Three triangles L1,L2,L3 are such that, L1 cuts the y-axis at y=5 and has a gradient of 2, L2 is perpendicular to L1 at the point where L1 cuts the x-axis, L3 is parallel to L2 and passes through point (1,2). (a) Find the equation in the form y=mx+c of, (i) L1 (ii) L2 (iii) L3 (b) Determine the coordinates of the point at which L3 is perpendicular to L1

(a) (i) The equation of a line in the form y = mx + c is determined by the gradient (m) and the y-intercept (c). For line L1, we are given that the gradient is 2 and it cuts the y-axis at y = 5. Therefore, the equation of L1 is y = 2x + 5.

(ii) Line L2 is perpendicular to L1 and it intersects L1 at the point where L1 cuts the x-axis. The gradient of a line perpendicular to another is the negative reciprocal of the original line's gradient. Therefore, the gradient of L2 is -1/2. To find the y-intercept, we need to find the x-coordinate where L1 cuts the x-axis. This is found by setting y = 0 in the equation of L1 and solving for x, which gives x = -5/2. Substituting this into the equation of L2 gives the y-intercept as 0. Therefore, the equation of L2 is y = -1/2x.

(iii) Line L3 is parallel to L2 and passes through the point (1,2). Lines that are parallel have the same gradient, so the gradient of L3 is also -1/2. To find the y-intercept, we substitute the coordinates of the point (1,2) into the equation y = mx + c and solve for c, which gives c = 2 + 1/2 = 5/2. Therefore, the equation of L3 is y = -1/2x + 5/2.

(b) The point at which L3 is perpendicular to L1 is found by setting the equations of L1 and L3 equal to each other and solving for x. This gives 2x + 5 = -1/2x + 5/2, which simplifies to 5/2x = -9/2, and x = -9/5. Substituting this into the equation of L1 gives y = 2*(-9/5) + 5 = -18/5 + 25/5 = 7/5. Therefore, the coordinates of the point at which L3 is perpendicular to L1 are (-9/5, 7/5).