# A is the point (2,-5) B is the point (4,-9) 1a. Show that the gradient of the straight line passing through a and b is -2 1b. C is the point (-305, 601) does c lie on a straight line passing through a and b? You must show your working?

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### Sajeel Qureshi

Skilled tutor and undergraduate at University of Liverpool

Gradient for a straight line is change in y/ change in x. So we get the difference in x values which is 2 - 4 =-2. We get change in y from -5 - (-9) = 4. So gradient is 4/-2 = -2 For 1b, using formula y=mx + c we work out equation of the line through A and B. M is the gradient. We sub in values of one of the known points. ( I’m gonna use A). So -5 = (-2)2 + c. If we rearrange we get c as -1. So equation of the line is y=-2x -1. We sub in x and y values of C and we check both sides of equal sign are equal. So we get 601=- -2(-305)-1. If we simply the rhs we get 601= 609 which isn’t true so answer is no c is not on line. P.S I rushed this in my head whilst on the bus so I could have made a mistake. But method should be right

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