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Let AB (Fig. 159) be the plan, IL being the intersecting line as before. As the plane on which the line AB is to lie equally inclines to the vertical and horizontal, its angle with either will be one of 45. Therefore, at any point in the IL, say O, draw a line OP at that angle with it. Now OP, we have before said, is a side view of the inclined plane on which the object in this case a line is to lie ; the view we shall obtain of the object in elevation will be one looking in the direc- tion of the arrow x, and the position of the YP with respect to OP can be either in front of or behind it. If in front, it would be represented by a line drawn through O perpendicular to the IL ; if behind, then a similar line through P. Generally, OP is assumed to be interposed between the VP and HP, and therefore a side view of the three planes would be correctly represented by the right-angled triangle PxO in the diagram; the vertical projection or elevation of the inclined object being made on Px or the VP, and the horizontal one, or plan, on Ox, or the HP. We can now proceed with the solution of the problem before us. Having the inclination of the plane on which the line AB is lying, to find a front elevation of it we require to know the exact position of its two ends with respect to the foot of the plane on which it is lying. Now, the intersection of two planes is a straight line ; therefore, as the inclined plane OP in the figure intersects the HP in O, the plan of that intersection will be a straight line parallel to the VP or the IL ; and assuming the VP to be as far from the foot of the inclined plane OP as x is from O, set off from the IL a distance equal to Ox, and through the point so found draw the line DL parallel to the IL, and it will be the plan of the intersection of the HP with the inclined plane OP. It