Points (6, -2) and (a, 6) are on the line with a slope of 4/3. What is the value of a ?

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(6,-2) and (a,6) are on the line with a slope 4/3.

The slope m of the line joining (x1,x2) and (y1, y2) is

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

Therefore slope of the line joining (6,2) and (a,2) is

m = (6- -2)/(a-6) equals 4/3 aas given.

8/(a-6) = 4/3

8*3 = 4(a-6)

24 = 4a-24

24+24 = 4a

4a = 48

a = 48/4 = 12

given, A(6,-2) and (a, 6) where the slope m = 4/3

We know that the fomula for the slope is:

m = (yB-yA)/(xB-xA)

==> 4/3 = (6--2)/(a-6)

==> 4/3 = 8/ (a-6)

==> coross multiply:

==> 3*8 = 4*(a-6)

Divide by 4:

==> 3*2 = a-6

==> 6 = a- 6

**==> a= 12**

Let us take the equation of the line as y= mx+ b, where m is the slope of the line and b is the y- intercept.

Now the slope of the line is 4/3. Therefore it can be written as y = (4/3)*x+ b

Point (6, -2) lies on the line, so -2 = (4/3)*6+ b

=> -2 = 24/3 + b

=> -2 = 8 + b

=> b = -10

So the equation of the line is y = (4/3)x -10

Now the point (a, 6) lies on the line, therefore 6= (4/3)a -10

=> 16 = (4/3)a

=> a= (16/4)*3

=> a = 12

**Therefore the value of a is 12.**

(6;-2) (a;6)

we have m=(x2-x1)/(y2-y1)=4/3

(a-6)/(6+2) =4/3

(a-6)/8=4/3

3(a-6)=32

3a-18=32 3a=50 a=50/3

We'll write the equation of the line in the standard form:

y = mx + n, where m is the slope and n is y intercept.

We know, from enunciation, that the line has the slope m= 4/3. We'll substitute the value of the slope in the equation of the line.

y = 4x/3 + n

The point (6, -2) is located on the line so it's coordinates verify the equation of the line:

-2 = 4*6/3 + n

-2 = 8 + n

We'll subtract 8 both sides:

n = -2 - 8

n = -10

The equation of the line is:

y = 4x/3 - 10

The point (a, 6) belongs to the same line also. So, it's coordinates belong to the line.

6 = 4a/3 - 10

We'll add 10 both sides:

4a/3 = 16

4a = 3*16

We'll divide by 4 both sides:

a = 3*4

**a = 12**

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