MATHS

Asked by BellaTo find the equation of a line with the points (3,-3) and (-2,2) we first need the gradient of the line. This is found by doing y2-y1/x2-x1 (or y1-y2/x1-x2). So let's call (x1,y1)=(3,-3) and (x2,y2)= (-2,2). So the gradient is given by 2-(-3)/-2-3 = 5/-5 = -1. We know the equation of a straight line is y=mx+c and we have m=-1 so y=-x+c. We can see that we need to find the value of c. By substituting one pair of coordinates into our equation we have -3 = -1*3 +c -3 = -3 +c and rearranging gives c=0. so the final equation of the line is y=-x

To find an equation of a line you must know it’s gradient (how steep the line is) to do this you use the formula Y2-Y1/X2-X1 2—3/-2-3 or change in y divided by change in X so that is -1=gradient to find the y intercept you use the formula y-y1=m(x-x1) this might look confusing but carry on reading so all you do is substitute the values in the formula y1 coordinate= -3 X1 coordinates = 3 m= gradient which was figured out above = -1 substituting —> y-(-3)=-1(x-(3)) Y+3=-x+3 rearranging into y=mx+c Y=-x

Y=mx*c Sub gives -3=3m+c Also 2=-2m+c Subtract both equations You get 5=-5m So m=-1 Sub into either equation C=0 Therefore y=-x Is equation of the straight line which is very clear from the points coordinates!

To find the equation of a straight line, we need to find the following information: 1) The gradient (denoted by the symbol, m) 2) The y-intercept (denoted by the symbol, c) To find the gradient of the line, we need to use the following equation: m = (y2 - y1)/(x2 - x1) , where (x1,y1) and (x2,y2) are two given coordinates. Let (x1,y1) = (3,-3) and (x2,y2) = (-2,2). Note that: The values that are assigned to the given coordinates does not make a difference to the answer that be obtained for m. For instance, assigning (x1,y1) as (-2,2) and (x2,y2) as (3,-3) will not change the value of m. Once the values of x1, y1, x2 and y2 are chosen, keep them unchanged throughout the calculation; consistency is key. Now substituting in the values of x1= 3, y1= -3, x2= -2 and y2= 2 into the equation for m (the gradient) yields: m = (y2 - y1)/(x2 - x1) = (2 - -3)/(-2 - 3) = (2 + 3)/(-2 - 3) (- and - make a +) = 5/-5 = -1 Now considering the form of a straight line equation: y = mx + c We know that m = -1 including the coordinates of the two points. To find c (the y-intercept), we can substitute the value of m and any of the two coordinates (as they both lie on the graph of the straight line). Substituting in m = -1 and the coordinates (3,-3), we find the value of c to be: y = mx + c -3 = -1(3) + c -3 = -3 + c (Add ‘3’ to both sides.) -3 + 3 = -3 + 3 + c 0 = 0 + c c = 0 Therefore, the equation of the line passing through points (3,-3) and (-2,2) is: y = (-1)x + 0 y = -x (This is the final answer.) I hope this helps.