The line l1 has equation 2x + 3y = 26 The line l2 passes through the origin O and is perpendicular to l1 (a) Find an equation for the line l2

First we make y the subject of the l1 equation. This is so we can have it in the form y=mx +c where m is the gradient and c is your y-intercept. For this example we would take away 2x from both sides leaving 3y=-2x +26. then dividing both sides by 3 gives us just y on its own: y=(-2/3)x+26/3. in order to find the gradient of a line we must differentiate the equation. we differentiate by multiplying anything in front of the x by the power of the x then taking away 1 from the power. so for (-2/3)x it would be just -2/3 and for 26/3 it would be 0 because we take x as having a power of 0 in front of 26/3. this gives a gradient of the line equal to -2/3.To find the perpendicular we can find the negative reciprocal of the gradient. The negative reciprocal is a number which when multiplied with the original number gives a result of -1. Essentially we are flipping the fraction and changing the sign. this becomes our new gradient and because the line goes through the origin the y intercept becomes 0 and so the equation of l2 is y=(3/2)x