Page 145 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes 3 1
Now cos 15° =
2 2
3 1
and sin 15° = (Why?)
2 2
By the normal form (6) above, the equation of the line is

3 1 3 1
x cos15 y sin15 4 or x y 4 or 3 1 x 3 1 y 8 2.
2 2 2 2
This is the required equation.

Example: The Fahrenheit temperature F and absolute temperature K satisfy a linear
equation. Given that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F
and find the value of F, when K = 0.
Solution:
Assuming F along x-axis and K along y-axis, we have two points (32, 273) and (212, 373) in
XY-plane. By two-point form, the point (F, K) satisfies the equation
373 273 100
K 273 = (F 32) or K 273 (F 32)
212 32 180

5
or K = (F 32) 273 …(1)
9

which is the required relation.

When K = 0, Equation (1) gives
5 273 9
0 (F 32) 273 or F 32 491.4 or F 459.4
9 5
Alternate method: We know that simplest form of the equation of a line is y = mx + c. Again
assuming F along x-axis and K along y-axis, we can take equation in the form
K = mF + c ... (1)
Equation (1) is satisfied by (32, 273) and (212, 373). Therefore

273 = 32m + c ... (2)
and 373 = 212m + c ... (3)
Solving (2) and (3), we get

5 2297
m = and c .
9 9
Putting the values of m and c in (1), we get

5 2297
K = F … (4)
9 9
which is the required relation. When K = 0, (4) gives F = – 459.4.

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