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499 Author(s): H. Freeman Source: The Mathematical Gazette, Vol. 8, No. 126 (Dec., 1916), p. 336Published by: Mathematical AssociationStable URL: http://www.jstor.org/stable/3602788Accessed: 29-12-2015 23:07 UTC

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336 336 THE MATHEMATICAL GAZETTE. THE MATHEMATICAL GAZETTE.

498. [K1. 6. a.] ilote on the Area of a Triangle in Plane Co-ordinate Geometry.

The determinant form of the expression for the area of a plane triangle in Cartesian rectangular co-ordinates is well known. The object of this note is to show that when the vertices of the triangle are given in polar co-ordinates, the area can also be expressed in the form of a determinant.

As given in the text-books, the area is r1r2 sin (1 - 02), i.e. 1r2r3^ {sin (01-02)/r3} or rlr2r3 rl-', sin 0, cos 0l,

where for reasons of space the first row of the determinant alone is given, both here and after.

It follows that the condition for three collinear points is that the above determinant vanishes.

In proving the property of Simson's Line (cf. Smith's Conic Sections, 1910, p. 97: ex. 4), the three points whose collinearity is to be shown have co- ordinates (2acos a cos /, a+/3), etc. These points can be shown to be collinear without actually finding the equation of the line. Thus, we have for the determinant

[ 1/(2a cos a cos /), sin(a + /), cos(a + /) , 1/2allcosax cosy, sin(a+ /), cos(a + ]) .

To facilitate reduction, let a+/3=x, 3+y =y, y +a=z, a + + y= u. Then the determinant becomes

cos (- x), sinx, cos x = 2 cos (u - x) sin (y - z)= cos y sin (/3 - .) = 0.

The condition derived above for the collinearity of three points can also be obtained from the polar equation of the straight line. Thus the line joining the points (r1, 01) and (s-2, 02) is

7Err,siin(0- -)=0 or 1r-1, cos 0, sinll =0, where r, 0, are current co-ordinates. If this line pass througli a third point (r3, 03), we have r1-l, sin 01, cos 01 =0.

New York City, U.S.A., CHARLES N. SCHMALL, B.A. July 15, 1915.

499. [A. 2. b.] Solve the equation 1 + x=7(1 + )4. Put 1 =r cos 4,

x=r sin ~, so that 1 + x2=r

x = r2 Cos 4 sin l.

Equation is r4 cos4 + + r4 sin4 = 7 (r cos q5 + r sin 4)4,

leading to 30 cos2 sin2 < + 28 cos / sin q + 6 =0, and giving cos > sin - = - , - 1.

. = _- r2 or - r2, i.e. 1+2 = - 5x or - 3x.

Thus the four values of x are found. The method may be adopted to solve equations of the type

(X4 + y4)(x2 +y2) = a, X6+y = b6

by putting x=r cos (, y=r sin ,

and proceeding on the above lines. H. FREEMAN.

* This determinant was first given, as far as he is aware, in the writer's First Course in Analytical Geometry (Blackie & Son), p. 66, ? 40.

498. [K1. 6. a.] ilote on the Area of a Triangle in Plane Co-ordinate Geometry.

The determinant form of the expression for the area of a plane triangle in Cartesian rectangular co-ordinates is well known. The object of this note is to show that when the vertices of the triangle are given in polar co-ordinates, the area can also be expressed in the form of a determinant.

As given in the text-books, the area is r1r2 sin (1 - 02), i.e. 1r2r3^ {sin (01-02)/r3} or rlr2r3 rl-', sin 0, cos 0l,

where for reasons of space the first row of the determinant alone is given, both here and after.

It follows that the condition for three collinear points is that the above determinant vanishes.

In proving the property of Simson's Line (cf. Smith's Conic Sections, 1910, p. 97: ex. 4), the three points whose collinearity is to be shown have co- ordinates (2acos a cos /, a+/3), etc. These points can be shown to be collinear without actually finding the equation of the line. Thus, we have for the determinant

[ 1/(2a cos a cos /), sin(a + /), cos(a + /) , 1/2allcosax cosy, sin(a+ /), cos(a + ]) .

To facilitate reduction, let a+/3=x, 3+y =y, y +a=z, a + + y= u. Then the determinant becomes

cos (- x), sinx, cos x = 2 cos (u - x) sin (y - z)= cos y sin (/3 - .) = 0.

The condition derived above for the collinearity of three points can also be obtained from the polar equation of the straight line. Thus the line joining the points (r1, 01) and (s-2, 02) is

7Err,siin(0- -)=0 or 1r-1, cos 0, sinll =0, where r, 0, are current co-ordinates. If this line pass througli a third point (r3, 03), we have r1-l, sin 01, cos 01 =0.

New York City, U.S.A., CHARLES N. SCHMALL, B.A. July 15, 1915.

499. [A. 2. b.] Solve the equation 1 + x=7(1 + )4. Put 1 =r cos 4,

x=r sin ~, so that 1 + x2=r

x = r2 Cos 4 sin l.

Equation is r4 cos4 + + r4 sin4 = 7 (r cos q5 + r sin 4)4,

leading to 30 cos2 sin2 < + 28 cos / sin q + 6 =0, and giving cos > sin - = - , - 1.

. = _- r2 or - r2, i.e. 1+2 = - 5x or - 3x.

Thus the four values of x are found. The method may be adopted to solve equations of the type

(X4 + y4)(x2 +y2) = a, X6+y = b6

by putting x=r cos (, y=r sin ,

and proceeding on the above lines. H. FREEMAN.

* This determinant was first given, as far as he is aware, in the writer's First Course in Analytical Geometry (Blackie & Son), p. 66, ? 40.

This content downloaded from 128.114.34.22 on Tue, 29 Dec 2015 23:07:05 UTCAll use subject to JSTOR Terms and Conditions