Let AB be the given straight line; C the given point; h, k its co-ordinates; B the given angle. Let the equation to AB be

y = mx + c.

Suppose CD and CE the two lines which can be drawn through C, each making an angle ẞ with AB. Then

46. The following particular cases of the preceding results may be noted.

(1) Suppose m = 0; then the given line is parallel to the axis of x. The required equations then are

(2) Suppose m = mo; then the given line is parallel to the

axis of y.

And since

1

we have when m∞, and therefore =0, for the equation

m

(3) Suppose mtan B. In this case the equation to CD

becomes

(y-k) (1-m tan B) = (m + tan B) (x − h),

and we see that when m = cot ẞ the left-hand side is zero; thus the required equation is then

(5) Suppose m=-tan B. Then the equation to CD

becomes

y-k=0,

(6) Suppose m-cot B. Then the equation to CD

The equation to CE may be written in the form

(y −k) (1 +m tan ẞ) = (m — tan B) (x − h),

and we see that when m— cot B the left-hand member is zero; thus the required equation is then

we have cot ẞ=0; thus the equation becomes

Similarly the equation to CE takes the same form; and thus the result agrees with that of Art. 44.

We have discussed these particular cases as an example of the manner in which the student should test his comprehension of the subject by applying the general formulæ to special examples. He will find it useful to illustrate these cases by figures.

47. To find the length of the perpendicular drawn from a given point upon a given straight line.

Let AB be the given straight line; D the given point; h, k its co-ordinates. Let the equation to AB be

The equation to the line through D perpendicular to AB is, by Art. 44,

Let x, y, be the co-ordinates of E; then, by Art. 9,

DE2 = (x, − h)2 + (y1 −k)2 .

(3).

It remains then to substitute for X1 and Y1 their values in (3). Now, since x, y, are the co-ordinates of E, which is the point where (1) and (2) meet, we have

The radical in the denominator may be taken with the positive or negative sign, according as the numerator is positive or negative, so as to give for DE a positive value.

We may also obtain the value of DE thus; draw the ordinate DM meeting the line AB in H; then

Now

OM=h; .. HM = mh+c, and DM = k;

Hence if on the line y-mx-c=0 a perpendicular be drawn from the point (h, k) and also a perpendicular from the point (h, k), the ratio of the first perpendicular to the second is equal to the numerical ratio of k-mh, — c to k2 — mh, — c.

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