Find the equation of the line which passes through A(3, -2) and B(-2, 4).?

i need the formula for the following 4 questions:








find the equation of the line which passes through A(3, -2) and B(-2, 4).








Find the slope of the line with equation 3x + 5y = -2








Find what "a" is on a right angle triangle with the coordinates being A(3, 2) B(-1,-3) C(7,a). With A being the point of the right angle.








Find the intersection of 2x+y=5 and 3x-2y=11.








thankyou.

Find the equation of the line which passes through A(3, -2) and B(-2, 4).?
slope = (4+2) /(-2-3) = 6 / -5


- 6/5 = (y + 2 ) / (x -3)


- 6 (x -3) = 5 (y + 2 )


18 - 6x = 5 y + 10


6x + 5y -8 = 0





3x + 5y = -2


5y = - 3x -2


y = - 3/5 x - 2/5


slope = - 3/5
Reply:line through (a, b) and (c, d) is


y -- b = [(d -- b)/(c - a)](x -- a)





slope of line ax + by = c is


m = --b/a





Formula to find a in your question is


(slope of AC)(slope of BC) = --1 and slope you found in question 1





Intersection of ax + by + c = 0 and


dx + ey + f = 0 is given by


[(bf --ce)/(ae -- bd), (cd -- af)/(ae -- bd)]
Reply:A(3,-2) and B(-2,4)





m = (4 - (-2))/(-2 - 3)


m = (4 + 2)/(-2 + (-3))


m = -6/5





A(3,-2), m = (-6/5)


-2 = (-6/5)(3) + b


-2 = (-18/5) + b


-10 = -18 + 5b


5b = 8


b = (8/5)





ANS : y = (-6/5)x + (8/5)





--------------------------------------...





3x + 5y = -2


5y = -3x - 2


y = (-3/5)x - (2/5)





ANS : m = (-3/5)





--------------------------------------...





A(3,2), B(-1,-3), C(7,a)





AB = sqrt((-1 - 3)^2 + (-3 - 2)^2)


AB = sqrt((-1 + (-3))^2 + (-3 + (-2))^2


AB = sqrt((-4)^2 + (-5)^2)


AB = sqrt(16 + 25)


AB = sqrt(41)





AC = sqrt((7 - 3)^2 + (a - 2)^2)


AC = sqrt(4^2 + (a - 2)^2)


AC = sqrt(16 + (a - 2)^2)


AC = sqrt(16 + ((a - 2)(a - 2)))


AC = sqrt(16 + (a^2 - 2a - 2a + 4))


AC = sqrt(16 + (a^2 - 4a + 4))


AC = sqrt(16 + a^2 - 4a + 4)


AC = sqrt(a^2 - 4a + 20)





BC = sqrt((7 - (-1))^2 + (a - (-3))^2)


BC = sqrt((7 + 1)^2 + (a + 3)^2)


BC = sqrt(8^2 + ((a + 3)(a + 3)))


BC = sqrt(64 + (a^2 + 6a + 9))


BC = sqrt(a^2 + 6a + 73)





(AC)^2 + (BC)^2 = (AC)^2





(sqrt(a^2-4a+20))^2 + (sqrt(a^2+6a+73))^2 = (sqrt(41))^2





a^2 - 4a + 20 + a^2 + 6a + 73 = 41


2a^2 + 2a + 93 = 41


2a^2 + 2a + 52 = 0


2(a^2 + a + 26) = 0


a^2 + a + 26 = 0





Sorry, i can't find a solution for this one.





--------------------------------------...





2x + y = 5


3x - 2y = 11





Multiply top by 2





4x + 2y = 10


3x - 2y = 11





7x = 21


x = 3





2(3) + y = 5


6 + y = 5


y = -1





ANS : (3,-1)
Reply:equation is ax+by+c = 0








3a-2b + c =0


-2a+4b+c = 0





subtract


5a - 6b = 0


so a:b = 6:5








allow a =6,b=5, then see what c has to be


(because we only need the ratio of a:b:c)





6x+5y+c = 0


18-10+c=0 c = -8





6x+5y+c=0


-12+20+c = 0





6x+5y-8 = 0
Reply:I'll do the last one for you.





For intersection questions you just have to make one of the equations in terms of either x or y (that means it says x = something or y=something) and then sub that into the other one.





2x+y=5 .... (1)





3x-2y=11 .... (2)





so with 1 you get y = 5-2x





sub that into 2





3x - 2(5-2x) = 11 (see how they're all x's now)


3x - 10 +4x = 11


7x = 21


x = 21/7


x=3





then if you sub that into either 1 or 2 you will get an answer for y





and that will give you a point (3,-1) that is their intesection














p.s you'd be better off asking each question seperate, more chance of getting each one answered
Reply:To find the equation of a line given two points (x1,y1), (x2,y2):





1. Find the slope m = (y1-y2)/(x1-x2)


In our case: m = (-2-4)/(3-(-2)) = -6/5





2. Point-slope form: Y-y1 = m(X-x1)


In our case: Y-(-2) = -6/5(X-3)





3. Solve for Y (slope-intercept form)


In our case:


Y = -6/5X + 18/5 -2


Y = -6/5X + 8/5





------------------------------





To find the slope, solve for y and then the coefficient of x will be the slope:





3x+5y=-2


5y = -3x - 2


y = -3/5 x - 2


Slope is -3/5





--------------------------------------...





First, note that angle BAC is right, as stated. Therefore, AB is perpendicular to AC.





The slope of AB = (2-(-3))/(3-(-1)) = 5/4.


The slope of AC = -4/5 (negative reciprocal because they are perpendicular)





We also know that the slope of AC is: (2-a)/(3-7). Thus:





(2-a)/(3-7) = -4/5


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


2-a = 16/5


a = 2-16/5


a = -6/5





----------------------------------





To find the intersection of two lines, you have to manipulate the two in such a way to eliminate a variable to solve for the other:





3x-2y=11


(2x+y=5) * 2 = 4x+2y = 10





Add 3x-2y=11 to 4x+2y=10:


7x = 21


x = 3





Now, plug in x=3 to either equation to solve for y:


3(3)-2y=11


9 - 2y = 11


-2y = 2


y = -1





Thus, the intersection is (x,y) = (3,-1)
Reply:Slope of line can be detemined with 2 points (x1,y1) %26amp; (x2, y2)


For A (3,-2) B(-2, 4)


slope m = (y2 - y1)/(x2-x1)


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


= -6/5





substitute into y = m x + b one of the points, say A





-2 = -6/5 * 3 + b





or -2 = -18/5 + b


b = -2 + 18/5 = 8/5





so equation of line is


y = -6/5 x + 8/5





Slope of 3x + 5y = -2


rearrange to give 5y = -3x -2


or y = -3/5 x - 2/5





Slope is -3/5





Line 1: 2x + y = 5


Line 2: 3x -2y = 11





Rearrange line 1 to terms of y


to give y = -2x + 5


Substitute into line 2 equation giving


3x - 2 ( -2x + 5) = 11


3x - (-4x +10) = 11


3x + 4x - 10 = 11


or 7x = 21 =%26gt; x = 3


substitute into y = -2 x + 5 gives y = -2 *3 + 5 = -6 + 5 = -1


so intersections is (3, -1)

teeth grinding