Because the straight line y kx b intersects the y axis at 3,0 then why ykx 2b intersects y axis at 6

y=kx+b (3,0)

Bring in 0=3k+b b=-3k

Since y=kx+b y=kx+2b is the same k, y=kx+2b=kx-2*3k=k(x-6) brings x=6 into y=0

So y=kx+2b intersects the y-axis (6,0).

Since the slope is all k, the intercept is the value of y at x=0.

Last intersect b(0,3).

The next 2b is not (0,6).

Let's draw a picture to understand better.

1）n=km+b, q=kp+b

n+q=k(m+p)+2b=2k+2b=2b+2b+4, get: k=b +2

Hence y=(b +2)x+b

Therefore y is a monotonic increasing function, so n2) a straight line parallel to the x-axis is y=n

The straight line parallel to the y-axis through b is x=p

If the intersection point is c(1,1), then there is n=1, p=1

a(m,1), b(1,q)

bc=|q-1|=4, gets: q=5 or -3, because k<0, so qac=|m-1|, ab=√[(m-1)²+q-1)²]=√[(m-1)²+16]

ABC Perimeter = 12 = 4 + |m-1|+ m-1) +16] shift, squared: 8 +(m-1) -16|m-1|=(m-1) +16 to get |m-1|=3

i.e. m = 4, or -2

Because m is therefore a(-2,1), b(1,-3).

k=(-3-1)/(1+2)=-4/3

The straight line is y=-4 3(x+2)+1=-4x3-5 3, i.e., b=-5, 3

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Summary. From the vector cross product formula, the area of the triangle obc is: $s = frac |\overrightarrow \times \overrightarrow|$ substituting the coordinates of the vectors ob and oc into the above equation yields:

s_=\frac |4 \times 10 - 0 \times m|=20$ Therefore, the area of δobc is 20.

Okay, hello, can you send this a** shot over?

Okay, thank you, dear.

According to the information given in the question, it can be known that the coordinates of point A are (4, 0) because the line intersects the x-axis at point A.

1) Since the line intersects the y-axis at the point b, the value of b is 4. Therefore, the analytic formula for a straight line is y=-2x+4.

2) Point o is the origin of the coordinate system, and the coordinates of point o are (0, 0). The coordinates of the shirt at point B are (4, 0) and the coordinates of point C are (m, 10). Thus, the coordinates of the vector ob are (4, 0) and the coordinates of the vector oc are (m, 10).

Isn't the second question a place to find area?

From the vector cross product formula, the quorum area of the triangle OBC is: $s = frac |\overrightarrow \times \overrightarrow|$ substituting the vectors ob and oc into the above equation yields: $s = frac |4 \times 10 - 0 \times m|=20$ Therefore, the area of δobc is changed to 20.

Thank you, my question.

Jiaozhong Liang dot finch is sold on the y-axis.

When x=0, y=2x+3=3

That is, the intersection point sits on the sale of the year's fortune mark is (0,3).

3=0+b, i.e. b=3

If the intersection of the line y=2x+b and the y-axis is below the x-axis, then b(<)0 and the coordinates of the intersection are (0,b).

In the sail straight line y= 3x+6, let x=0,y=6, let y=0,x=-2 3,a(-2 3,group good 0),b(0,6), let the tassel hail ma=mb=r, then om=6-r, in rtδoam, ma2=oa 2+om 2, r 2=12+(6-r) 2, 12r=48, r=4, om=6-4=2, m(0,2).