If we are given a point P(x₀,y₀) and the gradient m of a line, its "point gradient" equation is:  y = y₀ + m(x - x₀) .

This is shown in the next figure. Also the line passing through the point P(-2,1) with a gradient of m = -3/4 = -0.75 is calculated.

Using the buttons at the bottom of the figure you can vary the coordinates of the point P and the gradient m, to see how the graph and the equation of the line vary. This also appears in implicit form.

1.-Substitute the coordinates of P and the value of m into the POINT-GRADIENT equation, and find the implicit equation of the line. 

2.- In your exercise book write the POINT-GRADIENT equation of the line which passes through the point P(4,3) and has a gradient of m=1.8 

3.- Calculate its implicit equation. 

4.- Verify it in the figure.

1.- First you have to calculate its gradient, (you already know how) from two known points. 

2.- Apply the POINT-GRADIENT equation taking any of the known points, for example A (the final result will be the same if you take B). 

3.- Find the implicit equation. 

4.- Check it taking the point B instead of A, the implicit equation is the same.

5.- Calculate the equation of the line which passes through the points (-4,5) and (4,2) .

7.- Now the points are (3,8) and (3,-2). In this case x is always equal to 3, so the equation of the line is x=3. It is a line parallel to the y-axis. It has no gradient. It cannot be put into explicit form because there would be a division by zero. Try it in the figure. (Attention! It is not correct to say that the gradient is infinity). 

1.-Write down the equations of the lines represented in the diagram.

2.- To test the results, use this figure. The line y = 0 is drawn, that is the line with a gradient of m=0 and cutting at the origin where n=0. 

3.- When you give m and n the values obtained for the first line, this will be drawn. But the line y=0 will remain. 

You can draw all three lines required, without erasing one when another is introduced.