Find an equation oif the line that contains the two points. Write the equation in the general form Ax+By+C=0.
(-5,4)and(2,-5)
Can anyone please teach me how to do this???
I got no idea with writing the equation in the general form.
There is two points given (-5,4)&(2,-5)
find slope, m = (-5-4)/(2--5)
=-9/7
y=mx +b
4=(-9/7)(-5) +b
b=-17/7
y=(-9/7)x +(-17/7)
7y =-9x -17
9x+7y+17=0
William Karpatri on facebook
Sure! I'd be happy to explain how to find the equation of a line that contains two given points in the form Ax+By+C=0.
To find the equation of a line, we need two pieces of information - the slope of the line, and a point on the line. Once we have these, we can use either the point-slope form or the slope-intercept form to write the equation. However, in this case, we will use the point-slope form.
Let's start by finding the slope (m) of the line using the two given points (-5,4) and (2,-5).
The formula to find the slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
m = (y2 - y1) / (x2 - x1)
So, substituting the given points into the formula, we get:
m = (-5 - 4) / (2 - (-5))
Now, simplify the equation:
m = (-9) / (2 + 5)
m = -9 / 7
Now that we have the slope, we can choose one of the given points (-5,4) to plug in the values into the point-slope form.
Using the point-slope form y - y1 = m(x - x1), where (x1, y1) is a point on the line, we can replace (x1, y1) with (-5,4) and m with -9/7:
y - 4 = (-9/7)(x - (-5))
Now simplify further:
y - 4 = (-9/7)(x + 5)
Distribute (-9/7) to (x + 5):
y - 4 = (-9/7)x - 45/7
Finally, let's get the equation into general form Ax + By + C = 0 by moving all the terms to the left side:
(-9/7)x + y - 4 - (-45/7) = 0
Multiply through by 7 to clear the fractions:
-9x + 7y - 28 + 45 = 0
Simplifying:
-9x + 7y + 17 = 0
So the equation of the line that contains the points (-5,4) and (2,-5) is -9x + 7y + 17 = 0 in general form (Ax + By + C = 0).
I hope this explanation helps you understand how to find the equation of a line in general form! Let me know if you have any further questions.