Write the equation of the line in standard form that passes through (5,-2) and (-3,14).?

Question

Write the equation of the line in standard form that passes through (5,-2) and (-3,14).

Answer 1

slope = (14--2)/(-3-5)=16/-6=-2

(y+2)/(x-5)=-2
y+2=-2x+10
y+2x=8

Answer 2

(2,1) are the x and y coordinates. (x,y) Equation of a line is in the form: y= mx + c, where, m is the gradient or slope of the line c is the y-intercept "m" is known as -3/5 Next find y-intercept. substitue into the general form: y = mx+c we use co-ordinate (2, 1) 1 = -3/5 (2) + c 5/5 = -6/5 + c c = 11/5 Now, we have m = -3/5 and c = 11/5 Hence, general equation of the line is y = -3/5x + 11/5 multiply 5 on both sides: 5y = -3x +11 Re-arranging equation: 3x + 5y = 11

Answer 3

Equation of the line in standard form:

y = mx + c

where m is the slope
and c is the y-intercept

m = change in y / change in x

m = y2 - y1 / x2 - x1

The two points
(5, -2) ==> x1 = 5 , y1 = -2
(-3, 14) ==> x2 - -3 , y 2 = 14
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m = 14 - (-2) / -3 - 5
m = 16/ - 8
m = -2

y = -2x + c

Let's find c

Choose any point and substitute it in the equation

y = -2x + c

Let us take (5, -2)

-2 = -2(5) + c
-2 = -10 + c
-2 + 10 = c
c = 8
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Equation:

y = -2x + 8

Answer 4

(14+2)/(-3-5)=-2
y+2=-2(x-5)
y=-2x+8
2x+y=8