You are given the cordinates of the four points

A (6,2), B(2.4), C (-6, -2) and D (-2,-4).

Show that the equation of the line DA is 4y-3x = -10

help please!!

sub in the points A(6,2) and D(-2,-4)

into 4y - 3x = -10

for (6,2)
LS = 4(2) - 3(6) = -10 = RS
for (-2,-4)
LS = 4(-4) - 3(-2) = -10 = RS

YES , the equation of DA is 4y-3x = -10

Where do the other two points enter the picture ?

Could you please explain that in full, as im really struggling to understand. The other two points were for calculating gradients AB, CB, DC and DA but i got that.

Thanks reiny!

Is there a way of simplifying the equation 4y-3x=-10 to y=0.75x - 2.5, then proving that is the equation for DA?

I don't know how much further I can explain it.

Since both points satisfy the given equation, the equation must be the correct one
All I did was replace the x and y in the equation with the x and y of each of the points
LS stands for left side of the equation
RS stands for the right side.

in
4y - 3x = -10
4y = 3x - 10
y = (3/4)x - 10/4
or
y = .75x - 2.5

The last equation and the starting equation are really the same thing.
Subbing in either one gives you the same result

e.g.
sub in (6,2) into y = .75x - 2.5
LS = 2
RS = .75(6) - 2.5 = 4.5 - 2.5 = 2
since LS = RS , the point lies on the equation.

I will let you sub in the second point in the same way.
Let me know if you did NOT get it.

What is RS and LS?

Did you not read my reply ? I explained what they are.

Im sorry!!! Its so late and im half asleep. Worse time to do maths. I get it reiny!!!!!!!

Thank you
So the second point is -4 = -4
therefore it must be the correct equation as both points satisfy the equation?

You got it.

Your a star!!!!!!!!!!!!!

Many thanks

To show that the equation of the line DA is 4y-3x = -10, we can use the slope-intercept form of a linear equation. The slope-intercept form is given by:

y = mx + b

where m represents the slope of the line and b represents the y-intercept.

To find the slope of the line DA, we can use the formula:

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

where (x1, y1) and (x2, y2) are any two points on the line.

Using points D(-2, -4) and A(6, 2), we can calculate the slope:

m = (2 - (-4)) / (6 - (-2))
= 6 / 8
= 3 / 4

Now that we have the slope, we can substitute it into the slope-intercept form to find the equation of the line. Using the point A(6, 2), we have:

y = (3/4)x + b

Substituting the coordinates of point A into the equation, we can solve for b:

2 = (3/4)(6) + b
2 = 18/4 + b
2 = 9/2 + b
2 - 9/2 = b
4/2 - 9/2 = b
-5/2 = b

Therefore, the equation of the line DA is:

y = (3/4)x - 5/2

To convert this equation to the standard form (Ax + By = C), we can multiply both sides by 4 to eliminate the fraction:

4y = 3x - 10

Finally, rearranging the equation, we have the equation of line DA as:

4y - 3x = -10