The equation of the line passing through − 0.8 , 0.2 and 0.7 , 3.2 . The equation of line is x + 0.5 y + 0.9 = 0 . Calculation: Consider the points, x 1 , y 1 = − 0.8 , 0.2 , x 2 , y 2 = 0.7 , 3.2 Substitute all the values in the standard determinant x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 . This gives, x y 1 − 0.8 0.2 1 0.7 3.2 1 The determinant of the matrix along row 1 is, x y 1 − 0.8 0.2 1 0.7 3.2 1 = a 11 C 11 + a 12 C 12 + a 13 C 13 From the matrix a 11 = x , a 12 = y and a 13 = 1 . Substitute the value in the determinant, x y 1 − 0.8 0.2 1 0.7 3.2 1 = x . C 11 + y . C 12 + 1. C 13 The coefficient of the cofactors C 11 , C 12 and C 13 are given as, Compute the cofactor, C 11 as, C 11 = − 1 1 + 1 M 11 To compute M 11 , delete the first row and the first column, C 11 = − 1 1 + 1 0.2 1 3.2 1 = − 1 2 0.2 1 3.2 1 = 0.2 1 3.2 1 Now, expand using cofactors, 0.2 1 3.2 1 = 0.2 × 1 − 1 × 3.2 = 0.2 − 3.2 = − 3 To compute M 11 , delete the first row and the first column, C 12 = − 1 1 + 2 − 0.8 1 0.7 1 = − 1 3 − 0.8 1 0.7 1 = − 1 − 0.8 1 0.7 1 Now, expand using cofactors, − 0.8 1 0.7 1 = − 0.8 × 1 − 1 × 0.7 = − 0.8 − 0.7 = − 1.5 To compute M 11 , delete the first row and the first column, C 13 = − 1 1 + 3 − 0.8 0.2 0.7 3.2 = − 1 4 − 0.8 0.2 0.7 3.2 = − 0.8 0.2 0.7 3.2 Now, expand using cofactors, − 0.8 0.2 0.7 3.2 = − 0.8 × 3.2 − 0.2 × 0.7 = − 2.56 − 0.14 = − 2.7 Now, substitute the values in the determinant, x y 1 − 0.8 0.2 1 0.7 3.2 1 = x ⋅ − 3 + y ⋅ − 1.5 + 1 ⋅ − 2.7 0 = − 3 x − 1.5 y − 2.7 0 = x + 0.5 y + 0.9 The value of the determinant is 0 since the points are collinear. The equation of line is x + 0.5 y + 0.9 = 0 .

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Chapter 7, Problem 116RE

To determine

To calculate: The equation of the line passing through −0.8,0.2 and 0.7,3.2.

The equation of line is x+0.5y+0.9=0.

Calculation:

Consider the points,

x1,y1=−0.8,0.2,x2,y2=0.7,3.2

Substitute all the values in the standard determinant x1y11x2y21x3y31. This gives,

xy1−0.80.210.73.21

The determinant of the matrix along row 1 is,

xy1−0.80.210.73.21=a11C11+a12C12+a13C13

From the matrix a11=x,a12=y and a13=1.

Substitute the value in the determinant,

xy1−0.80.210.73.21=x.C11+y.C12+1.C13

The coefficient of the cofactors C11,C12 and C13 are given as,

Compute the cofactor, C11 as,

C11=−11+1M11

To compute M11, delete the first row and the first column,

C11=−11+10.213.21=−120.213.21=0.213.21

Now, expand using cofactors,

0.213.21=0.2×1−1×3.2=0.2−3.2=−3

To compute M11, delete the first row and the first column,

C12=−11+2−0.810.71=−13−0.810.71=−1−0.810.71

Now, expand using cofactors,

−0.810.71=−0.8×1−1×0.7=−0.8−0.7=−1.5

To compute M11, delete the first row and the first column,

C13=−11+3−0.80.20.73.2=−14−0.80.20.73.2=−0.80.20.73.2

Now, expand using cofactors,

−0.80.20.73.2=−0.8×3.2−0.2×0.7=−2.56−0.14=−2.7

Now, substitute the values in the determinant,

xy1−0.80.210.73.21=x⋅−3+y⋅−1.5+1⋅−2.70=−3x−1.5y−2.70=x+0.5y+0.9

The value of the determinant is 0 since the points are collinear.

The equation of line is x+0.5y+0.9=0.

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