How do you solve these maths questions ?

Question

Find the equation of the line though ( 0, -2) and (6, 1)

Find the angle of the line Y = 1/2x + 3

Can everyone please leave their working on how they solved these two questions cause they have really been bugging me. Thank You!:)

Answer 1

1)  The equation of any line is y=m*x+b where m is the slope and b is the y intercept.  So m= (y1-y2)/(x1-x2)= (-3)/(-6)=.5

y=.5x+b, now use a point to find b

-2=(.5)*0+b, b =-2, therefore y=.5x-2

2)  Like in question 1 y=mx+b, the slope is 1/2 or .5

So if you had a right triangle and you looked at the angle between the hypotenuse and the base leg the base leg would be 2 units long while the opposite leg would be 1 unit.  So tangent (theta) = (1/2)

theta then equals 26.565 degrees ot tan^(-1) (1/2)

Answer 2

1)    (0,-2)(6,1)=  (x1,y1)(x2,y2)

    M=y2-y1/x2-x1  =1-(-2)/6-0   1+2/6  3/6  M=gradient=0.5

  equation of a line=y-y1=M(x-x1) so  

     y-(-2)=0.5(x-0)  then y+2=0.5x-0  Y=0.5x-2.

2)Tantheta=m  so m=1/2,

    tan theta=0.5  theta=tan inverse of 0.5  theta=26.56degree.

Answer 3

1ST FIND THE SLOPE THAT IS

SLOPE=(Y2-Y1)/(X2-X1)

LETS DENOTE IT BY LETTER m

m=(1+2)/6=0.5

euatn of line=

(y-y1)=(x-x1)m

(y+2)=(x-0)0.5

y+2=0.5x

0.5x-y=2

tantheta=m

theta=tan inverse of 0.5=26.56degree