MATH SOLVE

2 months ago

Q:
# Write the point-slope form of the equation of the line through point (6, -2) that is perpendicular to the line y=2x-3

Accepted Solution

A:

Answer:The point - slope form of the equation of line 1 is y = (-1/2)x + 1 Step-by-step explanation:here, the given point on the equation 1 is A(6, -2).Now, equation of line 2 is y = 2x - 3Comparing it with the INTERCEPT SLOPE FORM : y = mx + CSlope of Line 2 = 2 (= m2)Now,as line 1 is perpendicular to line 1⇒ Slope of line 1 x Slope of line 2 = -1or, slope of line 1 = (-1/2)Now, by POINT SLOPE form of a equation:An equation with point (x0 , y0) and slope m is given as (y - y0)= m (x - x0)Here, the equation of line 1 with point (6, -2) and slope (-1/2) is given as:[tex]y - (-2) = \frac{-1}{2} (x -6)\\\implies 2( y + 2) = 6 -x\\or, 2y + 4 = 6 -x\\\implies 2y = -x + 2\\or, y = (-\frac{1}{2}) x + 1[/tex]Hence, the point - slope form of the equation of line 1 is y = (-1/2)x + 1