SOLUTION: A line passes through (1, 3) and (-4, 2). Write the equation of the line in a. Point-slope form b. Slope-intercept form c. Intercept form d. Normal form.

Algebra ->  Finance -> SOLUTION: A line passes through (1, 3) and (-4, 2). Write the equation of the line in a. Point-slope form b. Slope-intercept form c. Intercept form d. Normal form.      Log On


   

Question 1186114: A line passes through (1, 3) and (-4, 2). Write the equation of the line in
a. Point-slope form
b. Slope-intercept form
c. Intercept form
d. Normal form.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y-y%5B1%5D=m%28x-x%5B1%5D%29
use given (1, 3) and (-4, 2) to find a slope
m=%282-3%29%2F%28-4-1%29=-1%2F-5=1%2F5
y-y%5B1%5D=m%28x-x%5B1%5D%29 ....substitute m and the coordinates of the one point
y-3=%281%2F5%29%28x-1%29
y-3=%281%2F5%29x-1%2F5
y=%281%2F5%29x-1%2F5%2B3
y=%281%2F5%29x%2B14%2F5
b. Slope-intercept form
y=mx%2Bb.........substitute m and the coordinates of the one point
3=%281%2F5%29%2A1%2Bb
3-%281%2F5%29=b
b=14%2F5
y=%281%2F5%29x%2B14%2F5
c. Intercept form
The equation of a line which cuts off intercepts a+and b+respectively from the+x and y+axes is+x%2Fa+%2B+y%2Fb+=+1
y=%281%2F5%29x%2B14%2F5
-+x%2F5++%2By=14%2F5........multiply by 5
-+x++%2B5y=14.......divide by 14
-+x%2F14++%2By%2F%2814%2F5%29=1

d. Normal form.
y=%281%2F5%29x%2B14%2F5
-+x%2F5++%2By=14%2F5........multiply by 5
-+x++%2B5y=14
-+x++%2B5y-14=0