Question #4fa44

1 Answer
Jan 21, 2018

Equation of line passing through a first quadrant point #(1,1)# and pependicular to the line passing through #(3,-6) and (-2,4)# is #2y=x+1#.

Explanation:

The slope of the line passing through #(3,-6) and (-2,4)#

is #m_1= (y_2-y_1)/(x_2-x_1)= (4+6)/(-2-3)= -2#

The product of slopes of the pependicular lines is #m_1*m_2=-1#

#:.m_2=-1/-2=1/2#. Let the equation of the line in slope-

intercept form be #y=mx+c or y=1/2*x+c#. Say the point in

first quadrant is #(1,1)# .The point will satisfy the equation .So,

# 1= 1/2*1+c or c= 1-1/2=1/2#.Hence the equation of the line

is #y= 1/2x+1/2 or 2y=x+1#.Equation of line passing through

a point #(1,1)# in first quadrant and pependicular to the line

passing through #(3,-6) and (-2,4)# is #2y=x+1#. [Ans]