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A is obviously not on the line. -5 is the y intercept, thus if A is on the line, the coordinates would be (0, -5)To find out if B is on the line, substitute -5 as c. Then substitute x and y with the coordinates in B. If x and y is on the line, c=5 would be found.Grp 3
We can substitute the y and x in y=3x-5 with the x and y points in A and B. So these are the workings:A:y=3x-5-5=(3x2)-5-5=6-5-5=1Since -5 does not equal to 1, then point A is definitely not on the straight line.B:y=3x-52=(5x3)-52=15-52=10Since 2 does not equal to 10, then B is not on the line.Group 1
If A and B both lie on the straight line, all we need to do is to calculate the gradient using (2, -5) and (5, 2) to see if the result agrees with the equation.m = (-5-2)/(2-5) = 7/3The calculated gradient with the 2 points A and B is 7/3. But, the equation states that the gradient is 3, so the 2 points are not on the line.
To find out is A and B is on the line, we can just substitute in the coordinates into the equation.For Point A, the coordinates are (2,-5)So the equation is y=3x-5,therefore the equation will become: -5=3(2)-5-5=6-5-5≠1We can easily conclude that A does not lie on the graph.For B, we can also use the same method. B's coordinates are (5, -2)So the equation will become -2=3(5)-5-2=15-5-2≠10Again, we also can easily conclude that B is also not on the line.Group 4
How would you determine if the points A (2, -5) and B (5, 2) lie on the straight line y = 3x - 5?We can substitute the coordinates into the equation. In A, we can the substituted equation is -5=3(2)-5. The equation obviously does not match. Both sides of the equation are unequal. Thus, A is not the point on the line.Next, for B, the equation will be 2=3(5)-5 which is 2=10. The equation is invalid once again, and we can conclude that B is not a point on the graph.Group 2
we can subsitute the actual numbers to the equations it would look like "-5 = 6 - 5" and this is obviously wrong, hence A does not lie on the linefor B it will look like "2 = 15 - 5" which is once again wrong since it does not add up, hence B does not lie on the line group 1
In order to solve this, we just need to substitute the points into the equation… If the points satisfy the equation, the the points do sit on the straight line, y=3x-5.So here is my attempt to solve this question:For ordered pair (2,-5),y=3(2)-5 =6-5 =-1,Therefore, since (y) is (-1) and not (-5), this ordered pair does not satisfy the equation y=3x-5 and hence, this coordinate does not lie on the equation of the straight line.For ordered pair (5,2),y=3(5)-5 =15-5 =10,Therefore since (y) is (10) and not (2), this ordered pair does not satisfy the equation y=3x-5 and hence, this coordinate does not lie on the equation of the straight line.
Oh… and also I am from Group 2.
We can substitute y=mx+c with y=3x-5 If both points A and B lie on the same line, we can just find the gradient of a and b: m= (-5-2)/(2-5)= 2 1/3The calculated gradient with the 2 points A and B is 2 1/3. But the gradient stated in the question is 3 thus, point A and B are not on the straight line.
Group 2(forgot to mention it)
How would you determine if the points A (2, -5) and B (5, 2) lie on the straight line y = 3x - 5?Using y=mx +c ,m = gradient= y1-y2------------- x1-x2= -5 - 2------------ 2 - 5= -7------------- -3= 2.33 (3 sf)y= mx+cy = 2.33 x + -5Thus , points A and B are not on the same line.Group 4
A) y=mx+cIf (x,y) = (2,-5)-5=3(2)-5=1Since the answer is not 0 the coordinates do not lie on the line.B) if (x,y) = (5,2)then 2=3(5)-5=10 so it is also not on the line as any point that lies on the line must have a final answer of 0 after being simplified.Group:4
POINT A =(2,-5)So y=3x-5,thus -5=3(2)-5 -5=6-5 -5≠1SO obviously Point A is not on the LinePoint B(5, -2)So -2=3(5)-5 -2=15-5 -2≠10SO Point B is also not on the lineGrp 3 :D
Point A=(2,-5)Substitute x & y coordinates with x & y in y=3x-5Thus, y=3x-5=-5=6-5-5=1 which is wrong, so A is not on the linePoint B can also be done with this methody=3x-52=15-52=10 which is again wrongBoth points A and B are not on the line y=3x-5Group 1
In order for both points A and B to line on a straight line, their gradient* has to be the same as the given line. Hence, we calculate the gradient: (2 (-5)) / (5 - 2) = 7/3 As you can see, the gradient of the given line and *[the gradient of the line made when points A and B are joined] are not the same. Thus points A and B do not lie on the line.Group 3