y-y1 x-x1 -------- = ------- y2-y1 x2-x1
y-(-5) x-1 ------- = ---- 7-(-5) -3-1
y+5*(-4)=12*(x-1) -4y-20=12x-12 12x+4y+8=0 /4 3x+y+2=0
why: To form an linear equation for a line, you must remember 2 things
****First find its gradient, second, find the y-intercept.
General equation for a line: y = mx + c, where m = gradient, and c = y-intercept.
Gradient, m = {7 - (-5)} / {(-3) - 1} = - 3
To find the y-intercept, you must use one of the point (-3, 7) or (1, -5) and substitute the value of x and y from the coordinate u choose into the general equation y = mx + c.
I choose (1, -5), so -5 = -3(1) + c -5 = -3 + c -5 + 3 = c -2 = c
Thus,
the equation in slope-intercept form is y = -3x - 2.
the equation in standard form is y + 3x + 2 = 0
the equation in point-slope form is y + 3x = -2 divide both sides with -2 y / (-2) + 3x / (-2) = -2 / -2 - y / 2 - x / (2/3) = 1
The missing side length is 8 units
Hey there!
The answer to your question is ± 12
To solve the equation x²/-10 = 2, first multiply 10 to both sides:
x² = 20
Then, take the square root from both sides:
x = ± 20
Answer:
how long is the class
we need to know how long the class is to help you figure it out
