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  • A line L is passes through the points (1,1) and (2,0) find the equation of another line which is passing through (1,0) and perpendicular to line L.Option: 1</str

A line L is passes through the points (1,1) and (2,0) find the equation of another line which is passing through (1,0) and perpendicular to line L.

Option: 1

x-y+1=0


Option: 2

x+y-1=0


Option: 3

x+y+1=0


Option: 4

None of these


Answers (1)

best_answer

 

 

Foot of Perpendicular -

Foot of Perpendicular

 

P(x1, y1) is any point and M is the point of foot of perpendicular drawn from point P on the line AB: ax + by + c = 0. 

To find the coordinate of Point M, find the equation of PM which is perpendicular to the line AB: ax + by + c = 0  and passes through point P(x1, y1). 


 

OR

 

Foot of perpendicular of P(x1, y1) on the line AB : ax + by + c = 0 is  M is (x2, y2). then

Let the coordinate of foot of perpendicular be  M (x2, y2). Then, point M lies on the line AB.

\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;ax_2+b_2+c=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)}\\\mathrm{and, \;\;\;\;\;\because \;\; PM\perp AB}\\\text{then, (slope of PM)(slope of AB) = -1}\\\\\mathrm{\Rightarrow\;\;\;\;\;\;\;\;\;\;\; \left ( \frac{y_2-y_1}{x_2-x_1} \right )\left ( -\frac{a}{b} \right )=-1}\\\\\mathrm{or\;\;\;\;\;\;\;\;\;\;\;\;\;bx_2-ay_2=bx_1-ay_1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)}\\\\\text{After solving (i) and (ii) we get M} (x_2,y_2)
 

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Equation of Straight Lines Passing Through a Given Point and Making a given Angle with a Given Line -

Equation of Straight Lines Passing Through a Given Point and Making a given Angle with a Given Line

The equation of the straight lines which pass through a given point (x1 , y1) and make an angle α with the given straight line y = mx + c are

y-y_{1}=\tan (\theta \pm \alpha)\left(x-x_{1}\right)

Where, m = tan ?

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\\\text{Slope of Line L =}\frac{0-1}{2-1}=-1\\\\ \text{so slope of new line is }(m)=1\\ \text{Equation of new line is }\\ y-0=1(x-1)\\ x-y-1=0

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chirag

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