Question

What is the equation of the line that passes through (5, -2) and (-3, 4)?

Answer 1

Y= -4/3(x-7/2)

Explanation:

So first calculate the difference between them,

changes by 8 x units, and -6 y units.

Then substitute them into y/x to find gradient

-6/8 = -4/3

so now we have a part of the equation:

Y= -4/3(x-a)

substitute Y= -2 and x=5 (from (5,-2))

-2= -4/3(5-a)

-2= -20/3+4a/3

Multiply by 3 on both sides

-6= -20+4a

add 20 on both sides

14=4a

a=7/2

use this as the value of a

Y= -4/3(x-7/2)

Answer 2

y = (-3/4)x + 7/4

Explanation:

Step 1: Define general form of equation of line

An equation of a straight line on two-dimensional plane could be represented in form of: y = Mx + b, with M is slope and b is y-intercept

Step 2: Set up the system to solve for parameters of equation of line

(solve for M and b)

That equation passes 2 points, which are represented in form of (x, y), (5, -2) and (-3, 4).

Substitute these values of x and y into the original equation in step 1:

-2 = 5M + b

4 = -3M + b

Step 3: Solve the system of equations in step 2 for M and b

Subtract 1st equation from 2nd equation:

6 = -8M

=> M = -6/8 = -3/4

Substitute M back into 1st equation:

=> -2 = 5*(-3/4) + b

=> b = -2 + 15/4

=> b = 7/4

=> The equation of the line that passes through (5, -2) and (-3, 4):

y = (-3/4)x + 7/4

Hope this helps!

:)