The line l1 has equation y = −2x + 3. The line l2 is perpendicular to l1 and passes through the point (5, 6). (a) Find an equation for l2 in the form ax + by + c = 0, where a, b and c are integers.
The first thing to look at is l2 and l1 being perpendicular. This means the gradients of the two lines multiplied together = -1 . To determine the gradient a student could differentiate l1 but a slightly quicker way is using just using y = mx+c , spotting -2 is equivalent to m which is also equivalent to the gradient. Using l1's gradient and the fact the two lines are perpendicular l2 can be calculated to equal 0.5. This can then be placed into y = mx+c and then to find out c the point (5,6) will be subbed in. the final equation for l2 is 2y -x - 7 = 0
RP
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