Question

A triangle is formed by the straight line AB passing through (1, 1) and (2, 0), the straight line CD, perpendicular
to the line AB and passing through (1/2, 0) and the y-axis. Area of the triangle thus formed is:

a) 5/4
b)25/16
c) 4/5
d)16/25​

Answer 1

Area of the triangle thus formed = 25/16 is the straight line AB passing through (1, 1) and (2, 0)& CD ⊥ AB  passing through (1/2, 0) and the y-axis

Step-by-step explanation:

AB passing through (1, 1) and (2, 0),

=> slope =  (0-1)/(2 - 1)  = -1/1 = -1

line y = -x + c

0 = - 2 + c

=> c = 2

=> y = -x + 2

=> x + y  = 2

CD Line Perpendicular  to AB

will have slope = 1

y = x + c

0 = 1/2 + c

=> c = -1/2

y = x - 1/2

=> 2y = 2x - 1

=> 2x - 2y = 1

2x - 2y = 1

x + y  = 2

intersection points

4x = 5

=> x = 5/4

y = 3/4

(5/4 , 3/4)

at y axis x = 0

y = 2  & -1/2

vertex of triangle

(0, 2)  , (0 , -1/2) , (5/4 , 3/4)

Area of triangle = (1/2) | 0(-1/2 - 3/4) + 0(3/4 - 2) + 5/4(2 -(-1/2)|

= (1/2) | 0 + 0  + 25/8|

= 25/16

Area of the triangle thus formed = 25/16

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Answer 2

Area of triangle=$\frac{25}{16}$ square units

Step-by-step explanation:

Slope formula:$m=\frac{y_2-y_1}{x_2-x_1}$

By using the slope formula  

Slope of AB passing through the points (1,1) and (2,0) is given by  

$m=\frac{0-1}{2-1}=-1$

Point- slope form:$y-y_1=m(x-x_1)$

By using point-slope form

Equation of line AB passing through the point (2,0) with slope -1 is given by  

$y-0=-1(x-2)=-x+2$

$x+y=2$...(1)

Slope of CD=$-\frac{1}{-1}=1$

When two lines are perpendicular then  

Slope of one line=$-\frac{1}{slope\;of\;other\;line}$

Equation of line CD passing through the point (1/2,0) with slope 1

$y-0=1(x-\frac{1}{2})$

$y=x-\frac{1}{2}$

$y=\frac{2x-1}{2}$

$2y=2x-1$

$2x-2y=1$...(2)

and y- axis  means x=0

Equation (1) multiply by 2 and then adding to equation (2)

$4x=5$

$x=\frac{5}{4}$

Substitute x=5/4 in equation(1)

$\frac{5}{4}+y=2$

$y=2-\frac{5}{4}=\frac{3}{4}$

Equation (1) and equation (2) are intersect to each other at point (5/4,3/4).

Substitute x=0 in equation(1)  

$y=2$

Substitute x=0 in equation (2)

$-2y=1$

$y=-\frac{1}{2}$

Equation (1) and y- axis intersect at point (0,2) and equation (2) and y-axis intersect at point (0,-1/2)

Therefore,the vertices of triangle are (0,2),(5/4,3/4) and (0,-1/2)

Area of triangle =

$\frac{1}{2}\mid {0(\frac{3}{4}+\frac{1}{2})+\frac{5}{4}(-\frac{1}{2}-2)+0(2-\frac{3}{4})}\mid$

By using the formula of area of triangle =$\frac{1}{2}\mid(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\mid$

Area of triangle =$\frac{1}{2}\mid{0-\frac{25}{8}+0}\mid$

Area of triangle=$\frac{25}{16}$ square units

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