A is the point with coordinates (5, 9) B is the point with coordinates (d, 15) The gradient of the line AB is 3 Work out the value of d.
A straight line equation is defined as y=mx+c where m is the gradient and c is the intercept. Since the gradient is already said to be 3 we can substitute this in to mean the equation for AB is y=3x+c. Since we know point A we can substitute the values for x and y in to solve for c.9=3(5)+c9=15+cc= -6y=3x-6Now we have the equation of the line we can sub in point B to find what d is.15=3(d)-621=3dd=7
TA
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