equation (1),
[tex]y=\frac{2}{3} x+c[/tex] --- eqn 3
Now equation (3) passes through given two points that is (6, 10), (12,14), so on substituting x = 6 and y=10 in equation (3).
[tex]10=\frac{2}{3}(6)+c[/tex]
10=4+c
c=6
Now on substituting the value of c = 6 in equation (3),
[tex]y=\frac{2}{3} x+6[/tex]
Hence point-slope form an equation for the line through the points (6, 10) and (12, 14) is [tex]y=\frac{2}{3} x+6[/tex]
