Question

7. Find the equation of the straight line which passes through the midpoint of the line
segment joining (4, 2) and (3, 1) whose angle of inclination is 300

Answer 1

Finally, use the point-slope formula for the equation of a line through, say, ( x 3 , y 3 ) with slope : y − y 3 = m ( x − x 3 ) . So, for example, if you want the perpendicular bisector to the segment through P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) : The midpoint is ( x 1 + x 2 2 , y 1 + y 2 2 ) .

Answer 2

Solution :

Let A(4,2)= (x1,y1) and B(3,1)=(x2,y2)

i) P( x , y ) is the midpoint of AB.

coordinates of P=[(x1+x²)/2,(y1+y2)/2]

=[ (4+3)/2 , (2+1)/2 ]

= (7/2, 3/2 )

ii ) Equation of a line passing through

the point P( 7/2 , 3/2 ) = (x1,y1) and

whose inclination is x = 30°

y - y1 = m( x - x1 )

slope of a line ( m ) = tan x°

m = tan 30°

=> m = 1/√3

y - 3/2 = ( 1/√3 ) ( x - 7/2)

=> ( 2y - 3 )/2 = ( 1/√3)( 2x - 7)/2

=> 2y - 3 = (1/√3) ( 2x - 7 )

=> √3(2y - 3 ) = 2x - 7

=> 2√3y - 3√3 = 2x - 7

=> 2x - 2√3y + 3√3 - 7 = 0

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