On the portion of the straight line,x+2y =4 intercepted between the axes , a square a constructed on the side of the line away from the origin. Then point
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x + 2 y = 4
X- intercept at y=0, ie, at A(4,0)
Y-intercept at x=0, ie., at B(0,2)
Length of line segment AB = √(2²+4²) = 2√5
Let the square formed be ABCD.
Area of the square ABCD on AB : 20.
Slope of AB = - 1/2
Slope of BC = 2 = tanФ => SinФ = 2/√5
and Length of BC = AB = 2√5
Ordinate at C : BC sinФ = 4
Abscissa at C: BC CosФ + 4 = 2/5 + 4 = 22/5
So point C = (22/5 , 4)
Similarly, Point D = (BC cosФ, 2 + BCsinФ)
= (2/5 , 4)
X- intercept at y=0, ie, at A(4,0)
Y-intercept at x=0, ie., at B(0,2)
Length of line segment AB = √(2²+4²) = 2√5
Let the square formed be ABCD.
Area of the square ABCD on AB : 20.
Slope of AB = - 1/2
Slope of BC = 2 = tanФ => SinФ = 2/√5
and Length of BC = AB = 2√5
Ordinate at C : BC sinФ = 4
Abscissa at C: BC CosФ + 4 = 2/5 + 4 = 22/5
So point C = (22/5 , 4)
Similarly, Point D = (BC cosФ, 2 + BCsinФ)
= (2/5 , 4)
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