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COR. The general equation of a straight line parallel to (l, m, n) is

la + mß+ny = k (aa+bB + cy),

where k is an arbitrary constant.

16. To find the inclinations of a straight line, drawn through one of the angular points of the triangle of reference, to the sides which intersect in that point.

Let the equation of the straight line AP be

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and let be its inclination to AD, the internal bisector of the

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Hence, the inclinations of the given straight line to AB, AC, are determined.

17.

To find the condition that two given straight lines may be perpendicular to one another.

Let (l, m, n), (l', m', n') be the two given straight lines. Through A draw two straight lines parallel to them. These will be represented by the equations

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And these straight lines must be at right angles to one another.

If 0, 0' be the respective inclinations of these straight lines to the internal bisector of the angle A, then, by the result of the last article,

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And, if these be at right angles to one another,

Hence

1+tan 0 tan e' = 0.

(lc — na) (l'c — n'a) + (ma — lb) (m'a — l'b)

+ {(lc — na) (m'a — l'b) + (ma — lb) (l'c — n'a)} cos A = 0 ;

.. ll' (b2 + c2 — 2bc cos A) + mm'a2+ nn'a2

− (mn' + m'n) a2 cos A― (nl' + n'l) (ac — ab cos A)

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− (lm' + l'm) (ab — ac cos A) = 0,

which, since b+c2-2bc cos Aa2, c-b cos A= a cos B, b-c cos A =a cos C, reduces to

ll'+mm'+nn'—(mn'+m'n)cosA—(nl'+n'l)cosB—(lm'+l'm)cosC=0,

the required condition.

18. To find the perpendicular distance from a given point to a given straight line.

Let (f, g, h) be the given point, (l, m, n) the given straight line, Then, if q and r be the distance from A, of the points where this straight line meets AC, AB, respectively, we have shewn (Art. 7) that

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Now, let a denote the distance from (f, g, h) to (l, m, n).

Then

(q2 + r2 — 2qr cos A)3 a' +qg+rh = L" (af +bg+ch),

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bc

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" (If+mg+nh).

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с

lbc

a (l'cos B+m cos A— n)

h

==

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=

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a

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a (l cos C+ n cos A - m)

α

lbc

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l2b2 c2 (lc — na) (lcosB+mcosA—n)

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{l (c cos B+b cos C) + m2a+n3a − 2mn a cos A

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- m)}

-nl (c + a cos B-b cos A) – lm (b− c cos A+ a cos C)},

which, by reduction, is equal to

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(12 + m2 + n2 − 2mn cos A - 2nl cos B-2lm cos C).

Hence

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(12 + m2 + n3 − 2mn cos A – 2nl cos B – 2lm cos C)13

the required expression.

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It will be observed, that the numerator of this expression vanishes if the point (f, g, h) lie upon the line (l, m, n), as manifestly ought to be the case.

It will also be remarked, that the more nearly the ratios 7: mn approach to the ratios a: bc, the less does the denominator of the above fraction become, and the greater, therefore, the distance from the point to the line; which is in accordance with the remark made in Art. (14).

EXAMPLES.

1. Find the equation of the straight line joining the middle points of two sides of the triangle of reference; and thence prove that it is parallel to the third side.

2. Find the equations of the straight lines, drawn through the several angular points of the triangle of reference, respectively at right angles to

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and thence prove that they intersect in a point.

3. If ◊ be the angle between the two straight lines (l, m, n) (λ, μ, v), prove that

cote =

lλ+mμ+nv− (mv+nμ) cos A − (nλ +lv) cos B − (lμ + mλ) cos C (mv – nu) sin 4 + (n) − k)sin B + (u – mà) sin C

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4. On the sides of the triangle ABC, as bases, are constructed three triangles A'BC, AB'C, ABC, similar to each other, and so placed that the angle BA'C = B'AC = BAC', CB'A = C′BA = CBA', AC'B=A'CB= ACB. Prove that the straight lines A A', BB, CC intersect in one point.

5. Prove that the straight line, joining the centre of the circle inscribed in the triangle ABC, with the middle point of the side BC, is parallel to the straight line joining A with the point of contact of the circle touching BC externally and AB, AC produced.

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6. On the sides BC, CA, AB of the triangle ABC, respectively, pairs of points are taken, B,, C1; С2, A ̧; Ã ̧, B; such that the points of intersection of BC with BC, of CA with C, 4, and of AB with A,B, lie in a straight line; BC, CB, intersect in L; CA, AC, in M; AB1, BA, in N. Prove that AL, BM, CN intersect in one point.

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7. From the vertices of a triangle ABC, three straight lines. AP, BQ, CR are drawn to pass through one point, and three straight lines AP', BQ', CR' to pass through another point, the points P, P lying on BC, Q, Q on CA, R, R on AB; BQ, CR meet AP' in D, D2; CR, AP meet BQ' in E,, Eg; AP, BQ meet CR in F, F; CD,, BD, intersect in L; AE, CE, in M; BF1, AF in N. Prove that AL, BM, CN intersect in a point.

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