How do you find the point-slope form of the equation of the line passing through the points x(3, 2) and y(5,10)?

Question

Hazel Anglin · Accepted Answer

First we have to find the slope of the equation. To find the slope we have to do;
#(y2-y1)/(x2-x1)# For our question slope is;

#(10-2)/(5-3) = 8/2 = 4#

The main formula of a line is;
#y=ax+b#

We should use one point to find the real equation. If we use (3,2) point;
#y= ax+b => 2=3a+b#;
a is the slope of the equation, we found that as #4#;
#2=(3*4) + b => 2=12 +b => b=-10#;
So the equation of the line will be;
#y=ax+b => ul (y= 4x-10)#
We can check if our equation is right or not with other given point;
#(5,10) => y=4x-10 => 10=(4*5)-10 => 10=20-10 => 10=10#

Kennedy Adams · Answer

undefined To find the point-slope form of the equation of the line passing through the points x(3, 2) and y(5,10), first, calculate the slope using the formula: $m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the given points. Then, choose one of the given points (let's say x(3, 2)) and substitute its coordinates and the calculated slope into the point-slope formula: $y - y_1 = m(x - x_1)$.