What are the steps to solve these??

Question

Determine the equation of the straight line that passes by these pairs of points: (7, -1) and (3, -2)

AND

Determine the equation of the straight line, perpendicular to line "y = -(2/3)x +10/3" and passes by point (3, -4)

Answer 1

use two point form y-y(1)=(y(2)-y(1) / x(2)-x(1) ) ( x-x(1) )

for perpendicular lines m(1)*m(2)= -1 , m(2)= -2/3
use it to find slope then equation is y-y(1)=m(1)(x-x(1) )

Answer 2

a)put the x value and the y value of the 2 points(7;-1) and (3;-2) into the general equation: y=ax+b; then solve the set of equation you've just made.=> you'll come to the line equation: y=(2/3)x-(2+3/4).
b) set the unknown equation as (1);y=ax+b. (1) perpendicular to line 1=(-2/3)x+10/3, then a*(-2/3)=-1.=>a=3/2.
now put x,y value from the point(3;-4) into (1).
=> -4=3a+b. We've already have a=3/2
So -4=3*3/2+b
=> b= -9/2 -4+-17/2. There we've the equation of the line:
y=3/2x-17/2.

As this is done manually by myself with no help from the calculator, it may not be accurate.

Answer 3

First case:

You can use the two given points with the next equation:

y - yo = m(x - xo) where m = slope of the straight line.

1.- Let´s compute the slope:

m = (y1 - y2) / (x1 - x2)
m = (-4 -(-1)) / (3 - 7) = (-3 / - 4 ) = 3/4

2.- Substitute one of the points in the given equation:

(xo,yo) = (3,-2)
y - (-2) = (3/4) (x - 3)
y + 2 = (3/4) (x - 3)
y = (3/4)x -17/4.......That´s the requested equation!

Second case:

1.- To find the equation of a perpendicular line to another we need just to know that:

m1*m2 = -1 (product of slopes is equal to -1)
In this example, m1 = -2/3 so, m2 = 3/2 (this is the slope of the perpendicular line)

2.- Knowing that perpendicular psses by point (3,-4) we apply again the model we saw in first case:

y - yo = m(x - xo) in this case, (xo,yo) = (3,-4)
y -(-4) = (3/4) (x - 3)

y = (3/4) x + 7/4.......That's is the requested equation!

Good luck!

Answer 4

The vector equation of a straight line is

X = P + k V

X = (x,y) k real, V vector with direction of the line, P point in the line

vector V could be A-B where A and B are points in the line

A(7,-1) B(3,-2) V = B-A = (-4, -1)

Point P could be A(7,-1)

(x,y) = (7,-1) + k (-4, -1) => x= 7-4k and y= -1-k

eliminating k:

- (x-7) = -4 (y+1) thats the equation
-----------------------------------------------------------------------------------
Slope of the original line -2/3
Slope of a perpendicular line 3/2

Equation y = 3/2 x + B

this equation must be true for x = 3, y = -4 and you could find B

Answer 5

the equation can be found using the two point form of equation: or y-k =((q-k)/(p-h))(x-h) where h,k is one given point and p,q is the second given point:
that is; h=7, k=-1, p=3 and q= -2

therefore: y - (-1) =((-2 -(-1))/(3-7))(x-7)
y + 1 = 1/4(x-7)
4y + 4 = x-7
4y = x-11
y = .25x -2.75



first find the slope of the given line. since the equation is in the form y = mx +b, the slope is the coefficinent of x. that is the slope is -2/3. since the required line is perpendicualr ro the given line, its slope must be the negative reciprocal of the given slope or 3/2. also the coordinates of the given point (x1,y1) are 3,4

the slope of the line is ( y- y1)/(x- x1) = slope or
(y-4)/(x-3) = 3/2
simplifying: 2y -8 =3x -9
2y = 3x -1
y = 3/2x -1/2

Answer 6

(x-7)/(3-7) = (y+1)/ (-2+1)

=> (x-7)/ (-4) = (y+1)/ (-1)

=> (x-7) = 4 (y+1)

=> (x-7) = 4y+4

=> x-4y-11=0

2.

slope of y=-(2/3)x+10/3 is -2/3

the slope of line perpendicular to that line is 3/2

the passes from (3,-4)

so (y+4) = 3/2(x-3)

=> 2y+8 = 3x- 9

=> 3x-2y-17 =0

Answer 7

1)The slope formula:
m=(y2-y1)/(x2-x1)
m=(-2+1)/(3-7)
m=-1/-4=1/4

Point-slope form:
y-y1=m(x-x1)
y+1=1/4(x-7)
y+1=1/4x-7/4
y=1/4x-7/4-4/4
y=1/4x-11/4

2)The reciprocal of the slope -2/3 would be 3/2.

Use the point-slope form:
y+4=3/2(x-3)
y+4=3/2x-9/2
y=3/2x-9/2-8/2
y=3/2x-17/2

I hope this helps!